SBMPTN Zone : Integral Tentu

Berikut ini adalah kumpulan soal mengenai integral tentu tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

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No. 1

Jika \displaystyle\intop_{-4}^4f(x)(\sin x+1)\ dx=8, dengan f(x) fungsi genap dan \displaystyle\intop_{-2}^4f(x)\ dx=4, maka \displaystyle\intop_{-2}^0f(x)\ dx= ....

1. 0
2. 1
   
3. 2
   

4. 3
5. 4
   

SBMPTN 2017

$\begin{array}{rl}{\displaystyle \underset{-4}{\overset{4}{\int }}f(x)(\sin x+1)dx}& =8\\ {\displaystyle \underset{-4}{\overset{4}{\int }}f(x)\sin xdx+{\displaystyle \underset{-4}{\overset{4}{\int }}f(x)dx}}& =8\end{array}$
Karena f(x) genap dan \sin x ganjil, maka f(x)\sin x ganjil. Sehingga \displaystyle\int_{-4}^4f(x)\sin x\ dx=0.

$\begin{array}{rl}{\displaystyle \underset{-4}{\overset{4}{\int }}f(x)dx}& =8\\ 2{\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}& =8\\ {\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}& =4\end{array}$

$\begin{array}{rl}{\displaystyle \underset{-2}{\overset{4}{\int }}f(x)dx}& =4\\ {\displaystyle \underset{-2}{\overset{0}{\int }}f(x)dx+{\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}}& =4\\ {\displaystyle \underset{-2}{\overset{0}{\int }}f(x)dx+4}& =4\\ {\displaystyle \underset{-2}{\overset{0}{\int }}f(x)dx}& =0\end{array}$

Jadi, \displaystyle\intop_{-2}^0f(x)\ dx=0.
JAWAB: A

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No. 2

Diberikan fungsi f dan g yang memenuhi sistem
\displaystyle\int_0^1f(x)\ dx+\left(\int_0^2g(x)\ dx\right)^2=3
f(x)=3x^2+4x+\displaystyle\int_0^2g(x)\ dx, dengan {\displaystyle\int_0^2g(x)\ dx\neq0}
Nilai f(1)= ....

1. -6
2. -3
   
3. 0
   

4. 3
5. 6
   

Misal \displaystyle\intop_0^2g(x)\ dx=A

f(x)=3x^2+4x+A

$\begin{array}{rl}{\displaystyle \underset{0}{\overset{1}{\int }}f(x)dx+{\left({\displaystyle \underset{0}{\overset{2}{\int }}g(x)dx}\right)}^2}& =3\\ {\displaystyle \underset{0}{\overset{1}{\int }}(3x^2+4x+A)dx+A^2}& =3\\ {[x^3+2x^2+Ax]}_0^1+A^2& =3\\ 3+A+A^2& =3\\ A^2+A& =0\\ A(A+1)& =0\end{array}$
A=0 (TM) atau A=-1

$\begin{array}{rl}f(x)& =3x^2+4x-1\\ f(1)& =3(1{)}^2+4(1)-1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$

Jadi, f(1)=6.
JAWAB: E

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No. 3

f(x) adalah fungsi yang terdefinisi pada himpunan bilangan real dan f'(x) adalah turunan pertamanya. Jika {f(0)=f(2)=3} dan {f'(0)=f'(2)=-1} maka {\displaystyle\int_0^2x\cdot f''(x)\ dx=} ....

1. -3
2. -2
   
3. -1
   

4. 1
5. 2
   

Misal

$\begin{array}{rl}u& =x\\ du& =dx\end{array}$

$\begin{array}{rl}dv& =f^{″}(x)dx\\ v& =f^{\prime }(x)\end{array}$

$\begin{array}{rl}{\displaystyle \int udv}& =uv-{\displaystyle \int vdu}\\ {\displaystyle \underset{0}{\overset{2}{\int }}x\cdot f^{″}(x)dx}& ={[x{f}^{\prime }(x)-{\displaystyle \underset{0}{\overset{2}{\int }}{f}^{\prime }(x)dx}]}_0^2\\ & ={[x{f}^{\prime }(x)-f(x)]}_0^2\\ & =[(2)f^{\prime }(2)-f(2)]-[(0)f^{\prime }(0)-f(0)]\\ & =[(2)(-1)-3]-[0-3]\\ & =[-2-3]-[-3]\\ & =-5+3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -2}}}}\end{array}$

Jadi, \displaystyle\intop_0^2x\cdot f''(x)\ dx=-2.
JAWAB: B

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No. 4

Diberikan fungsi f(x) yang simetri terhadap sumbu y. Jika \displaystyle\intop_3^4f(x)\ dx=12. maka \displaystyle\intop_1^2f(5-x)\ dx= ....

2. 
   
3. 
   

5. 
   

Ganesha Operation

Misal 5-x=u maka
$\begin{array}{rl}-dx& =du\\ dx& =-du\end{array}$

Jika x=1 maka
5-1=4

Jika x=2 maka
5-2=3

$\begin{array}{rl}{\displaystyle \underset{1}{\overset{2}{\int }}f(5-x)dx}& ={\displaystyle \underset{4}{\overset{3}{\int }}f(u)(-du)}\\ & ={\displaystyle \underset{3}{\overset{4}{\int }}f(u)du}\\ & ={\displaystyle \underset{3}{\overset{4}{\int }}f(x)dx}\\ & =12\end{array}$

Jadi, \displaystyle\intop_1^2f(5-x)\ dx=12.
JAWAB: C

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No. 5

Jika (f\circ g)(x)=\dfrac{6x+3}{2x-5} dan g(x)=4x-11, maka hasil dari \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx adalah ....

1. 72\ln2-3
2. 36\ln 3-2
   
3. 36\ln 2-6
   

4. 36\ln 2 - 3
5. 72\ln 3-2
   

$\begin{array}{rl}(f\circ g)(x)& ={\displaystyle \frac{6x+3}{2x-5}}\\ f(g(x))& ={\displaystyle \frac{12x+6}{4x-10}}\\ f(4x-11)& ={\displaystyle \frac{3(4x-11)+39}{4x-11+1}}\\ f(x)& ={\displaystyle \frac{3x+39}{x+1}}\\ f^{-1}(x)& ={\displaystyle \frac{-x+39}{x-3}}\\ f^{-1}(x-1)& ={\displaystyle \frac{-(x-1)+39}{x-1-3}}\\ & ={\displaystyle \frac{-x+1+39}{x-4}}\\ & ={\displaystyle \frac{-x+40}{x-4}}\end{array}$

$\begin{array}{rl}{\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{f^{-1}(x-1)}{g(3)}}dx}& ={\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{{\displaystyle \frac{-x+40}{x-4}}}{4(3)-11}}dx}\\ & ={\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{{\displaystyle \frac{-x+4+36}{x-4}}}{12-11}}dx}\\ & ={\displaystyle \underset{5}{\overset{8}{\int }}{\displaystyle \frac{-1+{\displaystyle \frac{36}{x-4}}}{1}}dx}\\ & ={\displaystyle \underset{5}{\overset{8}{\int }}(-1+{\displaystyle \frac{36}{x-4}})dx}\\ & ={[-x+36\ln |x-4|]}_5^8\\ & =[-8+36\ln |8-4|]-[-5+36\ln |5-4|]\\ & =[-8+36\ln 4]-[-5+36\ln 1]\\ & =[-8+36\ln 2^2]-[-5-42(0)]\\ & =[-8+72\ln 2]-[-5]\\ & =-8+72\ln 2+5\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 72\ln 2-3}}}}\end{array}$

Jadi, \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx=72\ln2-3.
JAWAB: A

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No. 6

Jika diketahui \dfrac{df(x)}{dx}=g(x) kontinu pada interval p\leq x\leq q maka nilai dari {\displaystyle\intop_p^q f(x)g(x)\ dx=} ....

$\begin{array}{rl}{\displaystyle \frac{df(x)}{dx}}& =g(x)\\ df(x)& =g(x)dx\end{array}$

Misal
$\begin{array}{rl}u& =f(x)\\ du& =df(x)\\ & =g(x)dx\end{array}$

$\begin{array}{rl}{\displaystyle \underset{p}{\overset{q}{\int }}f(x)g(x)dx}& ={\displaystyle \underset{p}{\overset{q}{\int }}udu}\\ & ={\left[{\displaystyle \frac{1}{2}}{u}^2\right]}_p^q\\ & ={\left[{\displaystyle \frac{1}{2}}{[f(x)]}^2\right]}_p^q\\ & ={\displaystyle \frac{1}{2}}{[f(q)]}^2-{\displaystyle \frac{1}{2}}{[f(p)]}^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{{[f(q)]}^2-{[f(p)]}^2}{2}}}}}}\end{array}$

Jadi, \displaystyle\intop_p^q f(x)g(x)\ dx=\dfrac{\left[f(q)\right]^2-\left[f(p)\right]^2}2.
JAWAB: B

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No. 7

Jika \displaystyle\intop_{-4}^4f(x)(\sin x+1)\ dx=8, dengan f(x) fungsi genap dan {\displaystyle\intop_{-2}^4f(x)(\sin x+1)\ dx=4}, maka {\displaystyle\intop_{-2}^0f(x)(\sin x+1)\ dx=} ....

1. 0
2. 1
   
3. 2
   

4. 3
5. 4
   

ingat bahwa jika fungsi genap dikali fungsi ganjil hasilnya fungsi ganjil. \sin x merupakan fungsi ganjil sehingga f(x)\sin x merupakan fungsi ganjil.
$\begin{array}{rl}{\displaystyle \underset{-4}{\overset{4}{\int }}f(x)(\sin x+1)dx}& =8\\ {\displaystyle \underset{-4}{\overset{4}{\int }}(f(x)\sin x+f(x))dx}& =8\\ {\displaystyle \underset{-4}{\overset{4}{\int }}(f(x)\sin x)dx+{\displaystyle \underset{-4}{\overset{4}{\int }}f(x)dx}}& =8\\ 0+2{\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}& =8\\ {\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}& =4\end{array}$

$\begin{array}{rl}{\displaystyle \underset{-2}{\overset{4}{\int }}f(x)dx}& ={\displaystyle \underset{-2}{\overset{0}{\int }}f(x)dx+{\displaystyle \underset{0}{\overset{4}{\int }}f(x)dx}}\\ 4& ={\displaystyle \underset{-2}{\overset{0}{\int }}f(x)dx+4}\\ {\displaystyle \underset{-2}{\overset{0}{\int }}f(x)dx}& =\boxed{{\displaystyle \boxed{{\displaystyle 0}}}}\end{array}$

Jadi, \displaystyle\intop_{-2}^0f(x)(\sin x+1)\ dx=0.
JAWAB: A

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No. 8

Diketahui fungsi {f(x)=x^3+3x^2-5x+\displaystyle\intop_{-1}^1f(x)\ dx}. Nilai f(1)= ....

1. -3
2. -2
   
3. -1
   

4. 3
5. 4
   

Misal \displaystyle\intop_{-1}^1f(x)\ dx=c.
$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{1}{\int }}(x^3+3x^2-5x+c)dx}& =c\\ {[{\displaystyle \frac{1}{4}}{x}^4+x^3-{\displaystyle \frac{5}{2}}{x}^2+cx]}_{-1}^1& =c\\ [{\displaystyle \frac{1}{4}}(1{)}^4+(1{)}^3-{\displaystyle \frac{5}{2}}(1{)}^2+c(1)]-[{\displaystyle \frac{1}{4}}(-1{)}^4+(-1{)}^3-{\displaystyle \frac{5}{2}}(-1{)}^2+c(-1)]& =c\\ [{\displaystyle \frac{1}{4}}(1)+1-{\displaystyle \frac{5}{2}}(1)+c]-[{\displaystyle \frac{1}{4}}(1)+(-1)-{\displaystyle \frac{5}{2}}(1)-c]& =c\\ [{\displaystyle \frac{1}{4}}+1-{\displaystyle \frac{5}{2}}+c]-[{\displaystyle \frac{1}{4}}-1-{\displaystyle \frac{5}{2}}-c]& =c\\ {\displaystyle \frac{1}{4}}+1-{\displaystyle \frac{5}{2}}+c-{\displaystyle \frac{1}{4}}+1+{\displaystyle \frac{5}{2}}+c& =c\\ 2c+2& =c\\ c& =-2\end{array}$

$\begin{array}{rl}f(x)& =x^3+3x^2-5x-2\\ f(1)& =1^3+3(1{)}^2-5(1)-2\\ & =1+3(1)-5-2\\ & =1+3-5-2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -3}}}}\end{array}$

Jadi, f(1)=-3.
JAWAB: A

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No. 9

Misalkan f(x)=3x+b. Jika \displaystyle\intop_{-1}^1f(x)\ dx, \displaystyle\intop_{-1}^1\left[f(x)\right]^2\ dx, \displaystyle\intop_{-1}^1\left[f(x)\right]^3\ dx membentuk suatu barisan geometri, maka nilai b^2 adalah

1. 2
2. 3
   
3. 4
   

4. 5
5. 6
   

SBMPTN 2016, Kode 207

$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{1}{\int }}f(x)dx}& ={\displaystyle \underset{-1}{\overset{1}{\int }}(3x+b)dx}\\ & ={[{\displaystyle \frac{3}{2}}{x}^2+bx]}_{-1}^1\\ & =[{\displaystyle \frac{3}{2}}(1{)}^2+b(1)]-[{\displaystyle \frac{3}{2}}(-1{)}^2+b(-1)]\\ & =[{\displaystyle \frac{3}{2}}(1)+b]-[{\displaystyle \frac{3}{2}}(1)-b]\\ & =[{\displaystyle \frac{3}{2}}+b]-[{\displaystyle \frac{3}{2}}-b]\\ & ={\displaystyle \frac{3}{2}}+b-{\displaystyle \frac{3}{2}}+b\\ & =2b\end{array}$

$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{1}{\int }}{[f(x)]}^{2}dx}& ={\displaystyle \underset{-1}{\overset{1}{\int }}(3x+b{)}^{2}dx}\\ & ={\displaystyle \underset{-1}{\overset{1}{\int }}(9x^2+6bx+b^2)dx}\\ & ={[3x^3+3b{x}^2+b^{2}x]}_{-1}^1\\ & =[3(1{)}^3+3b(1{)}^2+b^2(1)]-[3(-1{)}^3+3b(-1{)}^2+b^2(-1)]\\ & =[3(1)+3b(1)+b^2]-[3(-1)+3b(1)-b^2]\\ & =[3+3b+b^2]-[-3+3b-b^2]\\ & =3+3b+b^2+3-3b+b^2\\ & =2b^2+6\end{array}$

$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{1}{\int }}{[f(x)]}^{3}dx}& ={\displaystyle \underset{-1}{\overset{1}{\int }}(3x+b{)}^{3}dx}\\ & ={\displaystyle \underset{-1}{\overset{1}{\int }}(3x+b{)}^3{\displaystyle \frac{d(3x+b)}{3}}}\\ & ={\displaystyle \frac{1}{3}}{[{\displaystyle \frac{1}{4}}(3x+b{)}^4]}_{-1}^1\\ & ={\displaystyle \frac{1}{12}}[(3(1)+b{)}^4-(3(-1)+b{)}^4]\\ & ={\displaystyle \frac{1}{12}}[(3+b{)}^4-(-3+b{)}^4]\\ & ={\displaystyle \frac{1}{12}}((3+b{)}^2+(-3+b{)}^2)((3+b{)}^2-(-3+b{)}^2)\\ & ={\displaystyle \frac{1}{12}}(9+6b+b^2+9-6b+b^2)((3+b)+(-3+b))((3+b)-(-3+b))\\ & ={\displaystyle \frac{1}{12}}(2b^2+18)(3+b-3+b)(3+b+3-b)\\ & ={\displaystyle \frac{1}{12}}(2b^2+18)\left(2b\right)\left(6\right)\\ & =2b^3+18b\end{array}$

Misal b^2=p
$\begin{array}{rl}{U_2}^2& =U_1\cdot U_3\\ {(2b^2+6)}^2& =(2b)(2b^3+18b)\\ 4b^4+24b^2+36& =4b^4+36b^2\\ 36& =12b^2\\ b^2& =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

Jadi, b^2=3.
JAWAB: B

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No. 10

Jika {\displaystyle\intop_0^1\dfrac{x}{1-x}dx=b}, maka {\displaystyle\intop_0^1\dfrac1{1-x}dx=}

1. 2b
2. b
   
3. {1-b}
   

4. {1+b}
5. b^2
   

$\begin{array}{rl}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \frac{1}{1-x}}dx}& ={\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \frac{1-x+x}{1-x}}dx}\\ & ={\displaystyle \underset{0}{\overset{1}{\int }}(1+{\displaystyle \frac{x}{1-x}})dx}\end{array}$

CARA BIASA

$\begin{array}{rl}{\displaystyle \underset{0}{\overset{1}{\int }}(1+{\displaystyle \frac{x}{1-x}})dx}& ={\displaystyle \underset{0}{\overset{1}{\int }}1dx+{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \frac{x}{1-x}}dx}}\\ & =[x{]}_0^1+b\\ & =1-0+b\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1+b}}}}\end{array}$

CARA CEPAT

\color{blue}{\displaystyle\intop_n^{n+1}c\ dx=c}, dengan c adalah konstanta
\displaystyle\intop_0^1\left(1+\dfrac{x}{1-x}\right)dx=\boxed{\boxed{1+b}}

Jadi, \displaystyle\intop_0^1\dfrac1{1-x}dx=1+b.
JAWAB: D

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