Contoh Soal 6 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 26.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Diketahui sebuah lingkaran berjari-jari}2\\ & \text{sebagaimana ilustrasi dibawah. Jika tali}\\ & \text{busur pada gambar berjarak 1 dari garis}\\ & \text{tengah, maka luas daerah di atas tali busur}\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{b}.{\displaystyle 2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{c}.{\displaystyle {\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{d}.{\displaystyle 2{\int }_0^1\sqrt{4-x^2}dx}\\ \text{e}.{\displaystyle 2{\int }_0^{\sqrt{3}}\sqrt{4-x^2}dx}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa sebuah lingkaran}\\ & \text{dengan}r=2,\text{artinya}{x}^2+y^2=r^2=2^2=4\\ & \text{Diketahui pula persamaan garis}y=1,\text{maka}\\ & x^2+1^2=4\iff x^2=4-1=3\iff x=\pm \sqrt{3}\end{array}$

$.\begin{array}{rl}& \text{Sehingga luas daerah di atas tali busur}\\ & \text{yaitu berupa}\mathbf{\text{temberang}}\text{adalah}\\ & ={\displaystyle {\int }_{-\sqrt{3}}^{\sqrt{3}}(f_2(x)-f_1(x))dx}\\ & ={\displaystyle {\int }_{-\sqrt{3}}^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ & \text{ingat fungsi genap pada pembahasan}\\ & \text{sebelumnya, karena}f(x)=\sqrt{4-x^2}=f(-x)\\ & =2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\end{array}$.

$\begin{array}{l}\\ 27.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Diketahui sebuah lingkaran berjari-jari}2\\ & \text{sebagaimana ilustrasi dibawah. Jika tali}\\ & \text{busur pada gambar berjarak 1 dari garis}\\ & \text{tengah, maka luas daerah di bawah tali busur}\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2\pi -2{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{b}.{\displaystyle 2\pi -2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{c}.{\displaystyle 4\pi -{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{d}.{\displaystyle 4\pi -2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{e}.{\displaystyle 4\pi -2{\int }_0^{\sqrt{3}}\sqrt{4-x^2}dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dengan ilustrasi sama dengan no.24 di atas}\\ & \text{Kita dapat tentukan luas wilayah di bawah}\\ & \mathbf{\text{tembereng}},\text{yaitu}:\\ & =\text{Luas lingkaran penuh itu}-\mathbf{\text{tembereng}}\\ & =\pi .r^2-2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\\ & =\pi .(2{)}^2-2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\\ & =4\pi -2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Luas daerah yang dibatasi oleh kurva}\\ & y=2-x^2\text{dan}y=\left|x\right|\text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2{\int }_{-1}^0(-x^2+x+2)dx}\\ \text{b}.{\displaystyle 2{\int }_{-1}^0(-x^2-x+2)dx}\\ \text{c}.{\displaystyle 2{\int }_{-1}^0(-x^2+x-2)dx}\\ \text{d}.{\displaystyle {\int }_{-1}^1(-x^2-x+2)dx}\\ \text{e}.{\displaystyle {\int }_0^1(-x^2+x+2)dx}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Kita tentukan poin batas-batasnya, yaitu}:\\ & \text{Titik potong kurva dengan sumbu-X},y=0\\ & y=2-x^2=0\iff (\sqrt{2}+x)(\sqrt{2}-x)=0\\ & \iff x=-\sqrt{2}\text{atau}x=\sqrt{2}\\ & \text{Titik potong kurva dengan garis, yaitu}:\\ & 2-x^2=\left|x\right|\iff x^2+\left|x\right|-2=0\\ & \bullet x\ge 0,\left|x\right|=x\\ & \iff x^2+x-2=(x+2)(x-1)=0\\ & \iff x=-2(\mathbf{\text{TM}})\text{atau}x=1\\ & \bullet x<0,\left|x\right|=-x\\ & \iff x^2-x-2=(x-2)(x+1)=0\\ & \iff x=2(\mathbf{\text{TM}})\text{atau}x=-1\\ & \text{Perhatikanlah ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Misalkan}{f}_1(x)=2-x^2\text{dan}{f}_2(x)=\left|x\right|\\ & \text{maka luas arsirannya adalah}:\\ & \mathbf{\text{Luas arsiran}}=2{\displaystyle {\int }_{-1}^0(f_1(x)-f_2(x))dx}\\ & =2{\displaystyle {\int }_{-1}^0(2-x^2-\left|x\right|)dx}\\ & =2{\displaystyle {\int }_{-1}^0(2-x^2-(-x))dx}\\ & =2{\displaystyle {\int }_{-1}^0(-x^2+x+2)dx}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 2}\\ \text{b}.{\displaystyle 4}\\ \text{c}.{\displaystyle 6}\\ \text{d}.{\displaystyle 8}\\ \text{e}.{\displaystyle 10}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

Jika diperjelas lagi

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^{3}f(x)dx={\displaystyle {\int }_{-1}^3\left[x\right]dx}}\\ & ={\displaystyle {\int }_{-1}^0-1dx+{\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx}\\ & =(-x){|}_{-1}^0+0+x{|}_1^2+2x{|}_2^3\\ & =(0-(-(-1)))+0+(2-1)+2(3-2)\\ & =-1+0+1+2\\ & =2\end{array}$.

$\begin{array}{l}\\ 30.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_a^b\left[x\right]dx+{\int }_a^b[-x]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle a+b}\\ \text{b}.{\displaystyle 0}\\ \text{c}.{\displaystyle a-b}\\ \text{d}.{\displaystyle 2a}\\ \text{e}.{\displaystyle 2b}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel berikut}\\ & \text{Misalkan}f(x)=\left[x\right],g(x)=[-x],\text{dan}\\ & h(x)=f(x)+g(x)=\left[x\right]+[-x]\end{array}$..

$.\begin{array}{|cccc|}\hline \text{Interval}& c\le f(x)<c+1& c\le g(x)<c+1& [h(x)]=c\\ ⋮& ⋮& ⋮& ⋮\\ -3\le x<-2& -3\le x<-2& 2<x\le 3& -3+2=-1\\ -2\le x<-1& -2\le x<-1& 1\le x<2& -2+1=-1\\ -1\le x<0& -1\le x<0& 0\le x<1& -1+0=-1\\ 0\le x<1& 0\le x<1& -1\le x<0& 0+-1=-1\\ 1\le x<2& 1\le x<2& -2\le x<-1& 1+-2=-1\\ 2\le x<3& 2\le x<3& -3\le x<-2& 2+-3=-1\\ ⋮& ⋮& ⋮& ⋮\\ a\le x<b& a\le x<b& -b\le x<-a& a-b\\ \hline\end{array}$.

Contoh Soal 5 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 21.& \text{Pada gambar berikut, inetgral yang}\\ & \text{menyatakan luas daerah yang diarsir adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^3(2x^2-8x+6)dx}\\ \text{b}.{\displaystyle {\int }_0^1(x^2-4x+3)dx-{\int }_1^3(x^2-4x+3)dx}\\ \text{c}.{\displaystyle {\int }_0^1(x^2+4x+3)dx+{\int }_1^3(x^2+4x+3)dx}\\ \text{d}.{\displaystyle {\int }_0^1(2x^2-8x+6)dx-{\int }_1^3(2x^2-8x+6)dx}\\ \text{e}.{\displaystyle {\int }_0^1(2x^2+8x+6)dx-{\int }_1^3(2x^2+8x+6)dx}\end{array}\\ \end{array}$.

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

$\begin{array}{l}\\ 22.& \text{Jika}f\text{fungsi ganjil, maka nilai}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}f(x)dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}f(x)dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perlu diketahui bahwa}\\ & f(x)=\{\begin{array}{ll}\text{fungsi genap}& \Rightarrow f(-x)=f(x)\\ \text{fungsi ganjil}& \Rightarrow f(-x)=-f(x)\end{array}\\ & \text{Untuk}\\ & \begin{array}{|cc|}\hline f\text{fungsi genap}& {\displaystyle {\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx}\\ f\text{fungsi ganjil}& {\displaystyle {\int }_{-a}^{a}f(x)dx=0}\\ \hline\end{array}\\ & \text{Karena}f(x)\text{fungsi ganjil, maka}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx=0}\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Jika}f\text{fungsi genap, maka nilai}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}f(x)dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}f(x)dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perlu diketahui bahwa}\\ & f(x)=\{\begin{array}{ll}\text{fungsi genap}& \Rightarrow f(-x)=f(x)\\ \text{fungsi ganjil}& \Rightarrow f(-x)=-f(x)\end{array}\\ & \text{Untuk}\\ & \begin{array}{|cc|}\hline f\text{fungsi genap}& {\displaystyle {\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx}\\ f\text{fungsi ganjil}& {\displaystyle {\int }_{-a}^{a}f(x)dx=0}\\ \hline\end{array}\\ & \text{Karena}f(x)\text{fungsi genap, maka}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx=2{\displaystyle {\int }_0^{2022}f(x)dx}}\end{array}\end{array}$.

 $\begin{array}{l}\\ 24.& \text{Nilai dari}{\displaystyle {\int }_{-2022}^{2022}\left|x\right|dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left|x\right|dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left|x\right|dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Karena}f(x)=\left|x\right|,\text{adalah fungsi genap},\\ & \text{maka}\\ & {\displaystyle {\int }_{-2022}^{2022}\left|x\right|dx={\displaystyle 2{\int }_0^{2022}\left|x\right|dx={2022}^2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Nilai dari}{\displaystyle {\int }_{-2022}^{2022}x\left|x\right|dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{-{2022}^3}{3}}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}x\left|x\right|dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}x\left|x\right|dx}\\ \text{e}.{\displaystyle \frac{{2022}^3}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Karena}f(x)=x\left|x\right|,\text{adalah fungsi ganjil},\\ & \text{maka}\\ & {\displaystyle {\int }_{-2022}^{2022}x\left|x\right|dx=0}\end{array}\end{array}$

Contoh Soal 4 Materi Integral Tentu Fungsi Aljabar

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

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Luas arsiran}=\text{Luas trapesium}-\text{luas II}\\ & \text{Proses penyelesaian 1}\\ & \text{Kurva}y=4x-x^2,\text{maka koordinat titik}\\ & \text{puncaknya adalah}:(p,q).\text{Pilih}y=(x)=0\\ & \iff y=f(x)=4x-x^2=x(4-x)=0\\ & \iff x_1=0\text{atau}{x}_2=4\\ & \text{dengan}p={\displaystyle \frac{x_1+x_2}{2}=\frac{0+4}{2}=2,\text{dan}}\\ & q=y=f(2)=4.2-2^2=8-4=4\\ & \text{Sehingga}(p,q)=(2,4)\\ & \text{Proses penyelesaian 2}\\ & \text{Untuk garis}y=-2x+8,\text{maka koordinat}\\ & \text{titik potongnya terhadap sumbu-X dan Y}:\\ & \begin{array}{|ccc|}\hline x& 0& 4\\ y& 8& 0\\ (x,y)& (0,8)& (4,0)\\ \hline\end{array}\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Luas arsiran}=\text{Luas trapesium}-\text{luas II}\\ & ={\displaystyle \frac{(8+4)\times 2}{2}-{\displaystyle {\int }_0^{2}4x-x^{2}dx}}\\ & =12-\left((2x^2-{\displaystyle \frac{1}{3}{x}^3}){|}_0^2\right)\\ & =12-((8-{\displaystyle \frac{8}{3}})-0)\\ & =12+{\displaystyle \frac{8}{3}-8=6\frac{2}{3}\text{satuan luas}}\end{array}$.

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

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Proses penyelesaian 1}\\ & \text{Kurva}y=4-x^2,\text{maka koordinat titik}\\ & \text{puncaknya adalah}:(p,q).\text{Pilih}y=(x)=0\\ & \iff y=f(x)=4-x^2=(2+x)(2-x)=0\\ & \iff x_1=-2\text{atau}{x}_2=2\\ & \text{dengan}p={\displaystyle \frac{x_1+x_2}{2}=\frac{-2+2}{2}=0,\text{dan}}\\ & q=y=f(0)=4-0^2=4\\ & \text{Sehingga}(p,q)=(0,4)\\ & \text{Proses penyelesaian 2}\\ & \text{Untuk garis}y=-x+2,\text{maka koordinat}\\ & \text{titik potongnya terhadap sumbu-X dan Y}:\\ & \begin{array}{|ccc|}\hline x& 0& 2\\ y& 2& 0\\ (x,y)& (0,2)& (2,0)\\ \hline\end{array}\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Luas arsiran}=\text{Dibawah kurva}-\text{segitiga}\\ & ={\displaystyle {\int }_0^{2}4-x^{2}dx-{\displaystyle \frac{1}{2}\times \text{alas}\times \text{tinggi}}}\\ & =(4x-{\displaystyle \frac{1}{3}{x}^3}){|}_0^2-{\displaystyle \frac{2\times 2}{2}}\\ & =8-{\displaystyle \frac{8}{3}-2=6-2{\displaystyle \frac{2}{3}=3\frac{1}{3}}}\\ & =\frac{10}{3}\text{satuan luas}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Luas daerah di bawah kurva}y=-x^2+8x\\ & \text{di atas}y=6x-24\text{dan terletak di }\\ & \text{kuadran I adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(x^2-2x-24)dx}\\ \text{b}.{\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+2x+24)dx}\\ \text{c}.{\displaystyle {\int }_0^6(-x^2+8x)dx+{\int }_6^8(-x^2+2x+24)dx}\\ \text{d}.{\displaystyle {\int }_4^6(6x-24)dx+{\int }_6^8(-x^2+8x)dx}\\ \text{e}.{\displaystyle {\int }_0^4(6x-24)dx+{\int }_0^6(-x^2+8x)dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Kita tentukan poin batas-batasnya dulu}\\ & \text{Titik potong kurva dengan sumbu-X, yaitu}:\\ & y=-x^2+8x=x(-x+8)=0\\ & \iff x=0\text{atau}x=8\\ & \text{Titik potong garis dengan sumbu-X, yaitu}:\\ & y=6x-24=0\iff x=6\\ & \text{Dan titik potong kurva dengan garis adalah}:\\ & y=y\iff -x^2+8x=6x-24\\ & \iff -x^2+2x+24=0\iff (-x-4)(x-6)=0\\ & \iff x_1=-4\text{atau}{x}_2=6\\ & \text{Sehingga luas arsiran}=\text{Luas I}+\text{Luas II}\\ & ={\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+8x)-(6x-24)dx}\\ & ={\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+2x+24)dx}\\ & \text{Perhatikan ulustrasi berikut}\end{array}$ 

$\begin{array}{l}\\ 19.& \text{Luas daerah yang diarsir pada gambar}\\ & \text{di bawah dapat dirumuskan dengan}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_b^{d}g(x)dx-{\int }_b^{c}f(x)dx}\\ \text{b}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_b^d(g(x)-f(x))dx}\\ \text{c}.{\displaystyle {\int }_a^d(f(x)-g(x))dx}\\ \text{d}.{\displaystyle {\int }_a^d(f(x)-g(x))dx-{\int }_c^{d}g(x)dx}\\ \text{e}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_c^d(g(x)-f(x))dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Cukup jelas bahwa}\\ & \text{pada gambar di atas pada selang}\\ & \bullet a\le x\le b:{\displaystyle {\int }_a^b(f(x)-g(x))dx}\\ & \bullet b\le x\le d:{\displaystyle {\int }_b^d(g(x)-f(x))dx}\\ & \text{dikurangi luasan yang}\\ & \text{berselang}b\le x\le c\end{array}$.

$\begin{array}{l}\\ 20.& \text{Daerah}\text{R}\text{di kuadran II, dibatasi oleh}\\ & \text{grafik}y=x^2,y=x+2\text{dan}y=0\\ & \text{Integral yang sesuai kondisi di atas adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_{-2}^{-1}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ \text{b}.{\displaystyle {\int }_{-1}^{-2}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ \text{c}.{\displaystyle {\int }_{-1}^{-2}{x}^{2}dx+{\int }_{-1}^0(x+2)dx}\\ \text{d}.{\displaystyle {\int }_0^2(x^2+x+2)dx}\\ \text{e}.{\displaystyle {\int }_0^2(-x^2+x+2)dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Dari gukup jelas bahwa}\\ & \text{daerah arsiran akan terformulasikan}\\ & {\displaystyle {\int }_{-2}^{-1}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ & \text{Berikut sedikit uraian prosesnya}\\ & \mathbf{\text{Sebagai langkah awal}},\text{tentukan dulu}\\ & \text{titik potong kura dengan garis, yaitu}:\\ & y=y\iff x^2=x+2\iff x^2-x-2=0\\ & \iff (x+1)(x-2)=0\\ & \iff x=-1\text{atau}x=2\\ & \mathbf{\text{Langkah berikutnya}}\\ & \text{Kita tentukan daerah arsiran, yaitu}\\ & \text{seperti terformulasikan di atas}\end{array}$.

Contoh Soal 3 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 11.& \text{Diketahui}{\displaystyle \int f(x)dx=a{x}^2+bx+c}\\ & \text{dengan}a\ne 0.\text{Jika}a,f(a),2b\text{membentuk}\\ & \text{barisan arimetika dan}f(b)=6,\text{maka}\\ & \text{nilai}{\displaystyle {\int }_0^{1}f(x)dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{17}{4}}\\ \text{b}.{\displaystyle \frac{21}{4}}\\ \text{c}.{\displaystyle \frac{25}{4}}\\ \text{d}.{\displaystyle \frac{13}{4}}\\ \text{e}.{\displaystyle \frac{11}{4}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int f(x)dx=a{x}^2+bx+c}\\ & \Rightarrow f(x)=2ax+b\\ & \Rightarrow f(a)=2a^2+b\\ & \Rightarrow f(b)=2ab+b=6\\ & \text{Karena}a,f(a),2b\text{membentuk barisan}\\ & \text{aritmetika, maka}f(a)={\displaystyle \frac{a+2b}{2}}\\ & \text{Sehingga}\\ & f(a)=2a^2+b={\displaystyle \frac{a+2b}{2}}\\ & \iff 4a^2-a=0\iff a(4a-1)=0\\ & \iff a=0\text{atau}a={\displaystyle \frac{1}{4}}\\ & \text{Dan karena}a\ne 0,\text{maka}a={\displaystyle \frac{1}{4}\Rightarrow b=4}\\ & \text{Selanjutnya}\\ & f(x)=2ax+b={\displaystyle \frac{1}{2}x+4}\\ & \text{Dari fakta di atas, maka}\\ & {\displaystyle {\int }_0^{1}f(x)dx={\int }_0^1{\displaystyle \frac{1}{2}x+4dx}}\\ & ={\displaystyle \frac{1}{4}{x}^2+4x{|}_0^1}\\ & =\left({\displaystyle \frac{1}{4}+4}\right)-0={\displaystyle \frac{17}{4}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 13.& \text{Jika}y={\displaystyle \frac{1}{3}(x^3+\frac{3}{x}),\text{maka}}\\ & {\displaystyle {\int }_1^2\sqrt{4+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{13}{14}}\\ \text{b}.{\displaystyle \frac{13}{6}}\\ \text{c}.{\displaystyle \frac{15}{6}}\\ \text{d}.{\displaystyle \frac{16}{6}}\\ \text{e}.{\displaystyle \frac{17}{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& y={\displaystyle \frac{1}{3}(x^3+\frac{3}{x})}\\ & \iff dy=(x^2-{\displaystyle \frac{1}{x^2}})dx\\ & \iff {\displaystyle \frac{dy}{dx}=(x^2-{\displaystyle \frac{1}{x^2}})}\\ & \iff {\left({\displaystyle \frac{dy}{dx}}\right)}^2={(x^2-{\displaystyle \frac{1}{x^2}})}^2={\displaystyle x^4-2+{\displaystyle \frac{1}{x^4}}}\\ & \text{maka}\\ & {\displaystyle {\int }_1^2\sqrt{4+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{4+\left({\displaystyle x^4-2+{\displaystyle \frac{1}{x^4}}}\right)}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{\left({\displaystyle x^4+2+{\displaystyle \frac{1}{x^4}}}\right)}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{{\left({\displaystyle x^2+{\displaystyle \frac{1}{x^2}}}\right)}^2}dx}\\ & ={\displaystyle {\int }_1^2{x}^2+{\displaystyle \frac{1}{x^2}dx}}\\ & ={\displaystyle \frac{1}{3}{x}^3-{\displaystyle \frac{1}{x}{|}_1^2}}\\ & =\left({\displaystyle \frac{8}{3}-\frac{1}{2}}\right)-\left({\displaystyle \frac{1}{3}-1}\right)\\ & ={\displaystyle \frac{16-3-2+6}{6}}\\ & ={\displaystyle \frac{17}{6}}\end{array}\end{array}$.

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

$.\begin{array}{rl}& \text{Luas arsiran}=12\\ & =4.a-{\displaystyle {\int }_1^4{\displaystyle \frac{a}{(4-1)(4-7)}(x-1)(x-7)dx}}\\ & =4a-{\displaystyle {\int }_1^4{\displaystyle -\frac{a}{9}(x^2-8x+7)dx}}\\ & =4a-{\displaystyle {\int }_1^4{\displaystyle \frac{a}{9}(-x^2+8x-7)dx}}\\ & =4a-(-{\displaystyle \frac{a}{27}{x}^3+\frac{4a}{9}{x}^2-\frac{7a}{9}x{|}_1^4})\\ & =4a+{\displaystyle \frac{64a}{27}-\frac{64a}{9}+\frac{28a}{9}-\frac{a}{27}+\frac{4a}{9}-\frac{7a}{9}}\\ & =4a-{\displaystyle \frac{18a}{9}=2a=12\iff a=6}\\ & \text{Berikut ilustrasi gambar lengkapnya}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Luas daerah yang diarsir pada}\\ & \text{gambar berikut dapat dinyatakan}\\ & \text{dengan}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^2(3x-x^2)dx}\\ \text{b}.{\displaystyle {\int }_0^2(x+3)dx-{\int }_0^2{x}^{2}dx}\\ \text{c}.{\displaystyle {\int }_0^1(x+3)dx-{\int }_0^2{x}^{2}dx}\\ \text{d}.{\displaystyle {\int }_0^1(x+3-x^2)dx-{\int }_1^2{x}^{2}dx}\\ \text{e}.{\displaystyle {\int }_0^1(x+3-x^2)dx-{\int }_1^2(4-x^2)dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \text{Daerah yang diarsir di atas dibatasi}\\ & \text{oleh 4 buah batas, yaitu}\\ & \bullet \text{Garis}y=4\\ & \bullet \text{Kurva}y=x^2\\ & \bullet \text{Garis}x=0,\text{dan}\\ & \bullet \text{Garis yang melalui titik}(0,3)\text{dan}(1,4)\\ & \begin{array}{|cc|}\hline \text{Titik}& \text{Garis}\\ \begin{array}{rl}& 0,3\\ & \text{dan}\\ & (1,4)\\ & \end{array}& \begin{array}{rl}{\displaystyle y}& =\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1\\ y& ={\displaystyle \frac{4-3}{1-0}(x-0)+3}\\ y& =x+3\end{array}\\ \hline\end{array}\\ & \text{Sehingga luas arisrannya adalah}:\\ & ={\displaystyle {\int }_0^1(x+3)-x^{2}dx+{\int }_1^{2}4-x^{2}dx}\end{array}$.

Contoh Soal 2 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 6.& \text{Hasil dari}{\displaystyle {\int }_2^4(-x^2+6x-8)dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{38}{3}}\\ \text{b}.{\displaystyle \frac{26}{3}}\\ \text{c}.{\displaystyle \frac{20}{3}}\\ \text{d}.{\displaystyle \frac{16}{3}}\\ \text{e}.{\displaystyle \frac{4}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_2^4(-x^2+6x-8)dx}\\ & =-{\displaystyle \frac{1}{3}{x}^3+3x^2-8x{|}_2^4}\\ & =(-{\displaystyle \frac{1}{3}(4{)}^3+3(4{)}^2-8(4)})-(-{\displaystyle \frac{1}{3}(2{)}^3+3(2{)}^2-8(2)})\\ & =(-{\displaystyle \frac{64}{3}+48-32})-(-{\displaystyle \frac{8}{3}+12-16})\\ & =-{\displaystyle \frac{56}{3}+20=-{\displaystyle \frac{56+60}{3}={\displaystyle \frac{4}{3}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Hasil dari}{\displaystyle {\int }_0^2{x}^2(x+2)dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 6}\\ \text{b}.6{\displaystyle \frac{1}{3}}\\ \text{c}.6{\displaystyle \frac{2}{3}}\\ \text{d}.9{\displaystyle \frac{1}{3}}\\ \text{e}.{\displaystyle 20}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\displaystyle {\int }_0^2{x}^2(x+2)dx={\displaystyle {\int }_0^2{x}^3+2x^{2}dx}}}\\ & ={\displaystyle \frac{1}{4}{x}^4+\frac{2}{3}{x}^3{|}_0^2}\\ & =(-{\displaystyle \frac{1}{4}(2{)}^4+\frac{2}{3}(2{)}^3})-\left(0\right)\\ & =4+{\displaystyle \frac{16}{3}={\displaystyle \frac{4}{3}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai dari}{\displaystyle {\int }_0^{2}18x\sqrt{3x^2+4}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 4}\\ \text{b}.16\\ \text{c}.112\\ \text{d}.128\\ \text{e}.{\displaystyle 168}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=3x^2+4\Rightarrow du=6xdx\\ & \text{Selanjutnya}\\ & {\displaystyle {\int }_0^{2}18x\sqrt{3x^2+4}dx={\displaystyle {\int }_0^{2}3\sqrt{u}dx}}\\ & ={\displaystyle {\int }_0^{2}3u^{{.}^{\frac{1}{2}}}dx=2u^{{.}^{\frac{3}{2}}}{|}_0^2}\\ & =2{(3x^2+4)}^{{.}^{\frac{3}{2}}}{|}_0^2\\ & =2{({3.2}^2+4)}^{{.}^{\frac{3}{2}}}-(2(0+4{)}^{{.}^{\frac{3}{2}}})\\ & =2.(16{)}^{{.}^{\frac{3}{2}}}={2.4}^3-{2.2}^3=128-16\\ & =112\end{array}\end{array}$ .

$\begin{array}{l}\\ 9.& \text{Hasil substitusi dari}u=x+1\text{pada}\\ & {\displaystyle {\int }_0^1{x}^2\sqrt{x+1}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{b}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{c}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{d}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{e}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x+1\Rightarrow x=u-1\Rightarrow dx=du\\ & \text{Selanjutnya}\\ & {\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Nilai dari}{\displaystyle {\int }_0^{1}5x(1-x{)}^{6}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{75}{56}}\\ \text{b}.{\displaystyle \frac{10}{56}}\\ \text{c}.{\displaystyle \frac{5}{56}}\\ \text{d}.-{\displaystyle \frac{7}{56}}\\ \text{e}.-{\displaystyle \frac{10}{56}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui soal bentuk}\mathbf{\text{integral parsial}}\\ & \text{dengan metode}\mathbf{\text{Tanzalin}}\text{diperoleh}\\ & \begin{array}{|cc|}\hline \text{Diturunkan}& \text{Diintegralkan}\\ 5x& (1-x{)}^6\\ 5& -{\displaystyle \frac{1}{7}(1-x{)}^7}\\ 0& {\displaystyle \frac{1}{56}(1-x{)}^8}\\ \hline\end{array}\\ & \text{Sehingga}\\ & {\displaystyle {\int }_0^{1}5x(1-x{)}^{6}dx}\\ & =(5x)(-{\displaystyle \frac{1}{7}(1-x{)}^7})-(5)\left({\displaystyle \frac{1}{56}(1-x{)}^8}\right){|}_0^1\\ & =(0-0)-(0-\left({\displaystyle \frac{5}{56}}\right))\\ & ={\displaystyle \frac{5}{56}}\end{array}\end{array}$ 

Contoh Soal 1 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 1.& \mathbf{\text{(UAN 2014)}}\\ & \text{Hasil}{\displaystyle {\int }_{-1}^2(x^3+3x^2+4x+5)dx=....}\\ & \begin{array}{ll}\text{a}.& 33{\displaystyle \frac{1}{4}}\\ \text{b}.& 33{\displaystyle \frac{3}{4}}\\ \text{c}.& 32{\displaystyle \frac{1}{4}}\\ \text{d}.& 31{\displaystyle \frac{3}{4}}\\ \text{e}.& 23{\displaystyle \frac{3}{4}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\int }_{-1}^2(x^3+3x^2+4x+5)dx\\ & ={\displaystyle \frac{x^4}{4}+x^3+2x^2+5x{|}_{-1}^2}\\ & =\left({\displaystyle \frac{2^4}{4}+2^3+{2.2}^2+5.2}\right)\\ & -(\frac{{(-1)}^4}{4}+{(-1)}^3+2.{(-1)}^2+5(-1))\\ & =33\frac{3}{4}\end{array}\end{array}$.

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

$.\begin{array}{rl}& \text{Alternatif 1}\\ & \begin{array}{rl}& {\displaystyle {\int }_0^4(g(x)-f(x))dx}\\ & ={\displaystyle {\int }_0^4(4x-x^2)-(x^2-4x)dx}\\ & ={\displaystyle {\int }_0^4(8x-2x^2)dx}\\ & ={\displaystyle {[4x^2-\frac{2}{3}{x}^3]}_0^4}\\ & ={\displaystyle ({4.4}^2-\frac{2}{3}{.4}^3)-({4.0}^2-\frac{2}{3}{.0}^3)}\\ & ={\displaystyle (64-\frac{2}{3}.64)-0}\\ & ={\displaystyle \frac{64}{3}=21\frac{1}{3}\text{satuan luas}}\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan menggunakan rumus}L={\displaystyle \frac{D\sqrt{D}}{6a^2}}\\ & \begin{array}{|c|}\hline \begin{array}{rl}& f(x)=g(x)\\ & f(x)=-f(x),\text{ingat}g(x)=-f(x)\\ & 2f(x)=0,\text{tidak boleh disederhanakan},\\ & 2\times ({(x-2)}^2-4)=0,\text{karena akan }\\ & 2\times (x^2-4x)=0\text{mempengaruhi hasil }\\ & 2x^2-8x=0,\text{akhir}\\ & \{\begin{array}{ll}a=2,b=-8& c=0\\ D=b^2-4ac,& D={(-8)}^2-4(2)(0)=64\end{array}\\ & L_{\mathbf{\text{daerah}}}={\displaystyle \frac{\mathbf{\text{D}}\sqrt{\mathbf{\text{D}}}}{6{\mathbf{\text{a}}}^2}}\\ & ={\displaystyle \frac{64\sqrt{64}}{6(2{)}^2}}\\ & ={\displaystyle \frac{64\times 8}{6\times 4}}\\ & ={\displaystyle \frac{64}{3}}\\ & =21{\displaystyle \frac{1}{3}}\end{array}\\ \hline\end{array}\end{array}$.

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

$.\begin{array}{l}\\ & \begin{array}{ll}\text{a}.& {\displaystyle {\int }_0^{4}4xdx-{\int }_2^4(2x-4)dx}\\ \text{b}.& {\displaystyle {\int }_0^{4}4xdx-{\int }_2^4(2x+4)dx}\\ \text{c}.& {\displaystyle {\int }_0^{4}2\sqrt{2}dx-{\int }_2^4(2x-4)dx}\\ \text{d}.& {\displaystyle {\int }_0^{4}2\sqrt{2}dx-{\int }_2^4(4-2x)dx}\\ \text{e}.& {\displaystyle {\int }_0^{4}2\sqrt{x}dx-{\int }_2^4(4+2x)dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\text{Luas}}_{\text{arsiran}}={\displaystyle {\int }_0^4{y}_{1}dx-{\int }_2^4{y}_{2}dx}\\ & ={\displaystyle {\int }_0^4\sqrt{4x}dx-{\int }_2^4(2x-4)dx}\\ & ={\displaystyle {\int }_0^{4}2\sqrt{x}dx-{\int }_2^4(2x-4)dx.}\end{array}\end{array}$.

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

$.\begin{array}{rl}& \text{Volumenya jika diputar mengelilingi Sumbu }\\ & \text{Y adalah}:\\ & V=\pi {\displaystyle {\int }_0^{\frac{1}{2}}{x}_1^{2}dy+\pi {\int }_{\frac{1}{2}}^1{x}_2^{2}dy}\\ & \iff V=\pi {\int }_0^{\frac{1}{2}}{\left(2\sqrt{3}{y}^2\right)}^{2}dy+\pi {\int }_{\frac{1}{2}}^1(1-y^2)dy\\ & \iff V=\frac{12}{5}{y}^5\pi {|}_0^{\frac{1}{2}}+(y-\frac{1}{3}{y}^3)\pi {|}_{\frac{1}{2}}^1\\ & \iff V=\pi (\frac{12}{5}\times \frac{1}{32}+(1-\frac{1}{3})-(\frac{1}{2}-\frac{1}{3}\times \frac{1}{8}))\\ & \iff V=\frac{17}{60}\pi .\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jika}{\displaystyle \frac{d}{dx}g(x)=f(x)\text{di mana}f(x)}\\ & \text{kontinu dari a sampai b, }\\ & \text{maka}{\displaystyle {\int }_a^{b}f(x).g(x)dx}\\ & \begin{array}{l}\\ \text{a}.0\\ \text{b}.f(b)-f(a)\\ \text{c}.g(b)-g(a)\\ \text{d}.{\displaystyle \frac{{[f(b)]}^2-{[f(a)]}^2}{2}}\\ \text{e}.{\displaystyle \frac{{[g(b)]}^2-{[g(a)]}^2}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_a^{b}f(x).g(x)dx}\\ & ={\displaystyle {\int }_a^{b}g(x).f(x)dx}\\ & ={\displaystyle {\int }_a^{b}g(x).{\displaystyle \frac{d(g(x))}{dx}dx,}}\\ & \text{ingat bahwa}{\displaystyle \frac{d}{dx}g(x)=f(x),}\\ & \text{f(x) kontinu dari a sampai b}\\ & ={\displaystyle {\int }_a^{b}g(x)d(g(x))}\\ & ={\left[{\displaystyle \frac{{(g(x))}^2}{2}}\right]}_a^b\\ & ={\displaystyle \frac{{[g(b)]}^2-{[g(a)]}^2}{2}}\end{array}\end{array}$.

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

$\begin{array}{|cc|}\hline \text{Di Atas Sumbu X}& \text{Di Bawah Sumbu X}\\ & -{\displaystyle {\int }_a^{b}f(x)dx}\\ {\displaystyle {\int }_a^{b}f(x)dx}& atau\\ & {\displaystyle {\int }_b^{a}f(x)dx}\\ \hline\end{array}$.

B. Volume Benda Putar

Adapun untuk volume diformulasikan dengan integral tentu berikut

$\boxed{{\displaystyle V=\pi {\displaystyle {\int }_a^b{(f(x))}^{2}dx=\pi {\displaystyle {\int }_a^b{y}^{2}dx}}}}$.

$\text{CONTOH SOAL}$.

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

$.\begin{array}{rl}& \begin{array}{rl}{L}_{\text{Arsiran}}& ={\displaystyle {\int }_1^{3}2xdx}\\ & ={\displaystyle {\left[x^2\right]}_1^3}\\ & ={\left(3\right)}^2-{\left(1\right)}^2\\ & =9-1\\ & =8\mathbf{\text{satuan luas}}\end{array}\\ & \begin{array}{rl}{V}_{\text{Benda putar}}& =\pi {\displaystyle {\int }_1^3{\left(y\right)}^{2}dx}\\ & =\pi {\displaystyle {\int }_1^3{\left(2x\right)}^{2}dx}\\ & =\pi {\displaystyle {\int }_1^{3}4x^{2}dx}\\ & =\pi {\left[{\displaystyle \frac{4x^3}{3}}\right]}_1^3\\ & =\pi \left({\displaystyle \frac{4\times 3^3}{3}}\right)-\pi \left({\displaystyle \frac{4\times 1^3}{3}}\right)\\ & =36\pi -{\displaystyle \frac{4}{3}\pi }\\ & =34{\displaystyle \frac{2}{3}\pi \mathbf{\text{satuan volum}}}\end{array}\end{array}$.

$$\begin{array}{l}\\ 2.& \text{Diketahui parabola}{f}_1(x)=a_1{x}^2+b_{1}x+c_1\\ & \text{dan}{f}_2(x)=a_2{x}^2+b_{2}x+c_2.\\ & \text{Titik potong kedua parabola tersebut }\\ & \text{dapat cari dengan}\\ & \\ & f_1(x)=f_2(x)\\ & \iff a_1{x}^2+b_{1}x+c_1=a_2{x}^2+b_{2}x+c_2\\ & \iff a{x}^2+bx+c=0.\\ & \\ & \text{Jika kedua parabola berpotongan di dua}\\ & \text{titik, tunjukkan bahwa luas daerah antara}\\ & \text{kedua parabola tersebut dapat }\\ & \text{dinyatakan dengan}{\displaystyle \mathbf{\text{L}}=\frac{\mathbf{\text{D}}\sqrt{\mathbf{\text{D}}}}{{\mathbf{\text{6a}}}^{\mathbf{\text{2}}}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & a{x}^2+bx+c=0\{\begin{array}{ll}& x_1={\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}}\\ & \\ & x_2={\displaystyle \frac{-b-\sqrt{b^2-4ac}}{2a}}\end{array}\\ & \begin{array}{rl}L& ={\displaystyle {\int }_{\frac{-b-\sqrt{b^2-4ac}}{2a}}^{\frac{-b+\sqrt{b^2-4ac}}{2a}}(a{x}^2+bx+c)dx={\left[{\displaystyle \frac{a{x}^3}{3}+\frac{b{x}^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]\\ & -\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}[({\sqrt{D}}^3-3{\sqrt{D}}^{2}b+3\sqrt{D}{b}^2-b^3)+({\sqrt{D}}^3+3{\sqrt{D}}^{2}b+3\sqrt{D}{b}^2+b^3)]}\\ & +{\displaystyle \frac{b}{8a^2}[(b^2-2b\sqrt{D}+{\sqrt{D}}^2)-(b^2+2b\sqrt{D}+{\sqrt{D}}^2)]+{\displaystyle \frac{c}{2a}[(-b+\sqrt{D})-(-b-\sqrt{D})]}}\\ & ={\displaystyle \frac{1}{24a^2}[2{\sqrt{D}}^3+6\sqrt{D}{b}^2]+{\displaystyle \frac{b}{8a^2}[-4b\sqrt{D}]+{\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}(D+3b^2-6b^2+12ac)}\\ & ={\displaystyle \frac{\sqrt{D}}{12a^2}[(b^2-4ac)-3b^2+12ac]}\\ & ={\displaystyle \frac{\sqrt{D}}{12a^2}[-2b^2+8ac]}\\ & =-{\displaystyle \frac{\sqrt{D}}{6a^2}[b^2-4ac]=-\frac{\sqrt{D}}{6a^2}\left[D\right]}\\ & =-\frac{D\sqrt{D}}{6a^2},\mathbf{\text{luas tidak mungkin negatif}}\\ & ={\displaystyle \frac{D\sqrt{D}}{6a^2}◼}\end{array}\end{array}$$.

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

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Menentukan batas}}\\ & \begin{array}{|rl|}\hline \begin{array}{rl}y& =y\\ x^2& =x\\ x^2-x& =0\\ x(x-1)& =0\\ x=0\text{atau}x& =1\\ & \\ & \\ & \end{array}& \begin{array}{rl}V& =\pi {\displaystyle {\int }_a^b(y_1^2-y_2^2)dx}\\ & =\pi {\displaystyle {\int }_0^1(x-x^2)dx}\\ & =\pi {\left[{\displaystyle \frac{1}{2}{x}^2-\frac{1}{3}{x}^3}\right]}_0^1\\ & =\pi \left[{\displaystyle \frac{1}{2}-\frac{1}{3}}\right]\\ V& ={\displaystyle \frac{1}{6}\pi }\end{array}\\ \hline\end{array}\\ & \mathbf{\text{Keterangan lanjutan}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa;}\\ & y^2=x\Rightarrow y=\sqrt{x},\text{dianggap sebagai}{y}_1\\ & \text{Sehingga}{y}_1-\text{nya adalah}\sqrt{x}\\ & \text{dan}y=x\text{dianggap sebagai}{y}_2\\ & (y_1^2-y_2^2)=({\left(\sqrt{x}\right)}^2-{\left(x\right)}^2)=x-x^2\end{array}\\ & \text{Jadi, volume dari benda putar tersebut}\\ & \text{dalam satuan volum adalah}{\displaystyle \frac{1}{6}\pi }\end{array}$.

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

$.\begin{array}{rl}& \text{Langkah-Langkah penyelesaiannya adalah:}\\ & \begin{array}{|cc|}\hline \text{Batas}& \text{Menentukan Volumenya}\\ x=1\text{dan}x=3& \begin{array}{rl}V& ={\displaystyle \pi {\int }_a^b(f^2(x)-g^2(x))dx}\\ & ={\displaystyle \pi {\int }_1^3({\left(2x\right)}^2-{\left(x\right)}^2)dx}\\ & ={\displaystyle \pi {\int }_1^{3}3x^{2}dx}\\ & ={\displaystyle \pi {\left[x^3\right]}_1^3}\\ & ={\displaystyle \pi \left(3^3\right)-\pi \left(1^3\right)}\\ & =27\pi -1\pi \\ V& =26\pi \mathbf{\text{Satuan Volum}}\end{array}\\ \hline\end{array}\end{array}$.

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

$.\begin{array}{rl}& \text{Langkah-Langkah penyelesaiannya adalah:}\\ & \begin{array}{|cc|}\hline \text{Batas}& \text{Menentukan Volumenya}\\ x=-2\text{sampai}x=2& \begin{array}{rl}V& ={\displaystyle \pi {\int }_a^b{y}^{2}dx}\\ & ={\displaystyle \pi {\int }_{-2}^2(4-x^2)dx}\\ & ={\displaystyle \pi {[4x-{\displaystyle \frac{x^3}{3}}]}_{-2}^2}\\ & ={\displaystyle \pi (8-{\displaystyle \frac{8}{3}})-\pi (-8+{\displaystyle \frac{8}{3}})}\\ & ={\displaystyle \pi (8+8-\frac{8}{3}-\frac{8}{3})}\\ V& ={\displaystyle \frac{32}{3}\pi \mathbf{\text{Satuan Volum}}}\end{array}\\ \hline\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

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

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

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

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

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:

${\displaystyle {\int }_a^{b}f(x)dx={[F(x)]}_a^b=F(x){|}_a^b=F(b)-F(a)}$.

$\text{CONTOH SOAL}$.

$\begin{array}{rl}1.& {\int }_1^3(2x-1)dx\\ & ={[x^2-x]}_1^3\\ & =(9-3)-(1-1)=6\\ 2.& {\int }_0^5|x-1|+|x-2|dx\\ & =\frac{1}{2}(x-1)|x-1|{|}_0^5+\frac{1}{2}(x-2)|x-2|{|}_0^5\\ & =(\frac{1}{2}.4.4+\frac{1}{2}.3.3)-(\frac{1}{2}.-1.1+\frac{1}{2}.-2.2)\\ & =8+4\frac{1}{2}+\frac{1}{2}+2=15\end{array}$.

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

B. Sifat-Sifat Integral Tentu

$\begin{array}{rl}1.& {\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx\\ 2.& {\displaystyle {\int }_a^{b}k.f(x)dx=k{\int }_a^{b}f(x)dx}\\ 3.& {\displaystyle {\int }_a^b(f(x)\pm g(x))dx}\\ & ={\displaystyle {\int }_a^{b}f(x)dx\pm {\int }_a^{b}g(x)dx}\\ 4.& {\displaystyle {\int }_a^{b}f(x)dx}\\ & ={\displaystyle {\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx,dengana<c<b}\\ 5.& {\displaystyle {\int }_a^{a}f(x)dx=0}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Hitunglah hasil dari integral}{\displaystyle {\int }_1^3(2y+1)dy}\\ \\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\int }_1^3(2y+1)dy& =y^2+y{|}_1^3\\ & =(3^2+3)-(1^2+1)\\ & =12-2\\ & =10\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Hitunglah hasil dari integral}{\displaystyle {\int }_{-1}^1{x}^3-1dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\int }_{-1}^1& (x^3-1)dx\\ & =-({\displaystyle \frac{1}{4}{x}^4-x{|}_{-1}^1)}\\ & =-\left({\displaystyle \frac{1}{4}.{\left(1\right)}^4-1}\right)+\left({\displaystyle \frac{1}{4}.{(-1)}^4-(-1)}\right)\\ & =-\left({\displaystyle \frac{1}{4}-1}\right)+\left({\displaystyle \frac{1}{4}+1}\right)\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Hitunglah hasil dari integral}{\displaystyle {\int }_{-2}^1{(x+3)}^{2}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\int }_{-2}^1& {(x+3)}^{2}dx\\ & ={\displaystyle \frac{1}{3}{x}^2+3x^2+9x{|}_{-2}^1}\\ & =\left({\displaystyle \frac{1}{3}.{\left(1\right)}^3+{3.1}^2+9.1}\right)-\left({\displaystyle \frac{1}{3}.{(-2)}^3+3.(-2{)}^2+9.(-2)}\right)\\ & =\left({\displaystyle \frac{1}{3}+12}\right)-(-{\displaystyle \frac{8}{3}+12-18})\\ & ={\displaystyle \frac{9}{3}+18=3+18=21}\end{array}\end{array}$ .

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ & \text{Hitunglah nilai tiap integral berikut ini}\\ & \begin{array}{l}\\ 1.& {\displaystyle {\int }_0^{3}2xdx}& 10.& {\displaystyle {\int }_1^2\sqrt{x^5}dx}\\ 2.& {\displaystyle {\int }_2^{3}2x^{2}dx}& 11.& {\displaystyle {\int }_1^2{\displaystyle \frac{1}{x^2}dx}}\\ 3.& {\displaystyle {\int }_3^7{\displaystyle \frac{1}{2}{x}^{3}dx}}& 12.& {\displaystyle {\int }_4^{9}3\sqrt{x}dx}\\ 4.& {\displaystyle {\int }_0^3(x^2+2x-1)dx}& {\displaystyle }\\ 5.& {\displaystyle {\int }_0^3{(x+3)}^{2}dx}& \\ 6.& {\displaystyle {\displaystyle {\int }_{-1}^2(x+2)(x-1)dx}}& \\ 7.& {\displaystyle {\int }_1^3(t+t^2-3t^3)dt}& \\ 8.& {\displaystyle {\int }_1^4\left({\displaystyle \frac{x^4-x^3+\sqrt{x}-1}{x^2}}\right)dx}& \\ 9.& {\displaystyle {\int }_4^6(\sqrt{t}+t^2-\sqrt[3]{t^2})dt}\end{array}\end{array}$.

Contoh Soal 4 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 16.& \mathbf{\text{(UN 2015 Matematika IPA)}}\\ & \text{Hasil}{\displaystyle \int 6x{(1-x^2)}^{4}dx\text{adalah ....}}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{3}{5}{(1+x^2)}^5+C}\\ \\ \text{b}.{\displaystyle \frac{2}{5}{(1+x^2)}^5+C}\\ \\ \text{c}.{\displaystyle -\frac{1}{5}{(1-x^2)}^5+C}\\ \\ \text{d}.-{\displaystyle \frac{2}{5}{(1-x^2)}^5+C}\\ \\ \text{e}.-{\displaystyle \frac{3}{5}{(1-x^2)}^5+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Alternatif 1}\\ & \text{Dengan integral substitusi}\\ & \begin{array}{c}\\ \begin{array}{rl}{\displaystyle }& \int 6x{(1-x^2)}^{4}dx\\ & =\int {(1-x^2)}^4.6xdx\\ & \{\begin{array}{ll}u& =1-x^2\\ & \\ du& =-2xdx\Rightarrow -3dx=6xdx\end{array}\\ & {\displaystyle \int u^4.(-3du)}\\ & =-3{\displaystyle \int u^{4}du}\\ & =-{\displaystyle \frac{3}{5}{u}^5+C}\\ & =-{\displaystyle \frac{3}{5}{(1-x^2)}^5+C}\end{array}\end{array}\\ & \text{Alternatif 2}\\ & \begin{array}{rl}& {\displaystyle \int {(1-x^2)}^{4}6xdx}\\ & =\int {(1-x^2)}^2.(-3).(-2xdx)\\ & =-3{\displaystyle \int {(1-x^2)}^4.(-2xdx)}\\ & =-3{\displaystyle \int {(1-x^2)}^4.d(1-x^2)}\\ & =-3\left[{\displaystyle \frac{{(1-x^2)}^5}{5}}\right]+C\\ & =-{\displaystyle \frac{3}{5}{(1-x^2)}^5+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \begin{array}{rl}{\displaystyle }& \int \frac{dx}{2022x+2023}=....\end{array}\\ & \begin{array}{l}\\ \text{a}.\ln |2022x+2023|+C\\ & \\ \text{b}.{\displaystyle \frac{1}{2021}\ln |2022x+2023|+C}\\ & \\ \text{c}.{\displaystyle \frac{1}{2022}\ln |2022x+2023|+C}\\ & \\ \text{d}.{\displaystyle \frac{1}{2023}\ln |2022x+2023|+C}\\ & \\ \text{e}.{\displaystyle \frac{2022}{2023}\ln |2022x+2023|+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int \frac{dx}{2022x+2023}}\\ & \text{Misalkan}\\ & u=2022x+2023\\ & \iff du=2022dx\iff {\displaystyle \frac{1}{2022}du=dx}\\ & \text{Sehingga}\\ & ={\displaystyle \int \frac{1}{u}.\left(\frac{1}{2022}du\right)={\displaystyle \frac{1}{2022}\int \frac{1}{u}du}}\\ & ={\displaystyle \frac{1}{2022}\ln |2022x+2023|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Diketahui}{\displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\text{jika}f(4)=19,}\\ & \text{ maka}f(1)=....\\ & \begin{array}{l}\\ \text{a}.2\\ \text{b}.3\\ \text{c}.4\\ \text{d}.5\\ \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perlu diingat bahwa}{\displaystyle \frac{d}{dx}f(x)=h(x)}\\ & \iff d(f(x))=h(x)dx\\ & \iff {\displaystyle \int d(f(x))={\displaystyle \int h(x)dx}}\\ & \text{Perhatikan kasus di pada soal di atas}\\ & {\displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\iff d(f(x))=3\sqrt{x}dx}\\ & \iff {\displaystyle \int d(f(x))=\int 3\sqrt{x}dx=\int 3x^{{.}^{\frac{1}{2}}}dx}\\ & \iff f(x)=2x^{{.}^{\frac{3}{2}}}+C=2x\sqrt{x}+C\end{array}\\ & \begin{array}{rl}& \text{Karena diketahui}\\ & f(4)=19,\text{maka}\\ & \iff 2(4)(\sqrt{4})+C=19\\ & \iff 16+C=19\\ & \iff C=3\\ & \text{Sehingga}f(x)=2x\sqrt{x}+3\\ & \text{maka}f(1)=2.1.\sqrt{1}+3=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \begin{array}{rl}{\displaystyle }& -\int \frac{dx}{x^2+3x+2}dx=....\end{array}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \ln \left|{\displaystyle \frac{x+1}{x+2}}\right|+C}\\ & \\ \text{b}.{\displaystyle \ln \left|{\displaystyle \frac{x+2}{x+1}}\right|+C}\\ & \\ \text{c}.{\displaystyle \ln |x^2+3x+2|+C}\\ & \\ \text{d}.\arctan 2\left({\displaystyle x+\frac{3}{2}}\right)+C\\ & \\ \text{e}.-{\displaystyle \arctan 2\left({\displaystyle x+\frac{3}{2}}\right)+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\displaystyle -\int \frac{dx}{x^2+3x+2}}& =-{\displaystyle \int \frac{1}{(x+1)(x+2)}dx}\\ & =-{\displaystyle \int \left({\displaystyle \frac{1}{x+1}-{\displaystyle \frac{1}{x+2}}}\right)dx}\\ & =-{\displaystyle \int \frac{1}{x+1}dx+{\displaystyle \int \frac{1}{x+2}dx}}\\ & ={\displaystyle \int \frac{1}{x+2}dx-{\displaystyle \int \frac{1}{x+1}dx}}\\ & ={\displaystyle \ln |x+2|-{\displaystyle \ln |x+1|+C}}\\ & ={\displaystyle \ln \left|{\displaystyle \frac{x+2}{x+1}}\right|+C}\end{array}\end{array}$.

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

Contoh Soal 3 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 11.& {\displaystyle \int d\left(2x\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x+C}\\ \text{b}.{\displaystyle 2x+C}\\ \text{c}.{\displaystyle 3x+C}\\ \text{d}.{\displaystyle 4x+C}\\ \text{e}.{\displaystyle 5x+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int d\left(2x\right)=\int 2dx=2x+C.}\end{array}$.

$\begin{array}{l}\\ 12.& {\displaystyle \int 5d\left(2x\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 5+C}\\ \text{b}.{\displaystyle \frac{5}{2}x+C}\\ \text{c}.{\displaystyle 5x+C}\\ \text{d}.{\displaystyle 10x+C}\\ \text{e}.{\displaystyle 25x+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int 5d\left(2x\right)=\int 5.2dx=\int 10dx=10x+C.}\end{array}$.

$\begin{array}{l}\\ 13.& {\displaystyle \int 6xd\left(3x\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 18x^2+C}\\ \text{b}.{\displaystyle 9x^2+C}\\ \text{c}.{\displaystyle 5x^2+C}\\ \text{d}.{\displaystyle 3x^2+C}\\ \text{e}.{\displaystyle 2x^2+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int 6xd\left(3x\right)=\int 6x.3dx=\int 18xdx}\\ & ={\displaystyle \frac{18x^2}{2}+C=9x^2+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& {\displaystyle \int (8x^3-2x)d(-2x)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -16x^4+4x^2+C}\\ \text{b}.{\displaystyle 16x^4-4x^2+C}\\ \text{c}.{\displaystyle -4x^4-2x^2+C}\\ \text{d}.{\displaystyle -4x^4+2x^2+C}\\ \text{e}.{\displaystyle 4x^4-2x^2+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int (8x^3-2x)d(-2x)}\\ & =\int (8x^3-2x).(-2dx)\\ & =\int (-16x^3+4x)dx\\ & ={\displaystyle -\frac{16x^4}{4}+\frac{4x^2}{2}+C}\\ & ={\displaystyle -4x^4+2x^2+C}\end{array}\end{array}$.

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

Contoh Soal 2 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 6.& {\displaystyle \int x^3\sqrt{x}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{2}{9}{x}^4\sqrt{x}+C}\\ \text{b}.{\displaystyle \frac{9}{2}{x}^4\sqrt{x}+C}\\ \text{c}.{\displaystyle \frac{1}{9}{x}^4\sqrt{x}+C}\\ \text{d}.{\displaystyle 9x^4\sqrt{x}+C}\\ \text{e}.{\displaystyle x^4\sqrt{x}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^3\sqrt{x}dx=\int x^3.x^{\frac{1}{2}}dx}\\ & =\int x^{3\frac{1}{2}}dx=\int x^{\frac{7}{2}}dx={\displaystyle \frac{x^{\frac{7}{2}+1}}{{\displaystyle \frac{7}{2}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{9}{2}}}{{\displaystyle \frac{9}{2}}}+C={\displaystyle \frac{2}{9}{x}^{4\frac{1}{2}}+C={\displaystyle \frac{2}{9}{x}^4\sqrt{x}+C}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& {\displaystyle \int x\sqrt{x\sqrt[3]{x^2}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{17}{6}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{b}.{\displaystyle \frac{6}{17}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{c}.{\displaystyle x^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{d}.{\displaystyle \frac{6}{17}x\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{e}.{\displaystyle \frac{1}{2}x\sqrt{x\sqrt[3]{x^2}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x\sqrt[3]{x^2}}dx}\\ & ={\displaystyle \int x{(x.x^{\frac{2}{3}})}^{\frac{1}{2}}dx}\\ & ={\displaystyle \int x^{1+\frac{1}{2}+\frac{2}{6}}dx}\\ & ={\displaystyle \int x^{\frac{11}{6}}dx={\displaystyle \frac{x^{\frac{11}{6}+1}}{{\displaystyle \frac{11}{6}+1}}+C}}\\ & ={\displaystyle \frac{x^{\frac{17}{6}}}{{\displaystyle \frac{17}{6}}}+C={\displaystyle \frac{6}{17}{x}^2.x^{\frac{5}{6}}+C}}\\ & ={\displaystyle \frac{6}{17}{x}^2{(x.x^{\frac{2}{3}})}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{17}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 9.& {\displaystyle \int x^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{b}.{\displaystyle \frac{6}{23}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{c}.{\displaystyle \frac{23}{6}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{d}.{\displaystyle \frac{23}{6}{x}^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{e}.{\displaystyle \frac{3}{4}{x}^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^3\sqrt{\frac{1}{x}\sqrt[3]{x^2}}dx=\int x^3{(x^{-1}.x^{\frac{2}{3}})}^{\frac{1}{2}}dx}\\ & =\int x^{3-\frac{1}{2}+\frac{1}{3}}dx=\int x^{\frac{17}{6}}dx\\ & ={\displaystyle \frac{x^{\frac{17}{6}+1}}{{\displaystyle \frac{17}{6}+1}}+C={\displaystyle \frac{x^{\frac{23}{6}}}{{\displaystyle \frac{23}{6}}}+C}}\\ & ={\displaystyle \frac{6}{23}{x}^{\frac{24-1}{6}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4.x^{-\frac{1}{6}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4.{\left(x^{-\frac{1}{3}}\right)}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4\sqrt{x^{-1+\frac{2}{3}}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4\sqrt{\frac{1}{x}\sqrt[3]{x^2}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& {\displaystyle \int x\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{b}.{\displaystyle \frac{13}{8}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{c}.{\displaystyle \frac{8}{13}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{d}.{\displaystyle \frac{1}{2}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{e}.{\displaystyle \frac{8}{5}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}dx}\\ & =\int x{\left(x^{-1}{\left(x{\left(x^{-1}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}dx\\ & =\int x(x^{-\frac{1}{2}}.x^{\frac{1}{4}}.x^{-\frac{1}{8}})dx\\ & =\int x^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}}dx=\int x^{\frac{5}{8}}dx\\ & ={\displaystyle \frac{x^{\frac{5}{8}+1}}{{\displaystyle \frac{5}{8}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{13}{8}}}{{\displaystyle \frac{13}{8}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^{\frac{16-3}{8}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2.x^{-\frac{3}{8}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{x^{-\frac{3}{4}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{x^{-1+\frac{1}{4}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}.{\left(x^{\frac{1}{2}}\right)}^{\frac{1}{2}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}\sqrt{x^{1-\frac{1}{2}}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C}\end{array}\end{array}$

Contoh Soal 1 Materi Integral Tak Tentu Fungsi Aljabar

$\begin{array}{l}\\ 1.& {\displaystyle \int x^{2}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{3}{x}^3+C}\\ \text{b}.{\displaystyle \frac{1}{4}{x}^6+C}\\ \text{c}.{\displaystyle \frac{1}{3}{x}^6+C}\\ \text{d}.{\displaystyle \frac{1}{6}{x}^3+C}\\ \text{e}.{\displaystyle \frac{2}{3}{x}^3+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{2}dx={\displaystyle \frac{x^{2+1}}{2+1}+C={\displaystyle \frac{1}{3}{x}^3+C}}}\end{array}$

$\begin{array}{l}\\ 2.& {\displaystyle \int x^{-2}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2x^{-1}+C}\\ \text{b}.{\displaystyle -x^{-1}+C}\\ \text{c}.{\displaystyle -\frac{1}{2}{x}^{-2}+C}\\ \text{d}.{\displaystyle -\frac{1}{3}{x}^{-3}+C}\\ \text{e}.{\displaystyle -3x^{-3}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{-2}dx={\displaystyle \frac{x^{-2+1}}{-2+1}+C={\displaystyle \frac{1}{-1}{x}^{-1}+C=-x^{-1}+C}}}\end{array}$.

$\begin{array}{l}\\ 3.& {\displaystyle \int x^{{.}^{\frac{1}{3}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{3}{4}{x}^{\frac{4}{3}}+C}\\ \text{b}.{\displaystyle x^{\frac{4}{3}}+C}\\ \text{c}.{\displaystyle \frac{3}{4}{x}^{\frac{2}{3}}+C}\\ \text{d}.{\displaystyle x^{-\frac{2}{3}}+C}\\ \text{e}.{\displaystyle \frac{3}{4}{x}^{-\frac{2}{3}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{{.}^{\frac{1}{3}}}dx={\displaystyle \frac{x^{\frac{1}{3}+1}}{{\displaystyle \frac{1}{3}+1}}+C={\displaystyle \frac{1}{{\displaystyle \frac{4}{3}}}{x}^{{.}^{\frac{4}{3}}}+C={\displaystyle \frac{3}{4}{x}^{{.}^{\frac{4}{3}}}+C.}}}}\end{array}$.

$\begin{array}{l}\\ 4.& {\displaystyle \int \frac{1}{x^3}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -\frac{1}{2x^2}+C}\\ \text{b}.{\displaystyle -\frac{2}{x^2}+C}\\ \text{c}.{\displaystyle \frac{1}{3x^4}+C}\\ \text{d}.{\displaystyle \frac{3}{x^4}+C}\\ \text{e}.{\displaystyle -\frac{1}{4x^3}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int \frac{1}{x^3}dx=\int x^{-3}dx}\\ & ={\displaystyle \frac{x^{-3+1}}{-3+1}+C=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^2}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& {\displaystyle \int \frac{1}{3}{x}^{3}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{3}{x}^4+C}\\ \text{b}.{\displaystyle \frac{1}{4}{x}^4+C}\\ \text{c}.{\displaystyle x^4+C}\\ \text{d}.{\displaystyle \frac{1}{12}{x}^4+C}\\ \text{e}.{\displaystyle \frac{4}{3}{x}^4+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int \frac{1}{3}{x}^{3}dx={\displaystyle \frac{1}{3}.\frac{x^{3+1}}{3+1}+C={\displaystyle \frac{1}{12}{x}^4+C}}}\end{array}$.

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.

$\begin{array}{rl}& \text{Misalkan}\\ & \bullet s\text{adalah menunjukkan posisi}\\ & \bullet v\text{adalah menunjukkan kecepatan}\\ & \bullet a\text{adalah menunjukkan percepatan}\\ & \bullet t\text{adalah menunjukkan waktu}\\ \\ & \mathbf{\text{Perhatikan hubungan berikut}}\\ & v={\displaystyle \frac{ds}{dt}\iff ds=vdt}\\ & \iff {\displaystyle \int ds=\int vdt}\\ & \iff s={\displaystyle \int vdt}\\ \\ & \mathbf{\text{Demikian juga hubungan berikut}}\\ & a={\displaystyle \frac{dv}{dt}\iff dv=adt}\\ & \iff {\displaystyle \int dv=\int adt}\\ & \iff v={\displaystyle \int adt}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan}y,\text{Jika}{\displaystyle \frac{dy}{dx}=2022x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & {\displaystyle \frac{dy}{dx}=2022x\iff dy=2022xdx}\\ & \iff {\displaystyle \int dy=\int 2022xdx}\\ & \iff y={\displaystyle \frac{2022}{2}{x}^2+C}\\ & \iff y=1011x^2+C\end{array}\end{array}$.

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

$\begin{array}{l}\\ 3.& \text{Diketahui bahwa}{f}^{″}(x)=x^2.\text{Jika}{f}^{\prime }(0)=6\\ & \text{dan}f(0)=3,\text{tentukanlah}f(x)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & {\displaystyle f^{″}(x)=x^2}\\ & \iff {\displaystyle \int f^{″}(x)dx={\displaystyle \int x^{2}dx}}\\ & \iff {\displaystyle f^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+C}}\\ & \text{Karena}{f}^{\prime }(x)=6,\text{maka}\\ & f^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+C\iff 6={\displaystyle \frac{1}{3}(0{)}^3+C}}\\ & \iff C=6\\ & \text{Sehingga}{f}^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+6}\\ & \text{Selanjutnya}{\displaystyle \int f^{\prime }(x)dx={\displaystyle \int {\displaystyle \frac{1}{3}{x}^3+6dx}}}\\ & f(x)={\displaystyle \frac{1}{12}{x}^4+6x+C}\\ & \text{Dan juga karena}f(0)=3,\text{maka}\\ & f(0)={\displaystyle \frac{1}{12}(0{)}^4+6(0)+C=3}\\ & C=3,\text{sehingga diperoleh}\\ & f(x)={\displaystyle \frac{1}{12}{x}^4+6x+3}\\ & \text{Jadi},f(x)={\displaystyle \frac{1}{12}{x}^4+6x+3}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui}{f}^{\prime }\left(x\right)=6x^2-2x+6\text{dan}\\ & \text{nilai}\text{fungsi}f\left(2\right)=-7.\text{Tentukanlah}\\ & \text{rumus}\text{fungsi}\text{tersebut}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f\left(x\right)& =\int f^{\prime }\left(x\right)dx\\ & =\int (6x^2-2x+6)dx\\ & =2x^3-x^2+6x+C\end{array}\\ \\ & \text{Karena}f\left(2\right)=-7,\text{maka}\\ \\ & \begin{array}{rl}f\left(2\right)& ={2.2}^3-2^2+6.2+C\\ \iff -7& =16-4+12+C\\ \iff C& =-31\end{array}\\ \\ & \text{Jadi},f\left(x\right)=2x^3-x^2+6x-31\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan}f(x),\text{Jika diketahui}\\ & \text{a}.f^{\prime }(x)=12x^3\text{dan}f(1)=8\\ & \text{b}.f^{\prime }(x)=x^2-2x+3\text{dan}f(3)=9\\ & \text{c}.f^{″}(x)=2,f^{\prime }(2)=2\text{dan}f(2)=10\\ & \text{d}.f^{″}(x)=x^2,f^{\prime }(0)=6\text{dan}f(0)=3\end{array}$.

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

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

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