Contoh Soal 5 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 21.&\textrm{Pada gambar berikut, inetgral yang}\\ &\textrm{menyatakan luas daerah yang diarsir adalah}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \int_{0}^{3}(2x^{2}-8x+6)dx\\ \textrm{b}.\quad \displaystyle \int_{0}^{1}(x^{2}-4x+3)dx-\int_{1}^{3}(x^{2}-4x+3)dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}(x^{2}+4x+3)dx+\int_{1}^{3}(x^{2}+4x+3)dx\\ \textrm{d}.\quad \color{red}\displaystyle \int_{0}^{1}(2x^{2}-8x+6)dx-\int_{1}^{3}(2x^{2}-8x+6)dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{1}(2x^{2}+8x+6)dx-\int_{1}^{3}(2x^{2}+8x+6)dx\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Kita tentukan poin batas-batasnya, yaitu}:\\&y=ax^{2}+bx+c=d(x-1)(x-3)\\ &\textrm{karena kurva melalui}\: \: (0,6),\: \textrm{maka}\\ &6=d(0-1)(0-3)\Rightarrow d=2\\&\textrm{Sehingga fungsi kurvanya adalah}:\\ &y=f(x)=2(x-1)(x-3)\\ &\Leftrightarrow y=2(x^{2}-4x+3)\\ &\Leftrightarrow y=2x^{2}-8x+6\\ &\textrm{Selanjutnya penentuan daerah arsir}\\ &\textrm{tinggal menyesuaikan dengan gambar}  \end{aligned}$.

$\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: f\: \: \textrm{fungsi ganjil, maka nilai}\\ &\displaystyle \int_{-2022}^{2022}f(x)\: dx\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}f(x)dx\\ \textrm{c}.\quad \color{red}\displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}f(x)dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perlu diketahui bahwa}\\ &f(x)=\begin{cases} \textrm{fungsi genap} &\Rightarrow f(-x)=f(x) \\  \textrm{fungsi ganjil} &\Rightarrow f(-x)=-f(x)  \end{cases}\\ &\textrm{Untuk}\\ &\begin{array}{|c|c|}\hline f\: \textrm{fungsi genap}&\displaystyle \int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\\\hline f\: \textrm{fungsi ganjil}&\displaystyle \int_{-a}^{a}f(x)dx=0\qquad\qquad\: \\\hline \end{array}\\ &\textrm{Karena}\: \: f(x)\: \: \textrm{fungsi ganjil, maka}\\ &\color{red}\displaystyle \int_{-2022}^{2022}f(x)\: dx=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Jika}\: \: f\: \: \textrm{fungsi genap, maka nilai}\\ &\displaystyle \int_{-2022}^{2022}f(x)\: dx\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}f(x)dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \color{red}\displaystyle 2\int_{0}^{2022}f(x)dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perlu diketahui bahwa}\\ &f(x)=\begin{cases} \textrm{fungsi genap} &\Rightarrow f(-x)=f(x) \\  \textrm{fungsi ganjil} &\Rightarrow f(-x)=-f(x)  \end{cases}\\ &\textrm{Untuk}\\ &\begin{array}{|c|c|}\hline f\: \textrm{fungsi genap}&\displaystyle \int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\\\hline f\: \textrm{fungsi ganjil}&\displaystyle \int_{-a}^{a}f(x)dx=0\qquad\qquad\: \\\hline \end{array}\\ &\textrm{Karena}\: \: f(x)\: \: \textrm{fungsi genap, maka}\\ &\color{red}\displaystyle \int_{-2022}^{2022}f(x)\: dx=2\displaystyle \int_{0}^{2022}f(x)dx \end{aligned} \end{array}$.

 $\begin{array}{ll}\\ 24.&\textrm{Nilai dari}\: \: \displaystyle \int_{-2022}^{2022}\left | x \right |\: dx\: =....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}\left | x \right |dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \color{red}\displaystyle 2\int_{0}^{2022}\left | x \right |dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Karena}\: \: f(x)=\left | x \right |,\: \: \textrm{adalah fungsi genap},\\ & \textrm{maka}\\ &\displaystyle \int_{-2022}^{2022}\left | x \right |\: dx\: =\color{red}\displaystyle 2\int_{0}^{2022}\left | x \right |dx=2022^{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Nilai dari}\: \: \displaystyle \int_{-2022}^{2022}x\left | x \right |\: dx\: =....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle\frac{ -2022^{3}}{3}\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}x\left | x \right |dx\\ \textrm{c}.\quad \color{red}\displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}x\left | x \right |dx\\ \textrm{e}.\quad \displaystyle \frac{2022^{3}}{3}\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Karena}\: \: f(x)=x\left | x \right |,\: \: \textrm{adalah fungsi ganjil},\\ & \textrm{maka}\\ &\displaystyle \int_{-2022}^{2022}x\left | x \right |\: dx\: =\color{red}0 \end{aligned} \end{array}$

Contoh Soal 4 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 16.&\textrm{Luas daerah yang batasi oleh parabola}\\ &y=4x-x^{2},\: y=-2x+8,\: \: \textrm{dan sumbu-Y}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle 4\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{b}.\quad \color{red}\displaystyle 6\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{c}.\quad \displaystyle 12\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{d}.\quad \displaystyle 20\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{e}.\quad \displaystyle 30\frac{2}{3}\: \: \textrm{satuan luas}\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Luas arsiran}=\textrm{Luas trapesium}-\textrm{luas II}\\&\color{blue}\textrm{Proses penyelesaian 1}\\ &\textrm{Kurva}\: \: y=4x-x^{2},\: \textrm{maka koordinat titik}\\ &\textrm{puncaknya adalah}:(p,q).\: \textrm{Pilih}\: y=(x)=0\\ &\Leftrightarrow y=f(x)=4x-x^{2}=x(4-x)=0\\ &\Leftrightarrow x_{1}=0\: \: \textrm{atau}\: \: x_{2}=4\\ &\textrm{dengan}\: \: p=\displaystyle \frac{x_{1}+x_{2}}{2}=\frac{0+4}{2}=2,\: \: \textrm{dan}\\ &q=y=f(2)=4.2-2^{2}=8-4=4\\ &\textrm{Sehingga}\: \: (p,q)=(2,4)\\ &\color{blue}\textrm{Proses penyelesaian 2}\\ &\textrm{Untuk garis}\: \: y=-2x+8,\: \: \textrm{maka koordinat}\\ &\textrm{titik potongnya terhadap sumbu-X dan Y}:\\ &\begin{array}{|c|c|c|}\hline x&0&4\\\hline y&8&0\\\hline (x,y)&(0,8)&(4,0)\\\hline \end{array}\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{aligned}$.

$.\: \qquad\begin{aligned}&\textrm{Luas arsiran}=\textrm{Luas trapesium}-\textrm{luas II}\\ &=\displaystyle \frac{(8+4)\times 2}{2}-\displaystyle \int_{0}^{2}4x-x^{2}\: dx\\ &=12-\left ( \left (2x^{2}-\displaystyle \frac{1}{3}x^{3}  \right )\: \: |_{0}^{2} \right )\\ &=12-\left ( \left (8-\displaystyle \frac{8}{3}  \right )-0 \right )\\ &=12+\displaystyle \frac{8}{3}-8=\color{red}6\frac{2}{3}\: \: \color{black}\textrm{satuan luas} \end{aligned}$.

$\begin{array}{ll}\\ 17.&\textrm{Luas daerah yang batasi oleh kurva}\: \: y=4-x^{2}\\ &y=-x+2,\: \: \textrm{dan}\: \: 0\leq x\leq 2\: \: \textrm{adalah}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \frac{8}{3}\: \: \textrm{satuan luas}\\ \textrm{b}.\quad \color{red}\displaystyle \frac{10}{3}\: \: \textrm{satuan luas}\\ \textrm{c}.\quad \displaystyle \frac{14}{3}\: \: \textrm{satuan luas}\\ \textrm{d}.\quad \displaystyle \frac{16}{3}\: \: \textrm{satuan luas}\\ \textrm{e}.\quad \displaystyle \frac{26}{3}\: \: \textrm{satuan luas}\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\color{blue}\textrm{Proses penyelesaian 1}\\&\textrm{Kurva}\: \: y=4-x^{2},\: \textrm{maka koordinat titik}\\ &\textrm{puncaknya adalah}:(p,q).\: \textrm{Pilih}\: y=(x)=0\\ &\Leftrightarrow y=f(x)=4-x^{2}=(2+x)(2-x)=0\\ &\Leftrightarrow x_{1}=-2\: \: \textrm{atau}\: \: x_{2}=2\\ &\textrm{dengan}\: \: p=\displaystyle \frac{x_{1}+x_{2}}{2}=\frac{-2+2}{2}=0,\: \: \textrm{dan}\\ &q=y=f(0)=4-0^{2}=4\\ &\textrm{Sehingga}\: \: (p,q)=(0,4)\\ &\color{blue}\textrm{Proses penyelesaian 2}\\ &\textrm{Untuk garis}\: \: y=-x+2,\: \: \textrm{maka koordinat}\\ &\textrm{titik potongnya terhadap sumbu-X dan Y}:\\ &\begin{array}{|c|c|c|}\hline x&0&2\\\hline y&2&0\\\hline (x,y)&(0,2)&(2,0)\\\hline \end{array}\\&\textrm{Perhatikan ilustrasi gambar berikut} \end{aligned}$.

$.\: \qquad\begin{aligned}&\textrm{Luas arsiran}=\textrm{Dibawah kurva}-\textrm{segitiga}\\&=\displaystyle \int_{0}^{2}4-x^{2}\: dx-\displaystyle \frac{1}{2}\times \textrm{alas}\times \textrm{tinggi}\\ &=\left (4x-\displaystyle \frac{1}{3}x^{3}  \right )|_{0}^{2}-\displaystyle \frac{2\times 2}{2}\\ &=8-\displaystyle \frac{8}{3}-2=6-2\displaystyle \frac{2}{3}=3\frac{1}{3}\\ &=\color{red}\frac{10}{3}\: \: \color{black}\textrm{satuan luas} \end{aligned}$.

$\begin{array}{ll}\\ 18.&\textrm{Luas daerah di bawah kurva}\: \: y=-x^{2}+8x\\ &\textrm{di atas}\: \: y=6x-24\: \: \textrm{dan terletak di }\\ &\textrm{kuadran I adalah}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(x^{2}-2x-24)\: dx\\ \textrm{b}.\quad \color{red}\displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(-x^{2}+2x+24)\: dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{6}(-x^{2}+8x)\: dx+\int_{6}^{8}(-x^{2}+2x+24)\: dx\\ \textrm{d}.\quad \displaystyle \int_{4}^{6}(6x-24)\: dx+\int_{6}^{8}(-x^{2}+8x)\: dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{4}(6x-24)\: dx+\int_{0}^{6}(-x^{2}+8x)\: dx\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Kita tentukan poin batas-batasnya dulu}\\&\textrm{Titik potong kurva dengan sumbu-X, yaitu}:\\ &y=-x^{2}+8x=x(-x+8)=0\\ &\Leftrightarrow x=0\: \: \textrm{atau}\: \: x=8\\ &\textrm{Titik potong garis dengan sumbu-X, yaitu}:\\ &y=6x-24=0\Leftrightarrow x=6\\ &\textrm{Dan titik potong kurva dengan garis adalah}:\\&y=y\Leftrightarrow -x^{2}+8x=6x-24\\ &\Leftrightarrow -x^{2}+2x+24=0\Leftrightarrow (-x-4)(x-6)=0\\ &\Leftrightarrow x_{1}=-4\: \: \textrm{atau}\: \: x_{2}=6\\ &\textrm{Sehingga luas arsiran}=\textrm{Luas I}+\textrm{Luas II}\\ &=\displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(-x^{2}+8x)-(6x-24)dx\\ &=\color{red}\displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(-x^{2}+2x+24)\: dx\\ &\textrm{Perhatikan ulustrasi berikut} \end{aligned}$ 

$\begin{array}{ll}\\ 19.&\textrm{Luas daerah yang diarsir pada gambar}\\ &\textrm{di bawah dapat dirumuskan dengan}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \color{red}\displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx+\int_{b}^{d}g(x)dx-\int_{b}^{c}f(x)dx\\ \textrm{b}.\quad \displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx+\int_{b}^{d}\left ( g(x)-f(x) \right )dx\\ \textrm{c}.\quad \displaystyle \int_{a}^{d}\left ( f(x)-g(x) \right )dx\\ \textrm{d}.\quad \displaystyle \int_{a}^{d}\left ( f(x)-g(x) \right )dx-\int_{c}^{d}g(x)\: dx\\ \textrm{e}.\quad \displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx+\int_{c}^{d}\left ( g(x)-f(x) \right )dx \end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Cukup jelas bahwa}\\ &\textrm{pada gambar di atas pada selang}\\ &\bullet\quad  a\leq x\leq b:\: \displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx\\ &\bullet\quad  b\leq x\leq d:\: \displaystyle \int_{b}^{d}\left (g(x)-f(x)  \right )dx\\ &\qquad\qquad\qquad \textrm{dikurangi luasan yang}\\&\qquad\qquad\qquad \textrm{berselang}\: \: b\leq x\leq c \end{aligned}$.

$\begin{array}{ll}\\ 20.&\textrm{Daerah}\: \: \textrm{R}\: \: \textrm{di kuadran II, dibatasi oleh}\\ &\textrm{grafik}\: \: y=x^{2},\: y=x+2\: \: \textrm{dan}\: \: y=0\\ &\textrm{Integral yang sesuai kondisi di atas adalah}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \color{red}\displaystyle \int_{-2}^{-1}\left ( x+2 \right )dx+\int_{-1}^{0}x^{2}dx\\ \textrm{b}.\quad \displaystyle \int_{-1}^{-2}\left ( x+2 \right )dx+\int_{-1}^{0}x^{2}dx\\ \textrm{c}.\quad \displaystyle \int_{-1}^{-2}x^{2}dx+\int_{-1}^{0}(x+2)dx\\ \textrm{d}.\quad \displaystyle \int_{0}^{2}\left (x^{2}+x+2 \right )dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{2}\left ( -x^{2}+x+2 \right )dx\end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{array}$.

$.\: \qquad\begin{aligned}&\textrm{Dari gukup jelas bahwa}\\ &\textrm{daerah arsiran akan terformulasikan}\\ &\displaystyle \int_{-2}^{-1}\left ( x+2 \right )dx+\int_{-1}^{0}x^{2}dx\\ &\textrm{Berikut sedikit uraian prosesnya}\\ &\textbf{Sebagai langkah awal},\: \textrm{tentukan dulu}\\ &\textrm{titik potong kura dengan garis, yaitu}:\\ &y=y\Leftrightarrow x^{2}=x+2\Leftrightarrow x^{2}-x-2=0\\ &\Leftrightarrow (x+1)(x-2)=0\\ &\Leftrightarrow x=-1\: \: \textrm{atau}\: \: x=2\\ &\textbf{Langkah berikutnya}\\ &\textrm{Kita tentukan daerah arsiran, yaitu}\\ &\textrm{seperti terformulasikan di atas} \end{aligned}$.

Contoh Soal 3 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 11.&\textrm{Diketahui}\: \: \displaystyle \int f(x)\: dx=ax^{2}+bx+c\\ &\textrm{dengan}\: \: a\neq 0.\: \textrm{Jika}\: a,\: f(a),\: 2b\: \: \textrm{membentuk}\\ &\textrm{barisan arimetika dan}\: \: f(b)=6,\: \: \textrm{maka}\\ &\textrm{nilai}\: \: \displaystyle \int_{0}^{1}f(x)\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \color{red}\displaystyle \frac{17}{4}\\ \textrm{b}.\quad \displaystyle \frac{21}{4}\\ \textrm{c}.\quad \displaystyle \frac{25}{4}\\ \textrm{d}.\quad \displaystyle \frac{13}{4}\\ \textrm{e}.\quad \displaystyle \frac{11}{4}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int f(x)\: dx=ax^{2}+bx+c\\ &\Rightarrow f(x)=2ax+b\\ &\Rightarrow f(a)=2a^{2}+b\\ &\Rightarrow f(b)=2ab+b=6\\ &\textrm{Karena}\: \: a,f(a),2b\: \: \textrm{membentuk barisan}\\ &\textrm{aritmetika, maka}\: \:  f(a)=\displaystyle \frac{a+2b}{2}\\ &\textrm{Sehingga}\\ &f(a)=2a^{2}+b=\displaystyle \frac{a+2b}{2}\\ &\Leftrightarrow 4a^{2}-a=0\Leftrightarrow a(4a-1)=0\\ &\Leftrightarrow a=0\: \: \textrm{atau}\: \: a=\displaystyle \frac{1}{4}\\ &\textrm{Dan karena}\: \: a\neq 0,\: \: \textrm{maka}\: \: a=\displaystyle \frac{1}{4}\Rightarrow b=4\\ &\textrm{Selanjutnya}\\ &f(x)=2ax+b=\displaystyle \frac{1}{2}x+4\\ &\textrm{Dari fakta di atas, maka}\\ &\displaystyle \int_{0}^{1} f(x)\: dx=\int_{0}^{1} \displaystyle \frac{1}{2}x+4\: dx\\ &=\displaystyle \frac{1}{4}x^{2}+4x|_{0}^{1}\\ &=\left ( \displaystyle \frac{1}{4}+4 \right )-0=\color{red}\displaystyle \frac{17}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Diketahui}\: \: f(x)=a+bx\: \: \textrm{dan}\: \: F(x)\\ &\textrm{adalah anti turunan}\: \: f(x).\: \textrm{Jika}\\  &F(1)-F(0)=3\: ,\: \textrm{maka}\: \: 2a+b\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle 10\\ \textrm{b}.\quad \color{red}\displaystyle 6\\ \textrm{c}.\quad \displaystyle 5\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 3\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Diketahui bahwa}\\ &f(x)=a+bx,\: \: \textrm{sedangkan}\\ &F(x)=\displaystyle \int f(x)\: dx=\int (a+bx)\: dx\\ &\qquad =ax+\displaystyle \frac{1}{2}bx^{2}+C\\ &\textrm{Dan diketahui pula}\\ &\begin{array}{ll} F(0)=a(1)+\displaystyle \frac{1}{2}b(1)^{2}+C&\\ F(1)=a(0)+\displaystyle \frac{1}{2}b(0)^{2}+C&-\\\hline \: \: \quad 3=a+\displaystyle \frac{1}{2}b \end{array}\\ &\textrm{Sehingga}\\ &2a+b=2\left ( a+\displaystyle \frac{1}{2}b \right )=2.3=\color{red}6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: y=\displaystyle \frac{1}{3}\left ( x^{3}+\frac{3}{x} \right ),\: \textrm{maka}\\ &\displaystyle \int_{1}^{2}\sqrt{4+\left (\displaystyle \frac{dy}{dx}  \right )^{2}}\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \frac{13}{14}\\ \textrm{b}.\quad \displaystyle \frac{13}{6}\\ \textrm{c}.\quad \color{red}\displaystyle \frac{15}{6}\\ \textrm{d}.\quad \displaystyle \frac{16}{6}\\ \textrm{e}.\quad \displaystyle \frac{17}{6}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&y=\displaystyle \frac{1}{3}\left ( x^{3}+\frac{3}{x} \right )\\ &\Leftrightarrow dy=\left (x^{2}-\displaystyle \frac{1}{x^{2}}  \right )\: dx\\ &\Leftrightarrow \displaystyle \frac{dy}{dx}=\left (x^{2}-\displaystyle \frac{1}{x^{2}}  \right )\\ &\Leftrightarrow \left ( \displaystyle \frac{dy}{dx} \right )^{2}=\left (x^{2}-\displaystyle \frac{1}{x^{2}}  \right )^{2}=\displaystyle x^{4}-2+\displaystyle \frac{1}{x^{4}}\\ &\textrm{maka}\\ &\displaystyle \int_{1}^{2}\sqrt{4+\left (\displaystyle \frac{dy}{dx}  \right )^{2}}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{4+\left ( \displaystyle x^{4}-2+\displaystyle \frac{1}{x^{4}} \right )}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{\left ( \displaystyle x^{4}+2+\displaystyle \frac{1}{x^{4}} \right )}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{\left ( \displaystyle x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}}\: dx\\ &=\displaystyle \int_{1}^{2}x^{2}+\displaystyle \frac{1}{x^{2}}\: dx\\&=\displaystyle \frac{1}{3}x^{3}-\displaystyle \frac{1}{x}|_{1}^{2}\\ &=\left ( \displaystyle \frac{8}{3}-\frac{1}{2} \right )-\left ( \displaystyle \frac{1}{3}-1 \right )\\ &=\displaystyle \frac{16-3-2+6}{6}\\&=\color{red}\displaystyle \frac{17}{6} \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Luas daerah yang diarsir pada}\\ &\textrm{gambar berikut adalah 12}\\ &\textrm{satuan luas, maka nilai}\: \: a=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle 4\\ \textrm{b}.\quad \displaystyle 5\\ \textrm{c}.\quad \color{red}\displaystyle 6\\ \textrm{d}.\quad \displaystyle 7\\ \textrm{e}.\quad \displaystyle 8\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textrm{Luas arsiran}=12\\ &=4.a-\displaystyle \int_{1}^{4}\displaystyle \frac{a}{(4-1)(4-7)}(x-1)(x-7)\: dx\\ &=4a-\displaystyle \int_{1}^{4}\displaystyle -\frac{a}{9}(x^{2}-8x+7)\: dx\\ &=4a- \displaystyle \int_{1}^{4}\displaystyle \frac{a}{9}(-x^{2}+8x-7)\: dx\\ &=4a-\left (-\displaystyle \frac{a}{27}x^{3}+\frac{4a}{9}x^{2}-\frac{7a}{9}x|_{1}^{4}  \right )\\ &=4a+\displaystyle \frac{64a}{27}-\frac{64a}{9}+\frac{28a}{9}-\frac{a}{27}+\frac{4a}{9}-\frac{7a}{9}\\ &=4a-\displaystyle \frac{18a}{9}=2a=12\Leftrightarrow a=\color{red}6\\ &\textrm{Berikut ilustrasi gambar lengkapnya} \end{aligned}$.

$\begin{array}{ll}\\ 15.&\textrm{Luas daerah yang diarsir pada}\\ &\textrm{gambar berikut dapat dinyatakan}\\&\textrm{dengan}\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \int_{0}^{2}(3x-x^{2})\: dx\\ \textrm{b}.\quad \displaystyle \int_{0}^{2}(x+3)\: dx-\int_{0}^{2}x^{2}\: dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}(x+3)\: dx-\int_{0}^{2}x^{2}\: dx\\ \textrm{d}.\quad \displaystyle \int_{0}^{1}(x+3-x^{2})\: dx-\int_{1}^{2}x^{2}\: dx\\ \textrm{e}.\quad \color{red}\displaystyle \int_{0}^{1}(x+3-x^{2})\: dx-\int_{1}^{2}\left (4-x^{2}  \right )\: dx\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textrm{Daerah yang diarsir di atas dibatasi}\\ &\textrm{oleh 4 buah batas, yaitu}\\ &\bullet \quad \textrm{Garis}\: \: y=4\\ &\bullet \quad \textrm{Kurva}\: \: y=x^{2}\\ &\bullet \quad \textrm{Garis}\: \: x=0,\: \: \textrm{dan}\\ &\bullet \quad \textrm{Garis yang melalui titik}\: (0,3)\: \textrm{dan}\: (1,4)\\ &\begin{array}{|c|c|}\hline \textrm{Titik}&\textrm{Garis}\\\hline \begin{aligned}&{0,3}\\ &\textrm{dan}\\ &(1,4)\\ & \end{aligned}&\begin{aligned}\displaystyle y&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})+y_{1}\\ y&=\displaystyle \frac{4-3}{1-0}(x-0)+3\\ y&=x+3 \end{aligned}\\\hline \end{array}\\ &\textrm{Sehingga luas arisrannya adalah}:\\ &=\color{red}\displaystyle \int_{0}^{1}(x+3)-x^{2}\: dx+\int_{1}^{2}4-x^{2}\: dx \end{aligned}$.

Contoh Soal 2 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 6.&\textrm{Hasil dari}\: \: \displaystyle \int_{2}^{4}\left ( -x^{2}+6x-8 \right )dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \frac{38}{3}\\ \textrm{b}.\quad \displaystyle \frac{26}{3}\\ \textrm{c}.\quad \displaystyle \frac{20}{3}\\ \textrm{d}.\quad \displaystyle \frac{16}{3}\\ \textrm{e}.\quad \color{red}\displaystyle \frac{4}{3} \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int_{2}^{4}\left ( -x^{2}+6x-8 \right )dx\\ &=-\displaystyle \frac{1}{3}x^{3}+3x^{2}-8x|_{2}^{4}\\ &=\left ( -\displaystyle \frac{1}{3}(4)^{3}+3(4)^{2}-8(4) \right )-\left ( -\displaystyle \frac{1}{3}(2)^{3}+3(2)^{2}-8(2) \right )\\ &=\left (-\displaystyle \frac{64}{3}+48-32 \right )-\left ( -\displaystyle \frac{8}{3}+12-16 \right )\\ &=-\displaystyle \frac{56}{3}+20=-\displaystyle \frac{56+60}{3}=\color{red}\displaystyle \frac{4}{3}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Hasil dari}\: \: \displaystyle \int_{0}^{2}x^{2}\left ( x+2 \right )dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle 6\\ \textrm{b}.\quad 6\displaystyle \frac{1}{3}\\ \textrm{c}.\quad 6\displaystyle \frac{2}{3}\\ \textrm{d}.\quad \color{red}9\displaystyle \frac{1}{3}\\ \textrm{e}.\quad \displaystyle 20 \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \displaystyle \int_{0}^{2}x^{2}\left ( x+2 \right )dx=\displaystyle \int_{0}^{2}x^{3}+2x^{2}\: \: dx\\ &=\displaystyle \frac{1}{4}x^{4}+\frac{2}{3}x^{3}|_{0}^{2}\\ &=\left ( -\displaystyle \frac{1}{4}(2)^{4}+\frac{2}{3}(2)^{3} \right )-\left ( 0 \right )\\ &=4+\displaystyle \frac{16}{3}=\color{red}\displaystyle \frac{4}{3}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: \displaystyle \int_{0}^{2}18x\sqrt{3x^{2}+4}\: \: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle 4\\ \textrm{b}.\quad 16\\ \textrm{c}.\quad \color{red}112\\ \textrm{d}.\quad 128\\ \textrm{e}.\quad \displaystyle 168 \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Misalkan}\\  &u=3x^{2}+4\Rightarrow du=6x\: dx\\ &\textrm{Selanjutnya}\\ &\displaystyle \int_{0}^{2}18x\sqrt{3x^{2}+4}\: \: dx=\displaystyle \int_{0}^{2}3\sqrt{u}\: \: dx\\ &=\displaystyle \int_{0}^{2}3u^{.^{\frac{1}{2}}}\: \: dx=2u^{.^{\frac{3}{2}}}\: |_{0}^{2}\\ &=2\left ( 3x^{2}+4 \right )^{.^{\frac{3}{2}}}|_{0}^{2}\\ &=2\left ( 3.2^{2}+4 \right )^{.^{\frac{3}{2}}}-(2(0+4)^{.^{\frac{3}{2}}})\\ &=2.(16)^{.^{\frac{3}{2}}}=2.4^{3}-2.2^{3}=128-16\\ &=\color{red}112 \end{aligned} \end{array}$ .

$\begin{array}{ll}\\ 9.&\textrm{Hasil substitusi dari}\: \: u=x+1\: \: \textrm{pada}\\ &\displaystyle \int_{0}^{1}x^{2}\sqrt{x+1}\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \color{red}\displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{b}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{d}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{e}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Misalkan}\\  &u=x+1\Rightarrow x=u-1\Rightarrow dx=du\\ &\textrm{Selanjutnya}\\ &\displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: \: du \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\: \: \displaystyle \int_{0}^{1}5x(1-x)^{6}\: dx=\: ....\\ &\begin{array}{ll}  \textrm{a}.\quad \displaystyle \frac{75}{56}\\ \textrm{b}.\quad \displaystyle \frac{10}{56}\\ \textrm{c}.\quad \color{red}\displaystyle \frac{5}{56}\\ \textrm{d}.\quad -\displaystyle \frac{7}{56}\\ \textrm{e}.\quad -\displaystyle \frac{10}{56}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Diketahui soal bentuk}\: \textbf{integral parsial}\\  &\textrm{dengan metode}\: \textbf{Tanzalin}\: \textrm{diperoleh}\\ &\begin{array}{|c|c|}\hline \textrm{Diturunkan}&\textrm{Diintegralkan}\\\hline \color{red}5x&(1-x)^{6}\\\hline \color{blue}5&\color{red}-\displaystyle \frac{1}{7}(1-x)^{7}\\\hline 0&\color{blue}\displaystyle \frac{1}{56}(1-x)^{8}\\\hline \end{array}\\ &\textrm{Sehingga}\\ &\displaystyle \int_{0}^{1}5x(1-x)^{6}\: dx\\ &=(5x)\left ( -\displaystyle \frac{1}{7}(1-x)^{7} \right )-(5)\left ( \displaystyle \frac{1}{56}(1-x)^{8} \right )\: |_{0}^{1}\\ &=(0-0)-\left ( 0-\left ( \displaystyle \frac{5}{56} \right ) \right )\\ &=\color{red}\displaystyle \frac{5}{56} \end{aligned} \end{array}$ 

Contoh Soal 1 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 1.&\textbf{(UAN 2014)}\\ &\textrm{Hasil}\: \: \displaystyle \int_{-1}^{2}\left ( x^{3}+3x^{2}+4x+5 \right )\: dx=\: ....\\ &\begin{array}{lll} \textrm{a}.&33\displaystyle \frac{1}{4}\\ \textrm{b}.&\color{red}33\displaystyle \frac{3}{4}\\ \textrm{c}.&32\displaystyle \frac{1}{4}\\ \textrm{d}.&31\displaystyle \frac{3}{4}\\ \textrm{e}.&23\displaystyle \frac{3}{4} \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\int_{-1}^{2}\left ( x^{3}+3x^{2}+4x+5 \right )\: dx\\ &=\displaystyle \frac{x^{4}}{4}+x^{3}+2x^{2}+5x\: |_{-1}^{2}\\ &=\left (\displaystyle \frac{2^{4}}{4}+2^{3}+2.2^{2}+5.2  \right )\\ &\quad -\left ( \frac{\left ( -1 \right )^{4}}{4}+\left ( -1 \right )^{3}+2.\left ( -1 \right )^{2}+5\left ( -1 \right ) \right )\\ &=\color{red}33\frac{3}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textbf{(UAN 2003)}\\ &\textrm{Jika}\: f(x)=\left ( x-2 \right )^{2}-4\: \: \textrm{dan}\: \: g(x)=-f(x),\\ &\textrm{maka luas daerah yang di batasi kurva }\\ &f\: \: \textrm{dan}\: \: g\: \:  \textrm{adalah}\: ....\\ &\begin{array}{lll} \textrm{a}.&10\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{b}.&\color{red}21\displaystyle \frac{1}{3}\: \: \textrm{satuan luas}\\ \textrm{c}.&22\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{d}.&42\displaystyle \frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{e}.&45\displaystyle \frac{1}{3}\: \: \textrm{satuan luas} \end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.

$.\qquad\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\displaystyle \int_{0}^{4}\left ( g(x)-f(x) \right )\: \: dx\\ &=\displaystyle \int_{0}^{4}\left ( 4x-x^{2} \right )-\left ( x^{2}-4x \right )\: \: dx\\ &=\displaystyle \int_{0}^{4}\left ( 8x-2x^{2} \right )\: \: dx\\ &=\displaystyle \left [4x^{2}-\frac{2}{3}x^{3} \right ]_{0}^{4}\\ &=\displaystyle \left ( 4.4^{2}-\frac{2}{3}.4^{3} \right )-\left ( 4.0^{2}-\frac{2}{3}.0^{3} \right )\\ &=\displaystyle \left ( 64-\frac{2}{3}.64 \right )-0\\ &=\displaystyle \frac{64}{3}=\color{red}21\frac{1}{3}\: \: \color{black}\textrm{satuan luas} \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan menggunakan rumus}\: \: L=\displaystyle \frac{D\sqrt{D}}{6a^{2}}\\ &\begin{array}{|l|}\hline \begin{aligned}&f(x)=g(x)\\ &f(x)=-f(x),\quad\textrm{ingat}\: \:  g(x)= -f(x)\\ &2f(x)=0,\quad\color{blue}\textrm{tidak boleh disederhanakan},\\ &2\times \left (\left ( x-2 \right )^{2}-4 \right )=0,\quad \color{blue}\textrm{karena akan }\\ &2\times \left ( x^{2}-4x \right )=0\quad \color{blue}\textrm{mempengaruhi hasil }\\ &2x^{2}-8x=0,\qquad\quad \color{blue}\textrm{akhir}\\ &\begin{cases} a=2,\: b=-8 & c=0 \\ D=b^{2}-4ac, & D=\left ( -8 \right )^{2}-4(2)(0)=64 \end{cases}\\ &L_{\: \textbf{daerah}}=\displaystyle \frac{\textbf{D}\sqrt{\textbf{D}}}{6\textbf{a}^{2}}\\ &\quad\qquad=\displaystyle \frac{64\sqrt{64}}{6(2)^{2}}\\ &\quad\qquad=\displaystyle \frac{64\times 8}{6\times 4}\\ &\quad\qquad=\displaystyle \frac{64}{3}\\ &\quad\qquad=\color{red}21\displaystyle \frac{1}{3} \end{aligned}\\\hline\end{array} \end{aligned}$.

$\begin{array}{ll}\\ 3.&\textbf{(UAN 2014)}\\ &\textrm{Luas daerah yang diarsir pada gambar }\\ &\textrm{dapat dinyatakan dengan rumus}  \end{array}$.

$.\: \quad\begin{array}{ll}\\  &\begin{array}{lll} \textrm{a}.&\displaystyle \int_{0}^{4}4x\: dx-\int_{2}^{4}\left ( 2x-4 \right )\: dx\\ \textrm{b}.&\displaystyle \int_{0}^{4}4x\: dx - \int_{2}^{4}\left ( 2x+4 \right )\: dx\\ \textrm{c}.&\color{red}\displaystyle \int_{0}^{4}2\sqrt{2}\: dx - \int_{2}^{4}\left ( 2x-4 \right )\: dx\\ \textrm{d}.&\displaystyle \int_{0}^{4}2\sqrt{2}\: dx - \int_{2}^{4}\left ( 4-2x \right )\: dx\\ \textrm{e}.&\displaystyle \int_{0}^{4}2\sqrt{x}\: dx - \int_{2}^{4}\left ( 4+2x \right )\: dx \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Luas}_{\color{blue}\textrm{arsiran}}=\displaystyle \int_{0}^{4}y_{1}\: dx-\int_{2}^{4}y_{2}\: dx\\ &=\displaystyle \int_{0}^{4}\sqrt{4x}\: dx-\int_{2}^{4}\left ( 2x-4 \right )\: dx\\ & =\displaystyle  \int_{0}^{4}2\sqrt{x}\: dx-\int_{2}^{4}\left ( 2x-4 \right )\: dx. \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 4.&\textbf{(UAN 2014)}\\ &\textrm{Volume benda putar yang terbentuk dari daerah}\\ &\textrm{yang dikuadran I yang dibatasi oleh kurva}\\ &x=2\sqrt{3}y^{2}\: ,\: \textrm{sumbu Y , dan lingkaran}\: x^{2}+y^{2}=1,\\ &\textrm{diputar mengelilingi sumbu Y adalah}\: ....\\ &\begin{array}{lll} \textrm{a}.&\displaystyle \frac{4}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{b}.&\color{red}\displaystyle \frac{17}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{c}.&\displaystyle \frac{23}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{d}.&\displaystyle \frac{44}{60}\pi \: \: \textrm{satuan volum}\\ \textrm{e}.&\displaystyle \frac{112}{60}\pi \: \: \textrm{satuan volum} \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

$.\: \quad\begin{aligned}&\textrm{Volumenya jika diputar mengelilingi  Sumbu }\\ &\textrm{Y adalah}:\\ & V=\pi \displaystyle \int_{0}^{\frac{1}{2}}x_{1}^{2}\: dy+\pi \int_{\frac{1}{2}}^{1}x_{2}^{2}\: dy\\ &\Leftrightarrow \: \: V=\pi \int_{0}^{\frac{1}{2}}\left ( 2\sqrt{3}y^{2} \right )^{2}\: dy+\pi \int_{\frac{1}{2}}^{1}\left ( 1-y^{2} \right )\: dy\\ &\Leftrightarrow \: \: V=\frac{12}{5}y^{5}\pi |_{0}^{\frac{1}{2}}+\left ( y-\frac{1}{3}y^{3} \right )\pi |_{\frac{1}{2}}^{1}\\ &\Leftrightarrow \: \: V=\pi \left ( \frac{12}{5}\times \frac{1}{32}+\left ( 1-\frac{1}{3} \right )-\left ( \frac{1}{2}-\frac{1}{3}\times \frac{1}{8} \right ) \right )\\ &\Leftrightarrow \: \: V=\color{red}\frac{17}{60}\pi.\end{aligned}$.

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \displaystyle \frac{d}{dx}g(x)=f(x)\: \: \textrm{di mana} \: f(x)\\ &\textrm{kontinu dari a sampai b, }\\ &\textrm{maka}\: \displaystyle \int_{a}^{b}f(x).g(x)\: dx\\ &\begin{array}{ll}\\ \textrm{a}.\quad 0\\ \textrm{b}.\quad f(b)-f(a)\\ \textrm{c}.\quad g(b)-g(a)\\ \textrm{d}.\quad \displaystyle \frac{\left [ f(b) \right ]^{2}-\left [ f(a) \right ]^{2}}{2}\\ \textrm{e}.\quad \color{red}\displaystyle \frac{\left [ g(b) \right ]^{2}-\left [ g(a) \right ]^{2}}{2}\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int_{a}^{b} f(x).g(x)\: dx\\ &=\displaystyle \int_{a}^{b} g(x).f(x)\: dx\\ &=\displaystyle \int_{a}^{b} g(x).\displaystyle \frac{d\left ( g(x) \right )}{dx}\: dx,\\ &\quad \textrm{ingat bahwa}\quad \displaystyle \frac{d}{dx}g(x)=f(x),\\ &\quad\textrm{f(x) kontinu dari a sampai b}\\ &=\displaystyle \int_{a}^{b} g(x)\: d\left ( g(x) \right )\\ &=\left [\displaystyle \frac{\left ( g(x) \right )^{2}}{2} \right ]_{a}^{b}\\ &=\displaystyle \frac{\left [ g(b) \right ]^{2}-\left [ g(a) \right ]^{2}}{2} \end{aligned} \end{array}$.

Penggunaan Integral Tentu Fungsi Aljabar

 A. Luas

Menghitung luas yang dibatasi oleh sebuah kurva dan sumbu X kita dapat menggunakan bantuan integral tentuk sebagaimana uraian sebelumnya

Perhatikan ilustrasi gambar berikut

$\begin{array}{|c|c|}\hline  \textrm{Di Atas Sumbu X}&\textrm{Di Bawah Sumbu X}\\\hline &-\displaystyle \int_{a}^{b}f(x)\: \: dx\\ \displaystyle \int_{a}^{b}f(x)\: \: dx&atau\\ &\displaystyle \int_{b}^{a}f(x)\: \: dx\\\hline \end{array}$.

B. Volume Benda Putar

Adapun untuk volume diformulasikan dengan integral tentu berikut

$\boxed{V=\pi \displaystyle \int_{a}^{b}\left ( f(x) \right )^{2}\: \: dx=\pi \displaystyle \int_{a}^{b}y^{2}\: \: dx}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah luas daerah bidang berikut dan }\\ &\textrm{tentukan pula volumenya seandainya bidang }\\ &\textrm{yang diarsir tersebut diputar terhadap sumbu X} \end{array}$.

$.\qquad\begin{aligned}&\begin{aligned}L_{\textrm{Arsiran}}&=\displaystyle \int_{1}^{3}2x\: dx\\ &=\displaystyle \left [ x^{2} \right ]_{1}^{3}\\ &=\left ( 3 \right )^{2}-\left ( 1 \right )^{2}\\ &=9-1\\ &=\color{red}8\quad \color{black}\textbf{satuan luas} \end{aligned}\\&\begin{aligned}V_{\textrm{Benda putar}}&=\pi \displaystyle \int_{1}^{3}\left ( y \right )^{2}\: dx\\ &=\pi \displaystyle \int_{1}^{3}\left ( 2x \right )^{2}\: dx\\ &=\pi \displaystyle \int_{1}^{3}4x^{2}\: dx\\ &=\pi \left [ \displaystyle \frac{4x^{3}}{3} \right ]_{1}^{3}\\ &=\pi \left ( \displaystyle \frac{4\times 3^{3}}{3} \right )-\pi \left ( \displaystyle \frac{4\times 1^{3}}{3} \right )\\ &=36\pi -\displaystyle \frac{4}{3}\pi \\ &=\color{red}34\displaystyle \frac{2}{3}\pi \quad \color{black}\textbf{satuan volum} \end{aligned}  \end{aligned}$.

\begin{array}{ll}\\ 2.&\textrm{Diketahui parabola}\: \: f_{1}(x)=a_{1}x^{2}+b_{1}x+c_{1}\\ &\textrm{dan}\: \: f_{2}(x)=a_{2}x^{2}+b_{2}x+c_{2}.\\ &\textrm{Titik potong kedua parabola tersebut }\\ &\textrm{dapat cari dengan}\\ &\\ &f_{1}(x)=f_{2}(x)\\ &\Leftrightarrow \: \: a_{1}x^{2}+b_{1}x+c_{1}=a_{2}x^{2}+b_{2}x+c_{2}\\ & \Leftrightarrow \: ax^{2}+bx+c=0.\\ &\\ &\textrm{Jika kedua parabola berpotongan di dua}\\ &\textrm{titik, tunjukkan bahwa luas daerah antara}\\ &\textrm{kedua parabola tersebut dapat }\\ &\textrm{dinyatakan dengan}\: \: \: \displaystyle \textbf{L}=\frac{\textbf{D}\sqrt{\textbf{D}}}{\textbf{6a}^{\textbf{2}}}\\\\ &\textbf{Bukti}:\\ &ax^{2}+bx+c=0\: \begin{cases} &x_{1}=\displaystyle \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \\ & \\ &x_{2}=\displaystyle \frac{-b- \sqrt{b^{2}-4ac}}{2a} \end{cases}\\ &\begin{aligned}L&=\displaystyle \int_{\frac{-b- \sqrt{b^{2}-4ac}}{2a}}^{\frac{-b+ \sqrt{b^{2}-4ac}}{2a}}\: \left ( ax^{2}+bx+c \right )\: \: dx=\left [ \displaystyle \frac{ax^{3}}{3}+\frac{bx^{2}}{2}+cx \right ]_{\frac{-b- \sqrt{b^{2}-4ac}}{2a}}^{\frac{-b+ \sqrt{b^{2}-4ac}}{2a}}\\ &=\left [ \displaystyle \frac{a}{3}\left ( \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \right )^{3}+\displaystyle \frac{b}{2}\left ( \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \right )^{2}+c\left ( \frac{-b+ \sqrt{b^{2}-4ac}}{2a} \right ) \right ]\\ &\quad -\left [ \displaystyle \frac{a}{3}\left ( \frac{-b- \sqrt{b^{2}-4ac}}{2a} \right )^{3}+\displaystyle \frac{b}{2}\left ( \frac{-b- \sqrt{b^{2}-4ac}}{2a} \right )^{2}+c\left ( \frac{-b- \sqrt{b^{2}-4ac}}{2a} \right ) \right ]\\ &=\displaystyle \frac{a}{24a^{3}}\left [ \left ( \sqrt{D}^{3}-3\sqrt{D}^{2}b+3\sqrt{D}b^{2}-b^{3} \right )+\left ( \sqrt{D}^{3}+3\sqrt{D}^{2}b+3\sqrt{D}b^{2}+b^{3} \right ) \right ]\\ &\quad +\displaystyle \frac{b}{8a^{2}}\left [ \left ( b^{2}-2b\sqrt{D}+\sqrt{D}^{2} \right )-\left ( b^{2}+2b\sqrt{D}+\sqrt{D}^{2} \right ) \right ]+\displaystyle \frac{c}{2a}\left [ \left ( -b+\sqrt{D} \right )-\left ( -b-\sqrt{D} \right ) \right ]\\ &=\displaystyle \frac{1}{24a^{2}}\left [ 2\sqrt{D}^{3}+6\sqrt{D}b^{2} \right ]+\displaystyle \frac{b}{8a^{2}}\left [ -4b\sqrt{D} \right ]+\displaystyle \frac{c}{2a}\left [ 2\sqrt{D} \right ]\\ &=\displaystyle \frac{\sqrt{D}^{3}}{12a^{2}}+\frac{b^{2}\sqrt{D}}{4a^{2}}-\frac{b^{2}\sqrt{D}}{2a^{2}}+\frac{c\sqrt{D}}{a}=\displaystyle \frac{D\sqrt{D}}{12a^{2}}+\frac{b^{2}\sqrt{D}}{4a^{2}}-\frac{b^{2}\sqrt{D}}{2a^{2}}+\frac{c\sqrt{D}}{a}\\ &=\displaystyle \frac{\sqrt{D}}{12a^{2}} \left (D+3b^{2}-6b^{2}+12ac  \right )\\ &=\displaystyle \frac{\sqrt{D}}{12a^{2}}\left [ \left ( b^{2}-4ac \right )-3b^{2}+12ac \right ]\\ &=\displaystyle \frac{\sqrt{D}}{12a^{2}}\left [ -2b^{2}+8ac \right ]\\ &=-\displaystyle \frac{\sqrt{D}}{6a^{2}}\left [ b^{2}-4ac \right ]=-\frac{\sqrt{D}}{6a^{2}}\left [ D \right ]\\ &=-\frac{D\sqrt{D}}{6a^{2}},\quad \textbf{luas tidak mungkin negatif}\\ &=\displaystyle \frac{D\sqrt{D}}{6a^{2}}\quad \blacksquare \end{aligned} \end{array}.

$\begin{array}{ll}\\ 3.&\textrm{Tentukan volume benda putar yang terbentuk, }\\ &\textrm{jika suatu daerah yang dibatasi oleh kurva}\\ &y^{2}=x\: \: \textrm{dan}\: \: y=x\: \textrm{diputar mengelilingi sumbu X} \\\end{array}$.

$.\: \quad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{aligned}$.

$.\: \quad\begin{aligned}&\textbf{Menentukan batas}\\ &\begin{array}{|r|l|}\hline  \begin{aligned}y&=y\\ x^{2}&=x\\ x^{2}-x&=0\\ x\left ( x-1 \right )&=0\\ x=0\: \: \textrm{atau}\: \: x&=1\\ &\\ &\\ & \end{aligned}&\begin{aligned}V&=\pi \displaystyle \int_{a}^{b}\left ( y_{1}^{2}-y_{2}^{2} \right )\: \: dx\\ &=\pi \displaystyle \int_{0}^{1}\left ( x-x^{2} \right )\: \: dx\\ &=\pi \left [ \displaystyle \frac{1}{2}x^{2}-\frac{1}{3}x^{3} \right ]_{0}^{1}\\ &=\pi \left [ \displaystyle \frac{1}{2}-\frac{1}{3} \right ]\\ V&=\displaystyle \frac{1}{6}\pi \end{aligned} \\\hline  \end{array}\\ &\textbf{Keterangan lanjutan}\\ &\begin{aligned}&\textnormal{Perhatikan bahwa;}\\&y^{2}=x\Rightarrow y=\sqrt{x},\: \textrm{dianggap sebagai}\: \: y_{1}\\ &\textnormal{Sehingga}\: y_{1}-\textrm{nya adalah}\: \: \sqrt{x}\\ &\textnormal{dan}\: \: y=x\: \: \textrm{dianggap sebagai}\: \: y_{2}\\ &\left ( y_{1}^{2}-y_{2}^{2} \right )=\left ( \left ( \sqrt{x} \right )^{2}-\left ( x \right )^{2} \right )=x-x^{2}\end{aligned}\\  &\textrm{Jadi, volume dari benda putar tersebut}\\ &\textrm{dalam satuan volum adalah}\: \: \color{red}\displaystyle \frac{1}{6}\pi \end{aligned}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukan volume benda putar yang terbentuk,}\\ &\textrm{jika suatu daerah yang dibatasi oleh kurva} \\ &y=2x\: ,\: y=x,\: x=1,\: \textrm{dan}\: \: x=3\\ &\textrm{diputar mengelilingi sumbu X}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{array}$.

$.\: \quad\begin{aligned}&\textrm{Langkah-Langkah penyelesaiannya adalah:}\\ &\begin{array}{|c|c|}\hline \textrm{Batas}&\textrm{Menentukan Volumenya}\\\hline x=1\: \: \textrm{dan}\: \: x=3&\begin{aligned}V&=\displaystyle \pi \int_{a}^{b}\left ( f^{2}(x)-g^{2}(x) \right )\: \: dx\\ &=\displaystyle \pi \int_{1}^{3}\left ( \left ( 2x \right )^{2}-\left ( x \right )^{2} \right )\: \: dx\\ &=\displaystyle \pi \int_{1}^{3}3x^{2}\: \: dx\\ &=\displaystyle \pi \left [ x^{3} \right ]_{1}^{3}\\ &=\displaystyle \pi \left ( 3^{3} \right )-\pi \left ( 1^{3} \right )\\ &=27\pi -1\pi \\ V&=26\pi\: \textbf{Satuan Volum} \end{aligned}\\\hline \end{array} \end{aligned}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukan volume daerah yang dibatasi}\\ &\textrm{oleh lingkaran}\: \: x^{2}+y^{2}=4\: ,\: \textrm{selang}\\ &-2\leq x\leq 2\: \: \textrm{dan}\: \: \textrm{diputar mengelilingi}\\ &\textrm{sumbu X}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \\\end{array}$.

$.\: \quad\begin{aligned}&\textrm{Langkah-Langkah penyelesaiannya adalah:}\\ &\begin{array}{|c|c|}\hline \textrm{Batas}&\textrm{Menentukan Volumenya}\\\hline x=-2\: \: \textrm{sampai}\: \: x=2&\begin{aligned}V&=\displaystyle \pi \int_{a}^{b}y^{2}\: \: dx\\ &=\displaystyle \pi \int_{-2}^{2}\left ( 4-x^{2} \right )\: \: dx\\ &=\displaystyle \pi \left [ 4x-\displaystyle \frac{x^{3}}{3} \right ]_{-2}^{2} \\ &=\displaystyle \pi \left ( 8-\displaystyle \frac{8}{3} \right )-\pi \left ( -8+\displaystyle \frac{8}{3} \right )\\ &=\displaystyle \pi \left ( 8+8-\frac{8}{3}-\frac{8}{3} \right )\\ V&=\displaystyle \frac{32}{3}\pi\: \: \textbf{Satuan Volum} \end{aligned}\\\hline \end{array} \end{aligned}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah luas daerah yang diarsir berikut} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa luas ellips}\: \: \displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ &\textrm{adalah}\: \: \pi ab \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah volume benda putar yang terjadi}\\ &\textrm{jika daerah dari hasil putar tersebut mengelilingi }\\ &\textrm{sumbu X serta dibatasi oleh}\\ &\begin{array}{ll} \textrm{a}.\quad y=x+3,\: \textrm{sumbu x, garis x = 2, dan x = 4}\\ \textrm{b}.\quad \displaystyle y=\frac{1}{2}x+2,\: \textrm{sumbu x, garis x = 0, dan x = 4}\\ \textrm{c}.\quad \displaystyle y=\sqrt{x+2} ,\: \textrm{sumbu x, garis x = 2, dan x = 4}\\ \textrm{d}.\quad y=4-2x,\: \textrm{sumbu x, garis x = 0, dan x = 4}\\ \textrm{e}.\quad \displaystyle x^{2}+y^{2}=16,\: \textrm{dan sumbu X} \end{array} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah volume benda putar yang}\\ &\textrm{terjadi jika daerah dari hasil putar tersebut }\\ &\textrm{mengelilingi sumbu X serta dibatasi oleh}\\ &\begin{array}{ll}  \textrm{a}.\quad y=2x-x^{2},\: \textrm{dan}\: \: y=0\\ \textrm{b}.\quad \displaystyle y^{2}=x,\: \textrm{dan}\: \: y=2\\ \textrm{c}.\quad \displaystyle y=x^{2} ,\: \textrm{dan}\: \: y=-x^{2}+4\\ \textrm{d}.\quad y=7-2x^{2},\: \textrm{dan}\: \: y=x^{2}+4\\ \textrm{e}.\quad \displaystyle y=x^{2},\: \textrm{dan}\: \: y^{2}=x \end{array} \end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007.  Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar ISI 2006. Jakarta: ESIS.

Inetgral Tentu Fungsi Aljabar

A. Integral Tentu

Sebelumnya sudah dibahasa mengenai integral tak tentu, di mana integral ini memiliki ciri selalu ada nilai konstantanya. Hal ini menunjukkan bahwa ada nilai yang belum terukur dengan jelas. Jika nantinya pada integral sudah ditentukan batas bawah dan batas atasnya, sehingga tidak akan diperlukan lagi nilai konstantanya. Adapun rumus formulasi dari integral tentu adalah sebagai berikut:

$\displaystyle \int_{a}^{b}f(x)\: dx=\left [ F(x) \right ]_{a}^{b}=F(x)|_{a}^{b}=F(b)-F(a)$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{aligned}1.\quad &\int_{1}^{3}\left ( 2x-1 \right )\: dx\\ &=\left [ x^{2}-x \right ]_{1}^{3}\\ &=(9-3)-(1-1)=6\\ 2.\quad &\int_{0}^{5}\left | x-1 \right |+\left | x-2 \right |\: dx\\ &=\frac{1}{2}(x-1)\left | x-1 \right ||_{0}^{5}\: +\frac{1}{2}(x-2)\left | x-2 \right ||_{0}^{5}\\ &=\left ( \frac{1}{2}.4.4+\frac{1}{2}.3.3 \right )-\left ( \frac{1}{2}.-1.1+\frac{1}{2}.-2.2 \right )\\ &=8+4\frac{1}{2}+\frac{1}{2}+2=15 \end{aligned}$.

Anda dapat mengecek jawaban di atas dengan melmperhatikan gambar kemudian menyelesaikannya dengan prinsip segitiga, persegi, dan atau persegipanjang.

B. Sifat-Sifat Integral Tentu

$\begin{aligned}1.\quad&\int_{a}^{b}f(x)\: dx=-\int_{b}^{a}f(x)\: dx\\ 2.\quad&\displaystyle \int_{a}^{b}k.f(x)\: dx=k\int_{a}^{b}f(x)\: dx\\ 3.\quad&\displaystyle \int_{a}^{b}\left ( f(x)\pm g(x) \right )\: dx\\ &=\displaystyle \int_{a}^{b}f(x)\: dx\: \pm\int_{a}^{b}g(x) \: dx\\ 4.\quad &\displaystyle \int_{a}^{b}f(x)\: dx\\ &=\displaystyle \int_{a}^{c}f(x)\: dx+\int_{c}^{b}f(x)\: dx\: ,\: \: dengan\: \: a<c<b\\ 5.\quad &\displaystyle \int_{a}^{a}f(x)\: dx=0 \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Hitunglah hasil dari integral}\: \: \displaystyle \int_{1}^{3}\left ( 2y+1 \right )\: dy\\ \\\\&\textbf{Jawab}:\\ &\begin{aligned}\int_{1}^{3}\left ( 2y+1 \right )\: dy&=y^{2}+y\: |_{1}^{3}\\ &=\left ( 3^{2}+3 \right )-\left ( 1^{2}+1 \right )\\ &=12-2\\ &=10 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Hitunglah hasil dari integral}\: \:  \displaystyle \int_{-1}^{1}x^{3}-1\: dx\\\\&\textbf{Jawab}:\\ &\begin{aligned}\int_{-1}^{1}&\left ( x^{3}-1 \right )\: dx\\ &=-(\displaystyle \frac{1}{4}x^{4}-x|_{-1}^{1})\\ &=-\left ( \displaystyle \frac{1}{4}.\left ( 1 \right )^{4}-1 \right )+\left ( \displaystyle \frac{1}{4}.\left ( -1 \right )^{4}-\left ( -1 \right ) \right )\\ &=-\left ( \displaystyle \frac{1}{4}-1 \right )+\left ( \displaystyle \frac{1}{4}+1 \right )\\ &=2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Hitunglah hasil dari integral}\: \:  \displaystyle \int_{-2}^{1}\left ( x+3 \right )^{2}\: dx\\\\&\textbf{Jawab}:\\ &\begin{aligned}\int_{-2}^{1}&\left ( x+3 \right )^{2}\: dx\\ &=\displaystyle \frac{1}{3}x^{2}+3x^{2}+9x|_{-2}^{1}\\ &=\left ( \displaystyle \frac{1}{3}.\left ( 1 \right )^{3}+3.1^{2}+9.1 \right )-\left ( \displaystyle \frac{1}{3}.\left ( -2 \right )^{3}+3.(-2)^{2}+9.(-2) \right )\\ &=\left ( \displaystyle \frac{1}{3}+12\right )-\left ( -\displaystyle \frac{8}{3}+12-18 \right )\\ &=\displaystyle \frac{9}{3}+18=3+18=21 \end{aligned} \end{array}$ .

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{array}{ll}\\ &\textrm{Hitunglah nilai tiap integral berikut ini}\\ &\begin{array}{lllll}\\ 1.&\displaystyle \int_{0}^{3}2x\: dx&10.&\displaystyle \int_{1}^{2}\sqrt{x^{5}}\: dx\\ 2.&\displaystyle \int_{2}^{3}2x^{2}\: dx&11.&\displaystyle \int_{1}^{2}\displaystyle \frac{1}{x^{2}}\: dx\\ 3.&\displaystyle \int_{3}^{7}\displaystyle \frac{1}{2}x^{3}\: dx&12.&\displaystyle \int_{4}^{9}3\sqrt{x}\: dx\\ 4.&\displaystyle \int_{0}^{3}\left ( x^{2}+2x-1 \right )\: dx&\displaystyle \\ 5.&\displaystyle \int_{0}^{3}\left ( x+3 \right )^{2}\: dx&\\ 6.&\displaystyle \displaystyle \int_{-1}^{2}\left ( x+2 \right )\left ( x-1 \right )\: dx&\\ 7.&\displaystyle \int_{1}^{3}\left ( t+t^{2}-3t^{3} \right )\: dt&\\ 8.&\displaystyle \int_{1}^{4}\left ( \displaystyle \frac{x^{4}-x^{3}+\sqrt{x}-1}{x^{2}} \right )\: dx&\\ 9.&\displaystyle \int_{4}^{6}\left ( \sqrt{t}+t^{2}-\sqrt[3]{t^{2}} \right )\: dt \end{array}  \end{array}$.

Contoh Soal 4 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 16.&\textbf{(UN 2015 Matematika IPA)}\\ &\textrm{Hasil}\: \: \displaystyle \int 6x\left ( 1-x^{2} \right )^{4}\: \: dx\: \: \textrm{adalah ....}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{3}{5}\left ( 1+x^{2} \right )^{5}+C\\\\ \textrm{b}.\quad \displaystyle \frac{2}{5}\left ( 1+x^{2} \right )^{5}+C\\\\ \textrm{c}.\quad \displaystyle -\frac{1}{5}\left ( 1-x^{2} \right )^{5}+C\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{5}\left ( 1-x^{2} \right )^{5}+C\\\\ \textrm{e}.\quad \color{red}-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C\end{array}\\\\&\textbf{Jawab}:\\&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Dengan integral substitusi}\\ &\begin{array}{l}\\ \begin{aligned}\displaystyle &\int 6x\left ( 1-x^{2} \right )^{4}\: dx\\ &=\int \left ( 1-x^{2} \right )^{4}.6x\: dx\\ &\begin{cases} u& =1-x^{2} \\ & \\ du& =-2x\: \: \: dx\quad \Rightarrow \quad -3\: dx=6x\: dx \end{cases}\\ &\displaystyle \int u^{4}.\left ( -3\: du \right )\\ &=-3\displaystyle \int u^{4}\: du\\ &=-\displaystyle \frac{3}{5}u^{5}+C\\ &=-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C \end{aligned} \end{array}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned} &\displaystyle \int \left ( 1-x^{2} \right )^{4}\: 6x\: dx\\ &=\int \left ( 1-x^{2} \right )^{2}.(-3).\left ( -2x\: dx \right )\\ &=-3\displaystyle \int \left ( 1-x^{2} \right )^{4}.\: \left ( -2x\: dx \right )\\ &=-3\displaystyle \int \left ( 1-x^{2} \right )^{4}\: .d\left ( 1-x^{2} \right )\\ &=-3\left [ \displaystyle \frac{\left ( 1-x^{2} \right )^{5}}{5} \right ]+C\\ &=-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C\end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 17.&\begin{aligned}\displaystyle &\int \frac{dx}{2022x+2023}=\: ....\end{aligned}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{b}.\quad \displaystyle \frac{1}{2021}\ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{c}.\quad \color{red}\displaystyle \frac{1}{2022}\ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{d}.\quad \displaystyle \frac{1}{2023}\ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{e}.\quad \displaystyle \frac{2022}{2023}\ln \left | 2022x+2023 \right |+C \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int \frac{dx}{2022x+2023}\\ &\textrm{Misalkan}\\ &u=2022x+2023\\ &\Leftrightarrow du=2022\: dx\Leftrightarrow \displaystyle \frac{1}{2022}\: du=\: dx\\ &\textrm{Sehingga}\\ &=\displaystyle \int \frac{1}{u}.\left ( \frac{1}{2022}\: du \right )=\displaystyle \frac{1}{2022}\int \frac{1}{u}\: du\\ &=\displaystyle \frac{1}{2022}\ln \left | 2022x+2023 \right |+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Diketahui}\: \: \displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\: \: \textrm{jika} \: f(4)=19,\\ &\textrm{ maka}\: \: f(1)=....\\ &\begin{array}{ll}\\ \textrm{a}.\quad 2\\ \textrm{b}.\quad 3\\ \textrm{c}.\quad 4\\ \textrm{d}.\quad \color{red}5\\ \textrm{e}.\quad 6\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perlu diingat bahwa}\: \: \displaystyle \frac{d}{dx}f(x)=h(x)\\ &\Leftrightarrow d(f(x))=h(x)\: dx\\ &\Leftrightarrow  \displaystyle \int d(f(x))=\displaystyle \int h(x)\: dx\\ &\textrm{Perhatikan kasus di pada soal di atas}\\ &\displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\Leftrightarrow d\left ( f(x) \right )=3\sqrt{x}\: dx\\ &\Leftrightarrow \displaystyle \int d\left ( f(x) \right )=\int 3\sqrt{x}\: dx=\int 3x^{.^{\frac{1}{2}}}\: dx\\&\Leftrightarrow f(x)=2x^{.^{\frac{3}{2}}}+C=2x\sqrt{x}+C \end{aligned}\\&\begin{aligned}&\textrm{Karena diketahui}\\ &f(4)=19,\: \: \textrm{maka}\\ &\Leftrightarrow 2(4)(\sqrt{4})+C=19\\ &\Leftrightarrow 16+C=19\\ &\Leftrightarrow C=3\\ &\textrm{Sehingga}\quad f(x)=2x\sqrt{x}+3\\ &\textrm{maka}\quad f(1)=2.1.\sqrt{1}+3=2+3=5 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\begin{aligned}\displaystyle &-\int \frac{dx}{x^{2}+3x+2}\: \: dx=\: .... \end{aligned}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \ln \left | \displaystyle \frac{x+1}{x+2} \right |+C\\ &\\ \textrm{b}.\quad \color{red}\displaystyle \ln \left | \displaystyle \frac{x+2}{x+1} \right |+C\\ &\\ \textrm{c}.\quad \displaystyle \ln \left | x^{2}+3x+2 \right |+C\\ &\\ \textrm{d}.\quad \arctan 2\left ( \displaystyle x+\frac{3}{2} \right )+C\\ &\\ \textrm{e}.\quad -\displaystyle \arctan 2\left (\displaystyle x+ \frac{3}{2}\right )+C \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}\displaystyle -\int \frac{dx}{x^{2}+3x+2}&=-\displaystyle \int \frac{1}{\left ( x+1 \right )\left ( x+2 \right )}\: dx\\ &=-\displaystyle \int \left ( \displaystyle \frac{1}{x+1}-\displaystyle \frac{1}{x+2} \right )\: dx\\ &=-\displaystyle \int \frac{1}{x+1}\: dx+\displaystyle \int \frac{1}{x+2}\: dx\\ &=\displaystyle \int \frac{1}{x+2}\: dx-\displaystyle \int \frac{1}{x+1}\: dx\\ &=\displaystyle \ln \left | x+2 \right |-\displaystyle \ln \left | x+1 \right |+C\\ &=\displaystyle \ln \left | \displaystyle \frac{x+2}{x+1} \right |+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Tentukanah integral berikut}?\\ &\textrm{a}.\quad \int dt\\&\textrm{b}.\quad \int 4\: dw\\ &\textrm{c}.\quad \int \left ( x^{3}+5 \right )\: dx\\ &\textrm{d}.\quad \int \left ( x^{\frac{3}{2}}-2\sqrt{x}+1 \right )dx\\ &\textrm{e}.\quad \int \left ( 5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}} \right )dx\\\\ &\textbf{Jawab}:\\ &\begin{array}{l}\\ \begin{aligned}\textrm{a}.\quad \int dt=t+c \end{aligned}\\ \begin{aligned}\textrm{b}.\quad \int 4\: dw=4w+C \end{aligned}\\ \begin{aligned}\textrm{c}.\quad \int (x^{3}+5)\: dx&=\displaystyle \frac{1}{3+1}x^{3+1}+5x+C\\ &=\displaystyle \frac{1}{4}x^{4}+5x+C \end{aligned} \end{array}\\ &\begin{aligned}\textrm{d}.\quad &\int (x^{\frac{3}{2}}+2\sqrt{x}+1)\: dx\\ &=\int \left (x^{\frac{3}{2}}+2x^{\frac{1}{2}}+1 \right )\: dx\\ &=\displaystyle \frac{1}{\displaystyle \frac{3}{2}+1}x^{\frac{3}{2}+1}+\frac{2}{\displaystyle \frac{1}{2}+1}x^{\frac{1}{2}+1} +x+C\\ &=\displaystyle \frac{1}{\displaystyle \frac{5}{2}}x^{\frac{5}{2}}+\frac{2}{\displaystyle \frac{3}{2}}x^{\frac{3}{2}}+x+C\\ &=\frac{2}{5}x^{\frac{5}{2}}+\frac{4}{3}x^{\frac{3}{2}}+x+C\\ &\quad \textrm{atau dapat juga kita menyatakan dengan}\\ &=\frac{2}{5}x^{2\frac{1}{2}}+\frac{4}{3}x^{1\frac{1}{2}}+x+C\\ &=\frac{2}{5}x^{2}\sqrt{x}+\frac{4}{3}x\sqrt{x}+x+C \end{aligned} \\ &\begin{aligned}\textrm{e}.\quad &\int \left ( 5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}} \right )dx\\ &=.....\\ &\quad \textrm{silahkan dicoba sendiri sebagai latihan}\end{aligned} \end{array}$.

Contoh Soal 3 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 11.&\displaystyle \int d\left ( 2x \right )=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle x+C\\ \textrm{b}.\quad \color{red}\displaystyle 2x+C\\ \textrm{c}.\quad \displaystyle 3x+C\\ \textrm{d}.\quad \displaystyle 4x+C\\ \textrm{e}.\quad \displaystyle 5x+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int d\left ( 2x \right )=\int 2\: \: dx=2x+C. \end{array}$.

$\begin{array}{ll}\\ 12.&\displaystyle \int 5\: \: d\left ( 2x \right )=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle 5+C\\ \textrm{b}.\quad \displaystyle \frac{5}{2}x+C\\ \textrm{c}.\quad \displaystyle 5x+C\\ \textrm{d}.\quad \color{red}\displaystyle 10x+C\\ \textrm{e}.\quad \displaystyle 25x+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int 5\: \: d\left ( 2x \right )=\int5. 2\: \: dx=\int 10\: dx=10x+C. \end{array}$.

$\begin{array}{ll}\\ 13.&\displaystyle \int 6x\: \: d\left ( 3x \right )=\: ....\\ &\begin{array}{ll}\\  \textrm{a}.\quad \displaystyle 18x^{2}+C\\ \textrm{b}.\quad \color{red}\displaystyle 9x^{2}+C\\ \textrm{c}.\quad \displaystyle 5x^{2}+C\\ \textrm{d}.\quad \displaystyle 3x^{2}+C\\ \textrm{e}.\quad \displaystyle 2x^{2}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int 6x\: \: d\left ( 3x \right )=\int 6x. 3\: \: dx=\int 18x\: dx\\ &=\displaystyle \frac{18x^{2}}{2}+C=9x^{2}+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\displaystyle \int \left ( 8x^{3}-2x \right )\: \: d\left ( -2x \right )=\: ....\\&\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle -16x^{4}+4x^{2}+C\\ \textrm{b}.\quad \displaystyle 16x^{4}-4x^{2}+C\\ \textrm{c}.\quad \displaystyle -4x^{4}-2x^{2}+C\\ \textrm{d}.\quad \color{red}\displaystyle -4x^{4}+2x^{2}+C\\ \textrm{e}.\quad \displaystyle 4x^{4}-2x^{2}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int \left (8x^{3}-2x \right )\: \: d\left ( -2x \right )\\ &=\int \left ( 8x^{3}-2x \right ).\left (-2\: \: dx \right )\\&=\int \left (-16x^{3}+4x \right )\: dx\\ &=\displaystyle -\frac{16x^{4}}{4}+\frac{4x^{2}}{2}+C\\ &=\displaystyle -4x^{4}+2x^{2}+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\displaystyle \int \left ( x+3 \right )\: \: d\left ( 2x+6 \right )=\: ....\\&\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle 4x^{2}+24x+C\\ \textrm{b}.\quad \displaystyle 2x^{2}+12x+C\\ \textrm{c}.\quad \color{red}\displaystyle x^{2}+6x+C\\ \textrm{d}.\quad \displaystyle 4x^{2}+C\\ \textrm{e}.\quad \displaystyle x^{2}+C\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|}\hline \color{blue}\textrm{Pertama}&\color{red}\textrm{Kedua}\\\hline \begin{aligned}&\displaystyle \int \left ( x+3 \right )\: d\left ( 2x+6 \right )\\ &=\displaystyle \frac{1}{2}\int \left ( 2x+6 \right )\: d\left ( 2x+6 \right )\\ &=\displaystyle \frac{1}{2}\left [ \frac{1}{2}\left ( 2x+6 \right )^{2} \right ]+C\\ &=\displaystyle \frac{1}{4}\left ( 4x^{2}+24x+36 \right )+C\\ &=\displaystyle \frac{4x^{2}}{4}+\frac{24x}{4}+\frac{36}{4}+C\\ &=x^{2}+6x+\underset{C}{\underbrace{9+C}}\\ &=x^{2}+6x+C \end{aligned}&\begin{aligned}&\displaystyle \int \left ( x+3 \right )\: d\left ( 2x+6 \right )\\ &=\int \left ( x+3 \right ).\left ( 2\: dx \right )\\ &=\int \left ( 2x+6 \right )\: dx\\ &=\displaystyle \frac{2x^{2}}{2}+6x+C\\ &=x^{2}+6x+C\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$ .

Contoh Soal 2 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 6.&\displaystyle \int x^{3}\sqrt{x}\: dx=\: ....\\ &\begin{array}{ll}\\  \textrm{a}.\quad \color{red}\displaystyle \frac{2}{9}x^{4}\sqrt{x}+C\\ \textrm{b}.\quad \displaystyle \frac{9}{2}x^{4}\sqrt{x}+C\\ \textrm{c}.\quad \displaystyle \frac{1}{9}x^{4}\sqrt{x}+C\\ \textrm{d}.\quad \displaystyle 9x^{4}\sqrt{x}+C\\ \textrm{e}.\quad \displaystyle x^{4}\sqrt{x}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int x^{3}\sqrt{x}\: dx=\int x^{3}.x^{\frac{1}{2}}\: dx\\ &=\int x^{3\frac{1}{2}}\: dx=\int x^\frac{7}{2}\: dx=\displaystyle \frac{x^{\frac{7}{2}+1}}{\displaystyle \frac{7}{2}+1}+C\\ &=\displaystyle \frac{x^{\frac{9}{2}}}{\displaystyle \frac{9}{2}}+C=\displaystyle \frac{2}{9}x^{4\frac{1}{2}}+C=\displaystyle \frac{2}{9}x^{4}\sqrt{x}+C  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\displaystyle \int x\sqrt{x\sqrt[3]{x^{2}}}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{17}{6}x^{2}\sqrt{x\sqrt[3]{x^{2}}}+C\\ \textrm{b}.\quad \color{red}\displaystyle \frac{6}{17}x^{2}\sqrt{x\sqrt[3]{x^{2}}}+C\\ \textrm{c}.\quad \displaystyle x^{2}\sqrt{x\sqrt[3]{x^{2}}}+C\\ \textrm{d}.\quad \displaystyle \frac{6}{17}x\sqrt{x\sqrt[3]{x^{2}}}+C\\ \textrm{e}.\quad \displaystyle \frac{1}{2}x\sqrt{x\sqrt[3]{x^{2}}}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int x\sqrt{x\sqrt[3]{x^{2}}}\: dx\\&=\displaystyle \int x\left ( x.x^{\frac{2}{3}} \right )^{\frac{1}{2}}\: dx\\ &=\displaystyle \int x^{1+\frac{1}{2}+\frac{2}{6}}\: dx\\ &=\displaystyle \int x^{\frac{11}{6}}\: dx=\displaystyle \frac{x^{\frac{11}{6}+1}}{\displaystyle \frac{11}{6}+1}+C\\ &=\displaystyle \frac{x^{\frac{17}{6}}}{\displaystyle \frac{17}{6}}+C=\displaystyle \frac{6}{17}x^{2}.x^{\frac{5}{6}}+C\\ &=\displaystyle \frac{6}{17}x^{2}\left ( x.x^{\frac{2}{3}} \right )^{\frac{1}{2}}+C\\&=\displaystyle \frac{6}{17}x^{2}\sqrt{x\sqrt[3]{x^{2}}}+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\displaystyle \int x^{2}\sqrt{x^{2}\sqrt[3]{x^{3}}}\: dx=\: ....\\&\begin{array}{ll}\\ \textbf{a}.\quad \displaystyle x^{3}\sqrt{x^{2}\sqrt[3]{x^{3}}}+C\\ \textbf{b}.\quad \displaystyle \frac{27}{6}x^{3}\sqrt{x^{2}\sqrt[3]{x^{3}}}+C\\ \textbf{c}.\quad \color{red}\displaystyle \frac{6}{27}x^{3}\sqrt{x^{2}\sqrt[3]{x^{3}}}+C\\ \textbf{d}.\quad \displaystyle \frac{6}{21}x^{2}\sqrt{x^{2}\sqrt[3]{x^{3}}}+C\\ \textbf{e}.\quad \displaystyle \frac{6}{21}x^{3}\sqrt{x^{2}\sqrt[3]{x^{3}}}+C\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int x^{2}\sqrt{x^{2}\sqrt[3]{x^{3}}}\: dx=\int x^{2}\left ( x^{2}.x^{1} \right )^{\frac{1}{2}}\: dx\\ &=\int x^{2+\frac{2}{2}+\frac{1}{2}}\: dx=\int x^{\frac{7}{2}}\: dx=\displaystyle \frac{x^{\frac{7}{2}+1}}{\displaystyle \frac{7}{2}+1}+C\\ &=\displaystyle \frac{x^{\frac{9}{2}}}{\displaystyle \frac{9}{2}}+C\\ &=\displaystyle \frac{2}{9}x^{\frac{9}{2}}+C\\ &=\displaystyle \frac{2.3}{9.3}x^{3}.x^{\frac{3}{2}}+C\\ &=\displaystyle \frac{6}{27}x^{3}\left ( x^{3} \right )^{\frac{1}{2}}+C\\ &=\displaystyle \frac{6}{27}x^{3}\sqrt{x^{2}.x}+C\\ &=\displaystyle \frac{6}{27}x^{3}\sqrt{x^{2}\sqrt[3]{x^{3}}}+C\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\displaystyle \int x^{3}\sqrt{\displaystyle \frac{1}{x}\sqrt[3]{x^{2}}}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}x^{4}\sqrt{\displaystyle \frac{1}{x}\sqrt[3]{x^{2}}}+C\\ \textrm{b}.\quad \color{red}\displaystyle \frac{6}{23}x^{4}\sqrt{\displaystyle \frac{1}{x}\sqrt[3]{x^{2}}}+C\\ \textrm{c}.\quad \displaystyle \frac{23}{6}x^{4}\sqrt{\displaystyle \frac{1}{x}\sqrt[3]{x^{2}}}+C\\ \textrm{d}.\quad \displaystyle \frac{23}{6}x^{3}\sqrt{\displaystyle \frac{1}{x}\sqrt[3]{x^{2}}}+C\\ \textrm{e}.\quad \displaystyle \frac{3}{4}x^{3}\sqrt{\displaystyle \frac{1}{x}\sqrt[3]{x^{2}}}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int x^{3}\sqrt{\frac{1}{x}\sqrt[3]{x^{2}}}\: dx=\int x^{3}\left ( x^{-1}.x^{\frac{2}{3}} \right )^{\frac{1}{2}}\: dx\\ &=\int x^{3-\frac{1}{2}+\frac{1}{3}}\: dx=\int x^{\frac{17}{6}}\: dx\\ &=\displaystyle \frac{x^{\frac{17}{6}+1}}{\displaystyle \frac{17}{6}+1}+C=\displaystyle \frac{x^{\frac{23}{6}}}{\displaystyle \frac{23}{6}}+C\\ &=\displaystyle \frac{6}{23}x^{\frac{24-1}{6}}+C\\ &=\displaystyle \frac{6}{23}x^{4}.x^{-\frac{1}{6}}+C\\ &=\displaystyle \frac{6}{23}x^{4}.\left ( x^{-\frac{1}{3}} \right )^{\frac{1}{2}}+C\\ &=\displaystyle \frac{6}{23}x^{4}\sqrt{x^{-1+\frac{2}{3}}}+C\\ &=\displaystyle \frac{6}{23}x^{4}\sqrt{\frac{1}{x}\sqrt[3]{x^{2}}}+C\end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 10.&\displaystyle \int x\sqrt{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle x^{2}\sqrt{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C\\\\ \textrm{b}.\quad \displaystyle \frac{13}{8}x^{2}\sqrt{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C\\\\ \textrm{c}.\quad \color{red}\displaystyle \frac{8}{13}x^{2}\sqrt{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C\\\\ \textrm{d}.\quad \displaystyle \frac{1}{2}x^{2}\sqrt{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C\\\\ \textrm{e}.\quad \displaystyle \frac{8}{5}x^{2}\sqrt{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int x\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}\: \: dx\\&=\int x\left ( x^{-1}\left ( x\left ( x^{-1} \right )^{\frac{1}{2}} \right )^{\frac{1}{2}} \right )^{\frac{1}{2}} \: dx\\ &=\int x\left ( x^{-\frac{1}{2}}.x^{\frac{1}{4}}.x^{-\frac{1}{8}} \right )\: dx\\ &=\int x^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}}\: dx=\int x^{\frac{5}{8}}\: dx\\ &=\displaystyle \frac{x^{\frac{5}{8}+1}}{\displaystyle \frac{5}{8}+1}+C\\ &=\displaystyle \frac{x^{\frac{13}{8}}}{\displaystyle \frac{13}{8}}+C\\ &=\displaystyle \frac{8}{13}x^{\frac{16-3}{8}}+C\\ &=\displaystyle \frac{8}{13}x^{2}.x^{-\frac{3}{8}}+C\\ &=\displaystyle \frac{8}{13}x^{2}\sqrt{x^{-\frac{3}{4}}}+C\\ &=\displaystyle \frac{8}{13}x^{2}\sqrt{x^{-1+\frac{1}{4}}}+C\\ &=\displaystyle \frac{8}{13}x^{2}\sqrt{\frac{1}{x}.\left ( x^{\frac{1}{2}} \right )^{\frac{1}{2}}}+C\\ &=\displaystyle \frac{8}{13}x^{2}\sqrt{\frac{1}{x}\sqrt{x^{1-\frac{1}{2}}}}+C\\ &=\displaystyle \frac{8}{13}x^{2}\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C\end{aligned} \end{array}$

Contoh Soal 1 Materi Integral Tak Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 1.&\displaystyle \int x^{2}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{1}{3}x^{3}+C\\ \textrm{b}.\quad \displaystyle \frac{1}{4}x^{6}+C\\ \textrm{c}.\quad \displaystyle \frac{1}{3}x^{6}+C\\ \textrm{d}.\quad \displaystyle \frac{1}{6}x^{3}+C\\ \textrm{e}.\quad \displaystyle \frac{2}{3}x^{3}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int x^{2}\: dx=\displaystyle \frac{x^{2+1}}{2+1}+C= \displaystyle \frac{1}{3}x^{3}+C \end{array}$

$\begin{array}{ll}\\ 2.&\displaystyle \int x^{-2}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle -2x^{-1}+C\\ \textrm{b}.\quad \color{red}\displaystyle -x^{-1}+C\\ \textrm{c}.\quad \displaystyle -\frac{1}{2}x^{-2}+C\\ \textrm{d}.\quad \displaystyle -\frac{1}{3}x^{-3}+C\\ \textrm{e}.\quad \displaystyle -3x^{-3}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int x^{-2}\: dx=\displaystyle \frac{x^{-2+1}}{-2+1}+C= \displaystyle \frac{1}{-1}x^{-1}+C=-x^{-1}+C \end{array}$.

$\begin{array}{ll}\\ 3.&\displaystyle \int x^{.^{\frac{1}{3}}}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{3}{4}x^{\frac{4}{3}}+C\\ \textrm{b}.\quad \displaystyle x^{\frac{4}{3}}+C\\ \textrm{c}.\quad \displaystyle \frac{3}{4}x^{\frac{2}{3}}+C\\ \textrm{d}.\quad \displaystyle x^{-\frac{2}{3}}+C\\ \textrm{e}.\quad \displaystyle \frac{3}{4}x^{-\frac{2}{3}}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int x^{.^{\frac{1}{3}}}\: dx=\displaystyle \frac{x^{\frac{1}{3}+1}}{\displaystyle \frac{1}{3}+1}+C= \displaystyle \frac{1}{\displaystyle \frac{4}{3}}x^{.^{\frac{4}{3}}}+C=\displaystyle \frac{3}{4}x^{.^{\frac{4}{3}}}+C. \end{array}$.

$\begin{array}{ll}\\ 4.&\displaystyle \int \frac{1}{x^{3}}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \color{red}\displaystyle -\frac{1}{2x^{2}}+C\\ \textrm{b}.\quad \displaystyle -\frac{2}{x^{2}}+C\\ \textrm{c}.\quad \displaystyle \frac{1}{3x^{4}}+C\\ \textrm{d}.\quad \displaystyle \frac{3}{x^{4}}+C\\ \textrm{e}.\quad \displaystyle -\frac{1}{4x^{3}}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int \frac{1}{x^{3}}\: dx=\int x^{-3}\: dx\\ &=\displaystyle \frac{x^{-3+1}}{-3+1}+C=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^{2}}+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\displaystyle \int \frac{1}{3}x^{3}\: dx=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{1}{3}x^{4}+C\\ \textrm{b}.\quad \displaystyle \frac{1}{4}x^{4}+C\\ \textrm{c}.\quad \displaystyle x^{4}+C\\ \textrm{d}.\quad \color{red}\displaystyle \frac{1}{12}x^{4}+C\\ \textrm{e}.\quad \displaystyle \frac{4}{3}x^{4}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int \frac{1}{3}x^{3}\: dx=\displaystyle \frac{1}{3}.\frac{x^{3+1}}{3+1}+C= \displaystyle \frac{1}{12}x^{4}+C \end{array}$.

Penggunaan Integral Tak Tentu

Penggunaan Integral Tak Tentu

Penggunaan integral tak tentu ini dapat digunakan dalam menentukan suatu fungsi jika turunan dari fungsi tersebut diberikan. Selain itu untuk menentukan posisi, kecepatan, percepatan suatu benda pada waktu tertentu.

$\begin{aligned}&\textrm{Misalkan}\\ &\bullet \quad s\: \: \textrm{adalah menunjukkan posisi}\\ &\bullet \quad v\: \: \textrm{adalah menunjukkan kecepatan}\\ &\bullet \quad a\: \: \textrm{adalah menunjukkan percepatan}\\ &\bullet \quad t\: \: \textrm{adalah menunjukkan waktu}\\\\ &\textbf{Perhatikan hubungan berikut}\\ &v=\displaystyle \frac{ds}{dt}\Leftrightarrow  ds=v\: dt\\ &\Leftrightarrow  \displaystyle \int ds =\int v\: dt\\ &\Leftrightarrow \: \: \quad s=\displaystyle \int v\: dt\\\\ &\textbf{Demikian juga hubungan berikut}\\ &a=\displaystyle \frac{dv}{dt}\Leftrightarrow  dv=a\: dt\\ &\Leftrightarrow  \displaystyle \int dv =\int a\: dt\\ &\Leftrightarrow \: \: \quad v=\displaystyle \int a\: dt \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukan}\: \: y,\: \textrm{Jika}\: \: \displaystyle \frac{dy}{dx}=2022x\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &\displaystyle \frac{dy}{dx}=2022x\Leftrightarrow  dy=2022x\: dx\\ &\Leftrightarrow \displaystyle \int dy=\int 2022x\: dx\\ &\Leftrightarrow y=\displaystyle \frac{2022}{2}x^{2}+C\\ &\Leftrightarrow y=1011x^{2}+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui fungsi turunan pertama kurva}\\ &\textrm{adalah}\: \:  \displaystyle \frac{dy}{dx}=2x-2\: .\:  \textrm{Jika kurva melalui}\\ &\textrm{titik}\: \: (3,2)\: ,\: \textrm{tentukan persamaan dari kurva}\\ &\textrm{tersebut}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\displaystyle \frac{dy}{dx}=2x-2\\ &\Leftrightarrow  dy=(2x-2)\: dx\\ &\Leftrightarrow \displaystyle \int dy=\int (2x-2)\: dx\\ &\Leftrightarrow \quad  y=x^{2}-2x+C\\ &\textrm{Karena kurva melalui}\: \: (3,2),\: \textrm{maka}\\ &(2)=(3)^{2}-2(3)+C\Leftrightarrow 2=3+C\\ &\Leftrightarrow C=-1\\ &\textrm{Jadi},\: y=x^{2}-2x-1\: \: \textrm{atau}\\ &\qquad f(x)=x^{2}-2x-1  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui bahwa}\: \: f''(x)=x^{2}.\: \textrm{Jika}\: \: f'(0)=6\\ &\textrm{dan}\: \:  f(0)=3\: ,\:  \textrm{tentukanlah}\: \: f(x)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\displaystyle f''(x)=x^{2}\\ &\Leftrightarrow  \displaystyle \int f''(x)\: dx=\displaystyle \int x^{2}\: dx\\ &\Leftrightarrow \displaystyle f'(x)=\displaystyle \frac{1}{3}x^{3}+C\\  &\textrm{Karena}\: \: f'(x)=6,\: \textrm{maka}\\ &f'(x)=\displaystyle \frac{1}{3}x^{3}+C\Leftrightarrow 6=\displaystyle \frac{1}{3}(0)^{3}+C\\ &\Leftrightarrow C=6\\ &\textrm{Sehingga}\: \: f'(x)=\displaystyle \frac{1}{3}x^{3}+6\\ &\textrm{Selanjutnya}\: \: \displaystyle \int f'(x)\: dx=\displaystyle \int \displaystyle \frac{1}{3}x^{3}+6\: dx\\ &f(x)=\displaystyle \frac{1}{12}x^{4}+6x+C\\ &\textrm{Dan juga karena}\: \: f(0)=3,\: \textrm{maka}\\ &f(0)=\displaystyle \frac{1}{12}(0)^{4}+6(0)+C=3\\ &C=3, \: \textrm{sehingga diperoleh}\\ &f(x)=\displaystyle \frac{1}{12}x^{4}+6x+3\\ &\textrm{Jadi},\: f(x)=\color{red}\displaystyle \frac{1}{12}x^{4}+6x+3  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: f'\left ( x \right )=6x^{2}-2x+6\: \textrm{dan}\\ & \textrm{nilai}\: \textrm{fungsi}\: f\left ( 2 \right )=-7.\: \textrm{Tentukanlah}\\ &\textrm{rumus}\: \textrm{fungsi}\: \textrm{tersebut}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f\left ( x \right )&=\int f'\left ( x \right )\: dx\\ &=\int \left ( 6x^{2}-2x+6 \right )\: dx\\ &=2x^{3}-x^{2}+6x+C \end{aligned}\\\\ &\textrm{Karena}\: f\left ( 2 \right )=-7,\: \textrm{maka}\\\\ &\begin{aligned}f\left ( 2 \right )&=2.2^{3}-2^{2}+6.2+C\\ \Leftrightarrow -7&=16-4+12+C\\ \Leftrightarrow C&=-31 \end{aligned}\\\\ &\textrm{Jadi},\: f\left ( x \right )=2x^{3}-x^{2}+6x-31 \end{array}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukan}\: \: f(x),\: \textrm{Jika diketahui}\\ &\textrm{a}.\quad f'(x)=12x^{3}\: \: \textrm{dan}\: \: f(1)=8\\ &\textrm{b}.\quad f'(x)=x^{2}-2x+3\: \: \textrm{dan}\: \: f(3)=9\\ &\textrm{c}.\quad f''(x)=2,\: \: f'(2)=2\: \: \textrm{dan}\: \: f(2)=10\\ &\textrm{d}.\quad f''(x)=x^{2},\: \: f'(0)=6\: \: \textrm{dan}\: \: f(0)=3 \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukan persamaan kurva}\: \: f(x)\\ &\textrm{disetiap titik}\: \: (x,y)\: \: \textrm{yang memenuhi}\\ &\textrm{syarat berikut}\\ &\textrm{a}.\quad \displaystyle \frac{dy}{dx}=4x+1\: \: \textrm{dan kurva melalui}\: \: (0,2)\\ &\textrm{b}.\quad \displaystyle \frac{dy}{dx}=\displaystyle \frac{1}{x^{2}}+3\: \: \textrm{dan kurva melalui}\: \: (1,4) \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: 2x-y=3\: \: \textrm{merupakan garis}\\ &\textrm{kurva di titik}\: \: (1,-1)\: .\: \textrm{Jika di tiap titik}\\ &\textrm{pada kurva berlaku}\: \: \displaystyle y''=2x^{2}-3x+1\\ &\textrm{tentukan persamaan kurva tersebut} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui kecepatan sebuah sepeda}\\ &\textrm{diformulasian dengan}\: \: \displaystyle \frac{ds}{dt}=3t^{2}+4t-1\\ &\textrm{Jika}\: \: s\: \: \textrm{menyatakan jarak yang ditempuh}\\ &\textrm{dalam satuan meter}\: ,\: t\: \textrm{menyatakan waktu}\\ &\textrm{dengan}\: \: s=5\: \: \textrm{untuk}\: \: t=2,\: \textrm{tentukanlah}\\ &\textrm{jarak yang ditempuh pengendara dalam}\\ &\textrm{waktu 5 menit} \end{array}$.

Teknik Pengintegralan (Bagian 2)

2. Integral Parsial

2. 1 Integral Parsial

Jika teknik pada no.1 pada pembahasan sebelumnya tidak dapat digunakan, maka kemungkinan adalah dengan menggunakan teknik yang satunya ini, yaitu teknik integral parsial. Adapun untuk teknik integral ini diformulasikan dengan bentuk rumus

$\displaystyle \int u\: dv=uv-\displaystyle \int v\: du$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Perhatian kembali soal berikut}\\ &\displaystyle \int x\sqrt{x-1}\: dx\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Misalkan}\qquad u=x-1\\ &du=1\quad dx\: \Leftrightarrow \: du=dx \end{aligned}\\  &\textrm{Dengan integral substitusi}\\ &\begin{aligned} &\displaystyle \int x\sqrt{x-1}\: dx\\&=\int \left ( x-1+1 \right )\sqrt{x-1}\: dx\\ &=\int \left ( \underset{u}{\underbrace{\left (x-1 \right )}}+1 \right )\sqrt{\underset{u}{\underbrace{x-1}}}\: dx \\ &=\int \left ( u+1 \right )\sqrt{u}\: du\\ &=\int \left (u\sqrt{u}+\sqrt{u}\: \right )\: du\\ &=\int \left (u^{\frac{3}{2}}+u^{\frac{1}{2}} \right )\: du\\ &=\displaystyle \frac{1}{\left ( \frac{3}{2}+1 \right )}u^{\left (\frac{3}{2}+1 \right )}+\displaystyle \frac{1}{\left (\frac{1}{2}+1 \right )}u^{\left (\frac{1}{2}+1 \right )}+C\\ &=\displaystyle \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}+C\\\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{array}{ll}\\ \displaystyle \int x\sqrt{x-1}\: dx=\int \underset{\displaystyle \underset{u}{\mid }}{x}.\underset{\displaystyle \underset{dv}{\mid }}{\underbrace{\sqrt{x-1}\: dx}}&\\ \begin{matrix} u=x &&& dv=\sqrt{x-1}\: dx\\ du=dx &&& \displaystyle \int dv=\int \sqrt{x-1}\: dx\\ &&&v=\int \left ( x-1 \right )^{\frac{1}{2}}\: dx\\ &&&v=\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}\\ \end{matrix} \end{array}\\ &\textrm{Dengan integral parsial}\\&\begin{aligned}&\displaystyle \int x\sqrt{x-1}\: dx\\ &=\int \underset{\displaystyle \underset{u}{\mid }}{x}.\underset{\displaystyle \underset{dv}{\mid }}{\underbrace{\sqrt{x-1}\: dx}}= u.v-\int v.du\\ &=x.\left ( \displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}} \right )-\int \left ( \displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}} \right )dx\\ &=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{2}{3}\times \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+C\\ &=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )^{\frac{5}{2}}+C \end{aligned} \\\end{aligned} \end{array}$.

$\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\begin{array}{|c|c|}\hline \textrm{Hasil dengan Substitusi}&\textrm{Hasil dengan Integral Parsial}\\\hline \displaystyle \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}+C&\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )^{\frac{5}{2}}+C\\\hline  \end{array}\\ &\textrm{Jika kita sejajarkan dengan ruas }\\ &\textrm{yang berbeda, maka}\\ &\begin{aligned}\displaystyle \frac{2}{5}\left ( x-1 \right )^{\frac{5}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}+C&=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )^{\frac{5}{2}}+C\\ \displaystyle \frac{2}{5}\left ( x-1 \right )\left ( x-1 \right )^{\frac{3}{2}}+\displaystyle \frac{2}{3}\left ( x-1 \right )^{\frac{3}{2}}&=\displaystyle \frac{2x}{3}\left ( x-1 \right )^{\frac{3}{2}}-\displaystyle \frac{4}{15}\left ( x-1 \right )\left ( x-1 \right )^{\frac{3}{2}}\\ \left ( \displaystyle \frac{2}{5}\left ( x-1 \right )+\frac{2}{3} \right )\left ( x-1 \right )^{\frac{3}{2}}&=\left ( \displaystyle \frac{2x}{3}-\frac{4}{15}\left ( x-1 \right ) \right )\left ( x-1 \right )^{\frac{3}{2}}\\ \displaystyle \frac{2x}{5}-\frac{2}{5}+\frac{2}{3}&=\displaystyle \frac{2x}{3}-\frac{4x}{15}+\frac{4}{15}\\ \displaystyle \frac{2x}{5}+\frac{4}{15}&=\displaystyle \frac{2x}{5}+\frac{4}{15}\\ \textrm{ruas kiri}\: &=\: \textrm{ruas kanan} \end{aligned} \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah integral dari}\: \: \int (x+2)\: dx\\ &\textrm{dengan cara}\\ &\textrm{a})\quad \textrm{substitusi}\\ &\textrm{b})\quad \textrm{parsial}\\\\&\textbf{Jawab}:\\ &\begin{aligned}&\textbf{Cara substitusi}\\ &\int (x+2)\: dx=........?\\ &\textrm{Misalkan}\: \: m=x+2\\ &\qquad\qquad \, dm=dx\\ &\textrm{maka},\\ &\int (x+2)\: dx=\int \underset{\begin{matrix} |\\ \textbf{m} \end{matrix}}{\underbrace{(x+2)}}.\underset{\begin{matrix} |\\ \textbf{dm} \end{matrix}}{\underbrace{dx}}\\ &=\frac{1}{2}m^{2}+C=\frac{1}{2}(x+2)^{2}+C\\ &=\frac{1}{2}\left ( x^{2}+4x+4 \right )+C=\frac{1}{2}x^{2}+2x+\underset{\begin{matrix} |\\ \textbf{C} \end{matrix}}{\underbrace{2+C}}\\ &=\frac{1}{2}x^{2}+2x+C \end{aligned}\\\\ &\begin{aligned}&\textbf{Cara parsial}\\ &\int (x+2)\: dx=\int \underset{\begin{matrix} |\\ \textbf{u} \end{matrix}}{\underbrace{1}}.\underset{\begin{matrix} |\\ \textbf{dv} \end{matrix}}{\underbrace{(x+2)\: dx}}\\ &\left\{\begin{matrix} u=1\quad \rightarrow \quad du=0\qquad\qquad\qquad\qquad\qquad \\ \\ v=\underset{\begin{matrix} |\\ =\frac{1}{2}x^{2}+2x+C \end{matrix}}{\int dv} \leftarrow  dv=(x+2)dx \end{matrix}\right.\\ &=\textbf{u.v}-\int \textbf{v.du}\\ &=1.\left (\frac{1}{2}x^{2}+2x+C \right )-\int \left (\frac{1}{2}x^{2}+2x+C \right ).0\\ &=\frac{1}{2}x^{2}+2x+C \end{aligned} \end{array}$.

2. 2 Aturan Tanzalin

Sumber Referensi

$\begin{aligned}&\textrm{Tentukanlah hasil integral berikut}\\ &\displaystyle \int 2x^{2}(x-4)^{2}dx\\\\ &\textbf{Jawab}:\\ &\textrm{Bentuk}\: \: \displaystyle \int 2x^{2}(x-4)^{2}dx\: \: \textrm{dianggap sebagai}\\ &\color{red}\displaystyle \int u\: dv\: \: \color{black}\textrm{dengan}\: \: \color{red}u\: \: \color{black}\textrm{adalah bagian yang mudah}\\ &\textrm{kita diferensialkan, maka}\: \: u\: \: \textrm{kita pilihkan}\\ &\textrm{yaitu}\: \: u=\color{red}2x^{2}\\ &\textrm{Selanjutnya dengan}\: \: \textbf{aturan Tanzalin}\\ &\textrm{sebagai berikut} \end{aligned}$.

$\begin{array}{|c|c|}\hline \textrm{Diderensialkan}&\textrm{Diintegralkan}\\\hline \begin{aligned}&+\: \color{red}2x^{2}\\ &-\: \color{blue}4x\\ &+\: \color{magenta}4\\ &-\: 0\\ & \end{aligned}&\begin{aligned}&(x-4)^{4}\\ &\color{red}\displaystyle \frac{1}{5}(x-4)^{5}\\  &\color{blue}\displaystyle \frac{1}{30}(x-4)^{6}\\ &\color{magenta}\displaystyle \frac{1}{210}(x-4)^{7} \end{aligned}\\\hline  \end{array}$.

Hasil dari integral teknik ini adalah:

$\begin{aligned}&\displaystyle \int 2x^{2}(x-4)^{4}\: dx\\ &=(+2x^{2})\left ( \displaystyle \frac{1}{5}(x-4)^{5} \right )+(-4x)\left ( \displaystyle \frac{1}{30}\left ( x-4 \right )^{6} \right )\\ &\quad +(+4)\left ( \displaystyle \frac{1}{210}(x-4)^{7} \right )+C\\ &=\displaystyle \frac{2}{5}x^{2}(x-4)^{5}-\displaystyle \frac{2}{15}x(x-4)^{6}+\displaystyle \frac{2}{105}(x-4)^{7}+C \end{aligned}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{aligned}&\textrm{1. Selesaikan soal berikut ini}\\ &\begin{array}{llllll}\\ &\textrm{a}.&\displaystyle \int \displaystyle x^{2}\sqrt{x+7}\: \: dx&\textrm{d}.&\displaystyle \int -(2x^{2}+1)\sqrt{3-x}\: \: dx\\ &\textrm{b}.&\displaystyle \int 2x^{2}\sqrt{3-x}\: \: dx&\textrm{e}.&\displaystyle \int  x^{5}(2x+1)^{6}\: dx\\ &\textrm{c}.&\displaystyle \int 3x^{2}\sqrt{2x+1}\: dx&\textrm{f}.&\displaystyle \int \displaystyle \frac{2x^{6}}{\sqrt[3]{3x-1}}\: \: dx  \end{array} \end{aligned}$ .

$\begin{aligned}&\textrm{2. Selesaikan soal berikut dengan Metode Tanzalin}\\ &\begin{array}{llllll}\\ &\textrm{a}.&\displaystyle \int -(2x^{2}+1)\sqrt{3-x}\: \: dx\\ &\textrm{b}.&\displaystyle \int  x^{5}(2x+1)^{6}\: dx\\ &\textrm{c}.&\displaystyle \int \displaystyle \frac{2x^{6}}{\sqrt[3]{3x-1}}\: \: dx\\ &\textrm{d}.&\displaystyle \int \displaystyle 3(x-2)^{4}.\sqrt[3]{x}\: \: dx\\ &\textrm{e}.&\displaystyle \int \displaystyle \frac{(x^{4}-3)}{\sqrt{1-x}}\: \: dx\\ &\textrm{f}.&\displaystyle \int \displaystyle x^{3}\sqrt{1-2x}\: \: dx  \end{array} \end{aligned}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.