Exercise Zone : Integral Tentu [2]

Berikut ini adalah kumpulan soal mengenai Integral Tentu. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Jika nilai {\displaystyle\intop_1^2f(x)\ dx=10}, maka nilai {\displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx} adalah

1. 3
2. 4
   
3. 5
   

4. 6
5. 7
   

Grup Telegram

Misal u=x^2+1
$$\begin{array}{rl}du& =2xdx\\ {\displaystyle \frac{1}{2}}du& =xdx\end{array}$$

x=0\rightarrow u=0^2+1=1
x=1\rightarrow u=1^2+1=2

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{1}{\int }}x\cdot f(x^2+1)dx}& ={\displaystyle \frac{1}{2}}{\displaystyle \underset{1}{\overset{2}{\int }}f(u)du}\\ & ={\displaystyle \frac{1}{2}}(10)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5}}}}\end{array}$$

Jadi, \displaystyle\intop_0^1x\cdot f\left(x^2+1\right)\ dx=5. JAWAB: C

No.

Hasil dari {\displaystyle\intop_2^42x^3+6x^2-2x-5\ dx=} ....

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$$\begin{array}{rl}{\displaystyle \underset{2}{\overset{4}{\int }}2x^3+6x^2-2x-5dx}& ={[{\displaystyle \frac{2}{4}}{x}^4+{\displaystyle \frac{6}{3}}{x}^3-{\displaystyle \frac{2}{2}}{x}^2-5x]}_2^4\\ & ={[{\displaystyle \frac{1}{2}}{x}^4+2x^3-x^2-5x]}_2^4\\ & =({\displaystyle \frac{1}{2}}(4{)}^4+2(4{)}^3-(4{)}^2-5(4))-({\displaystyle \frac{1}{2}}(2{)}^4+2(2{)}^3-(2{)}^2-5(2))\\ & =(8+128-16-20)-(8+16-4-10)\\ & =100-20\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 80}}}}\end{array}$$

Jadi, \displaystyle\intop_2^42x^3+6x^2-2x-5\ dx=80.

No.

Nilai {\displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx=} ....

1. 2
2. 4
   
3. 6
   

4. 7
5. 8
   

GRUP TELEGRAM

$$\begin{array}{rl}{\displaystyle \underset{1}{\overset{4}{\int }}(3\sqrt{x}-2)dx}& ={\displaystyle \underset{1}{\overset{4}{\int }}(3x^{\frac{1}{2}}-2)dx}\\ & ={3\cdot {\displaystyle \frac{2}{3}}{x}^{\frac{3}{2}}-2x|}_1^4\\ & ={2x\sqrt{x}-2x|}_1^4\\ & =(2(4)\sqrt{4}-2(4))-(2(1)\sqrt{1}-2(1))\\ & =(8(2)-8)-(2(1)-2)\\ & =(16-8)-(2-2)\\ & =8-0\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$$

Jadi, \displaystyle\intop_1^4\left(3\sqrt{x}-2\right)\ dx=8. JAWAB: E

No.

Nilai {\displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx=} ....

1. -12
2. -4
3. -3

4. 2
5. \dfrac32

GRUP TELEGRAM

$$\begin{array}{rl}{\displaystyle \underset{1}{\overset{4}{\int }}{\displaystyle \frac{2}{x\sqrt{x}}}dx}& ={\displaystyle \underset{1}{\overset{4}{\int }}{\displaystyle \frac{2}{x\cdot x^{\frac{1}{2}}}}dx}\\ & ={\displaystyle \underset{1}{\overset{4}{\int }}{\displaystyle \frac{2}{x^{\frac{3}{2}}}}dx}\\ & ={\displaystyle \underset{1}{\overset{4}{\int }}2x^{-\frac{3}{2}}dx}\\ & ={2(-{\displaystyle \frac{2}{1}}{x}^{-\frac{1}{2}})|}_1^4\\ & ={-{\displaystyle \frac{4}{x^{\frac{1}{2}}}}|}_1^4\\ & ={-{\displaystyle \frac{4}{\sqrt{x}}}|}_1^4\\ & =(-{\displaystyle \frac{4}{\sqrt{4}}})-(-{\displaystyle \frac{4}{\sqrt{1}}})\\ & =-{\displaystyle \frac{4}{2}}+{\displaystyle \frac{4}{1}}\\ & =-2+4\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

Jadi, \displaystyle\intop_1^4\dfrac2{x\sqrt{x}}\ dx=2. JAWAB: D

No.

Hasil {\displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx=} ....

1. 9\dfrac13
2. 9
3. 8

4. \dfrac{10}3
5. 3

GRUP TELEGRAM

$$\begin{array}{rl}{\displaystyle \underset{1}{\overset{3}{\int }}(x^2+{\displaystyle \frac{1}{6}})dx}& ={{\displaystyle \frac{1}{3}}{x}^3+{\displaystyle \frac{1}{6}}x|}_1^3\\ & =({\displaystyle \frac{1}{3}}(3{)}^3+{\displaystyle \frac{1}{6}}(3))-({\displaystyle \frac{1}{3}}(1{)}^3+{\displaystyle \frac{1}{6}}(1))\\ & =(9+{\displaystyle \frac{1}{2}})-({\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{6}})\\ & =9+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{3}{6}}\\ & =9+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 9}}}}\end{array}$$

Jadi, \displaystyle\intop_1^3\left(x^2+\dfrac16\right)\ dx=9. JAWAB: B

No.

Hasil {\displaystyle\intop_0^2x^2\left(x+2\right)\ dx=} ....

1. 6
2. 6\dfrac13
3. 6\dfrac23

4. 9\dfrac13
5. 20

GRUP TELEGRAM

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{2}{\int }}{x}^2(x+2)dx}& ={\displaystyle \underset{0}{\overset{2}{\int }}(x^3+2x^2)dx}\\ & ={{\displaystyle \frac{1}{4}}{x}^4+{\displaystyle \frac{2}{3}}{x}^3|}_0^2\\ & =({\displaystyle \frac{1}{4}}(2{)}^4+{\displaystyle \frac{2}{3}}(2{)}^3)-({\displaystyle \frac{1}{4}}(0{)}^4+{\displaystyle \frac{2}{3}}(0{)}^3)\\ & =(4+{\displaystyle \frac{16}{3}})-0\\ & =4+5{\displaystyle \frac{1}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 9{\displaystyle \frac{1}{3}}}}}}\end{array}$$

Jadi, \displaystyle\intop_0^2x^2\left(x+2\right)\ dx=9\dfrac13. JAWAB: D

No.

Hasil dari \displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx=

1. -\dfrac{120}{147}
2. -\dfrac{157}{120}
3. -\dfrac{147}{120}

4. \dfrac{147}{120}
5. \dfrac{120}{147}

Grup Telegram

CARA 1: SUBSTITUSI

Misal u=1+x\rightarrow x=u-1
du=dx

$$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{0}{\int }}3x\sqrt[7]{1+x}dx}& ={\displaystyle \underset{-1}{\overset{0}{\int }}3(u-1)\sqrt[7]{u}du}\\ & ={\displaystyle \underset{-1}{\overset{0}{\int }}3(u-1)u^{\frac{1}{7}}du}\\ & ={\displaystyle \underset{-1}{\overset{0}{\int }}(3u^{\frac{8}{7}}-3u^{\frac{1}{7}})du}\\ & ={[3\cdot {\displaystyle \frac{7}{15}}{u}^{\frac{15}{7}}-3\cdot {\displaystyle \frac{7}{8}}{u}^{\frac{8}{7}}]}_{-1}^0\\ & ={[{\displaystyle \frac{7}{5}}(1+x{)}^{\frac{15}{7}}-{\displaystyle \frac{21}{8}}(1+x{)}^{\frac{8}{7}}]}_{-1}^0\\ & =[{\displaystyle \frac{7}{5}}(1+0{)}^{\frac{15}{7}}-{\displaystyle \frac{21}{8}}(1+0{)}^{\frac{8}{7}}]-[{\displaystyle \frac{7}{5}}(1+(-1){)}^{\frac{15}{7}}-{\displaystyle \frac{21}{8}}(1+(-1){)}^{\frac{8}{7}}]\\ & =[{\displaystyle \frac{7}{5}}(1{)}^{\frac{15}{7}}-{\displaystyle \frac{21}{8}}(1{)}^{\frac{8}{7}}]-[{\displaystyle \frac{7}{5}}(0{)}^{\frac{15}{7}}-{\displaystyle \frac{21}{8}}(0{)}^{\frac{8}{7}}]\\ & =[{\displaystyle \frac{7}{5}}(1)-{\displaystyle \frac{21}{8}}(1)]-[{\displaystyle \frac{7}{5}}(0)-{\displaystyle \frac{21}{8}}(0)]\\ & =[{\displaystyle \frac{7}{5}}-{\displaystyle \frac{21}{8}}]-[0-0]\\ & =-{\displaystyle \frac{49}{40}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{147}{120}}}}}}\end{array}$$

CARA 2: PARSIAL

u	dv
3x	\sqrt[7]{1+x}=(1+x)^{\frac17}
3	\dfrac78(1+x)^{\frac87}
0	\dfrac78\cdot\dfrac7{15}(1+x)^{\frac{15}7}=\dfrac{49}{120}(1+x)^{\frac{15}7}

$$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{0}{\int }}3x\sqrt[7]{1+x}dx}& ={[3x\cdot {\displaystyle \frac{7}{8}}(1+x{)}^{\frac{8}{7}}-3\cdot {\displaystyle \frac{49}{120}}(1+x{)}^{\frac{15}{7}}]}_{-1}^0\\ & ={[{\displaystyle \frac{21}{8}}x(1+x{)}^{\frac{8}{7}}-\cdot {\displaystyle \frac{147}{120}}(1+x{)}^{\frac{15}{7}}]}_{-1}^0\\ & =[{\displaystyle \frac{21}{8}}(0)(1+0{)}^{\frac{8}{7}}-\cdot {\displaystyle \frac{147}{120}}(1+0{)}^{\frac{15}{7}}]-[{\displaystyle \frac{21}{8}}(-1)(1+(-1){)}^{\frac{8}{7}}-\cdot {\displaystyle \frac{147}{120}}(1+(-1){)}^{\frac{15}{7}}]\\ & =[0-\cdot {\displaystyle \frac{147}{120}}(1{)}^{\frac{15}{7}}]-[-{\displaystyle \frac{21}{8}}(0{)}^{\frac{8}{7}}-\cdot {\displaystyle \frac{147}{120}}(0{)}^{\frac{15}{7}}]\\ & =[-\cdot {\displaystyle \frac{147}{120}}(1)]-0\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{147}{120}}}}}}\end{array}$$

Jadi, \displaystyle\intop_{-1}^03x\sqrt[7]{1+x}\ dx=-\dfrac{147}{120}. JAWAB: C

No.

Hasil dari \displaystyle\intop_0^1\dfrac{10(18x+6)\ dx}{\sqrt{9x^2+6x+1}}=

1. 40
2. 50
3. 60

4. 70
5. 80

SKMP Juli 2020

Misal u=9x^2+6x+1
du=(18x+6)\ dx

x=0\rightarrow u=9(0)^2+6(0)+1=1
x=1\rightarrow u=9(1)^2+6(1)+1=16

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \frac{10(18x+6)dx}{\sqrt{9x^2+6x+1}}}}& ={\displaystyle \underset{1}{\overset{16}{\int }}{\displaystyle \frac{10du}{\sqrt{u}}}}\\ & ={\displaystyle \underset{1}{\overset{16}{\int }}{\displaystyle \frac{10du}{u^{\frac{1}{2}}}}}\\ & ={\displaystyle \underset{1}{\overset{16}{\int }}10u^{-\frac{1}{2}}du}\\ & ={[10\cdot 2u^{\frac{1}{2}}]}_1^{16}\\ & ={\left[20\sqrt{u}\right]}_1^{16}\\ & =20\sqrt{16}-20\sqrt{1}\\ & =20(4)-20(1)\\ & =80-20\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 60}}}}\end{array}$$

Jadi, \displaystyle\intop_0^1\dfrac{10(18x+6)\ dx}{\sqrt{9x^2+6x+1}}=60. JAWAB: C

No.

\displaystyle\intop_0^3x\sqrt{1+x}\ dx= ....

Grup Telegram

Misal
u=1+x\to x=u-1
du=dx

x=0\to u=1+0=1
x=3\to u=1+3=4

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{3}{\int }}x\sqrt{1+x}dx}& ={\displaystyle \underset{1}{\overset{4}{\int }}(u-1)\sqrt{u}du}\\ & ={\displaystyle \underset{1}{\overset{4}{\int }}(u-1)u^{\frac{1}{2}}du}\\ & ={\displaystyle \underset{1}{\overset{4}{\int }}(u^{\frac{3}{2}}-u^{\frac{1}{2}})du}\\ & ={[{\displaystyle \frac{2}{5}}{u}^{\frac{5}{2}}-{\displaystyle \frac{2}{3}}{u}^{\frac{3}{2}}]}_1^4\\ & =({\displaystyle \frac{2}{5}}(4{)}^{\frac{5}{2}}-{\displaystyle \frac{2}{3}}(4{)}^{\frac{3}{2}})-({\displaystyle \frac{2}{5}}(1{)}^{\frac{5}{2}}-{\displaystyle \frac{2}{3}}(1{)}^{\frac{3}{2}})\\ & =({\displaystyle \frac{2}{5}}(32)-{\displaystyle \frac{2}{3}}(8))-({\displaystyle \frac{2}{5}}(1)-{\displaystyle \frac{2}{3}}(1))\\ & =({\displaystyle \frac{64}{5}}-{\displaystyle \frac{16}{3}})-({\displaystyle \frac{2}{5}}-{\displaystyle \frac{2}{3}})\\ & =({\displaystyle \frac{192}{15}}-{\displaystyle \frac{80}{15}})-({\displaystyle \frac{6}{15}}-{\displaystyle \frac{10}{15}})\\ & =\left({\displaystyle \frac{112}{15}}\right)-(-{\displaystyle \frac{4}{15}})\\ & ={\displaystyle \frac{112}{15}}+{\displaystyle \frac{4}{15}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{116}{15}}}}}}\end{array}$$

Jadi, \displaystyle\intop_0^3x\sqrt{1+x}\ dx=\dfrac{116}{15}.