Contoh Soal 4 Materi Integral Tentu Fungsi Aljabar

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

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

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

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

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

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

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

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

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

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

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

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