Contoh Soal 2 Materi Integral Tentu Fungsi Aljabar

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

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

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

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

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