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

\(\begin{aligned} \displaystyle\intop_{-4}^4f(x)(\sin x+1)\ dx&=8\\ \displaystyle\intop_{-4}^4f(x)\sin x\ dx+\displaystyle\intop_{-4}^4f(x)\ dx&=8 \end{aligned}\)
Karena f(x) genap dan \sin x ganjil, maka f(x)\sin x ganjil. Sehingga \displaystyle\int_{-4}^4f(x)\sin x\ dx=0.

\(\begin{aligned} \displaystyle\intop_{-4}^4f(x)\ dx&=8\\ 2\displaystyle\intop_0^4f(x)\ dx&=8\\ \displaystyle\intop_0^4f(x)\ dx&=4 \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_{-2}^4f(x)\ dx&=4\\ \displaystyle\intop_{-2}^0f(x)\ dx+\displaystyle\intop_0^4f(x)\ dx&=4\\ \displaystyle\intop_{-2}^0f(x)\ dx+4&=4\\ \displaystyle\intop_{-2}^0f(x)\ dx&=0 \end{aligned}\)

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

\(\begin{aligned} \displaystyle\intop_0^1f(x)\ dx+\left(\displaystyle\intop_0^2g(x)\ dx\right)^2&=3\\ \displaystyle\intop_0^1\left(3x^2+4x+A\right)\ dx+A^2&=3\\ \left[x^3+2x^2+Ax\right]_0^1+A^2&=3\\ 3+A+A^2&=3\\ A^2+A&=0\\ A(A+1)&=0 \end{aligned}\)
A=0 (TM) atau A=-1

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

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

\(\begin{aligned} u&=x\\ du&=dx \end{aligned}\)

\(\begin{aligned} dv&=f''(x)\ dx\\ v&=f'(x) \end{aligned}\)

\(\begin{aligned} \displaystyle\intop u\ dv&=uv-\displaystyle\intop v\ du\\ \displaystyle\intop_0^2x\cdot f''(x)\ dx&=\left[xf'(x)-\displaystyle\intop_0^2f'(x)\ dx\right]_0^2\\ &=\left[xf'(x)-f(x)\right]_0^2\\ &=\left[(2)f'(2)-f(2)\right]-\left[(0)f'(0)-f(0)\right]\\ &=\left[(2)(-1)-3\right]-\left[0-3\right]\\ &=\left[-2-3\right]-\left[-3\right]\\ &=-5+3\\ &=\boxed{\boxed{-2}} \end{aligned}\)

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
\(\begin{aligned} -dx&=du\\ dx&=-du \end{aligned}\)

Jika x=1 maka
5-1=4

Jika x=2 maka
5-2=3

\(\begin{aligned} \displaystyle\intop_1^2f(5-x)\ dx&=\displaystyle\intop_4^3f(u)\ (-du)\\ &=\displaystyle\intop_3^4f(u)\ du\\ &=\displaystyle\intop_3^4f(x)\ dx\\ &=12 \end{aligned}\)

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
   

\(\begin{aligned} (f\circ g)(x)&=\dfrac{6x+3}{2x-5}\\[10pt] f(g(x))&=\dfrac{12x+6}{4x-10}\\[10pt] f(4x-11)&=\dfrac{3(4x-11)+39}{4x-11+1}\\[10pt] f(x)&=\dfrac{3x+39}{x+1}\\[10pt] f^{-1}(x)&=\dfrac{-x+39}{x-3}\\[10pt] f^{-1}(x-1)&=\dfrac{-(x-1)+39}{x-1-3}\\[10pt] &=\dfrac{-x+1+39}{x-4}\\[10pt] &=\dfrac{-x+40}{x-4} \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx&=\displaystyle\intop_5^8\dfrac{\dfrac{-x+40}{x-4}}{4(3)-11}\ dx\\ &=\displaystyle\intop_5^8\dfrac{\dfrac{-x+4+36}{x-4}}{12-11}\ dx\\ &=\displaystyle\intop_5^8\dfrac{-1+\dfrac{36}{x-4}}1\ dx\\ &=\displaystyle\intop_5^8\left(-1+\dfrac{36}{x-4}\right)\ dx\\ &=\left[-x+36\ln|x-4|\right]_5^8\\ &=\left[-8+36\ln|8-4|\right]-\left[-5+36\ln|5-4|\right]\\ &=\left[-8+36\ln4\right]-\left[-5+36\ln1\right]\\ &=\left[-8+36\ln2^2\right]-\left[-5-42(0)\right]\\ &=\left[-8+72\ln2\right]-\left[-5\right]\\ &=-8+72\ln2+5\\ &=\boxed{\boxed{72\ln2-3}} \end{aligned}\)

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=} ....

\(\begin{aligned} \dfrac{df(x)}{dx}&=g(x)\\ df(x)&=g(x)\ dx \end{aligned}\)

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

\(\begin{aligned} \displaystyle\intop_p^q f(x)g(x)\ dx&=\displaystyle\intop_p^q u\ du\\ &=\left[\dfrac12u^2\right]_p^q\\ &=\left[\dfrac12\left[f(x)\right]^2\right]_p^q\\ &=\dfrac12\left[f(q)\right]^2-\dfrac12\left[f(p)\right]^2\\ &=\boxed{\boxed{\dfrac{\left[f(q)\right]^2-\left[f(p)\right]^2}2}} \end{aligned}\)

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.
\(\begin{aligned} \displaystyle\intop_{-4}^4f(x)(\sin x+1)\ dx&=8\\ \displaystyle\intop_{-4}^4(f(x)\sin x+f(x))\ dx&=8\\ \displaystyle\intop_{-4}^4(f(x)\sin x)\ dx+\displaystyle\intop_{-4}^4f(x)\ dx&=8\\ 0+2\displaystyle\intop_0^4f(x)\ dx&=8\\ \displaystyle\intop_0^4f(x)\ dx&=4 \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_{-2}^4f(x)\ dx&=\displaystyle\intop_{-2}^0f(x)\ dx+\displaystyle\intop_0^4f(x)\ dx\\ 4&=\displaystyle\intop_{-2}^0f(x)\ dx+4\\ \displaystyle\intop_{-2}^0f(x)\ dx&=\boxed{\boxed{0}} \end{aligned}\)

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.
\(\begin{aligned} \displaystyle\intop_{-1}^1\left(x^3+3x^2-5x+c\right)\ dx&=c\\ \left[\dfrac14x^4+x^3-\dfrac52x^2+cx\right]_{-1}^1&=c\\ \left[\dfrac14(1)^4+(1)^3-\dfrac52(1)^2+c(1)\right]-\left[\dfrac14(-1)^4+(-1)^3-\dfrac52(-1)^2+c(-1)\right]&=c\\ \left[\dfrac14(1)+1-\dfrac52(1)+c\right]-\left[\dfrac14(1)+(-1)-\dfrac52(1)-c\right]&=c\\ \left[\dfrac14+1-\dfrac52+c\right]-\left[\dfrac14-1-\dfrac52-c\right]&=c\\ \dfrac14+1-\dfrac52+c-\dfrac14+1+\dfrac52+c&=c\\ 2c+2&=c\\ c&=-2 \end{aligned}\)

\(\begin{aligned} f(x)&=x^3+3x^2-5x-2\\ f(1)&=1^3+3(1)^2-5(1)-2\\ &=1+3(1)-5-2\\ &=1+3-5-2\\ &=\boxed{\boxed{-3}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\intop_{-1}^1f(x)\ dx&=\displaystyle\intop_{-1}^1(3x+b)\ dx\\ &=\left[\dfrac32x^2+bx\right]_{-1}^1\\ &=\left[\dfrac32(1)^2+b(1)\right]-\left[\dfrac32(-1)^2+b(-1)\right]\\ &=\left[\dfrac32(1)+b\right]-\left[\dfrac32(1)-b\right]\\ &=\left[\dfrac32+b\right]-\left[\dfrac32-b\right]\\ &=\dfrac32+b-\dfrac32+b\\ &=2b \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_{-1}^1\left[f(x)\right]^2\ dx&=\displaystyle\intop_{-1}^1(3x+b)^2\ dx\\ &=\displaystyle\intop_{-1}^1\left(9x^2+6bx+b^2\right)\ dx\\ &=\left[3x^3+3bx^2+b^2x\right]_{-1}^1\\ &=\left[3(1)^3+3b(1)^2+b^2(1)\right]-\left[3(-1)^3+3b(-1)^2+b^2(-1)\right]\\ &=\left[3(1)+3b(1)+b^2\right]-\left[3(-1)+3b(1)-b^2\right]\\ &=\left[3+3b+b^2\right]-\left[-3+3b-b^2\right]\\ &=3+3b+b^2+3-3b+b^2\\ &=2b^2+6 \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_{-1}^1\left[f(x)\right]^3\ dx&=\displaystyle\intop_{-1}^1(3x+b)^3\ dx\\ &=\displaystyle\intop_{-1}^1(3x+b)^3\ \dfrac{d(3x+b)}3\\ &=\dfrac13\left[\dfrac14(3x+b)^4\right]_{-1}^1\\ &=\dfrac1{12}\left[(3(1)+b)^4-(3(-1)+b)^4\right]\\ &=\dfrac1{12}\left[(3+b)^4-(-3+b)^4\right]\\ &=\dfrac1{12}\left((3+b)^2+(-3+b)^2\right)\left((3+b)^2-(-3+b)^2\right)\\ &=\dfrac1{12}\left(9+6b+b^2+9-6b+b^2\right)\left((3+b)+(-3+b)\right)\left((3+b)-(-3+b)\right)\\ &=\dfrac1{12}\left(2b^2+18\right)\left(3+b-3+b\right)\left(3+b+3-b\right)\\ &=\dfrac1{12}\left(2b^2+18\right)\left(2b\right)\left(6\right)\\ &=2b^3+18b \end{aligned}\)

Misal b^2=p
\(\begin{aligned} {U_2}^2&=U_1\cdot U_3\\ \left(2b^2+6\right)^2&=(2b)\left(2b^3+18b\right)\\ 4b^4+24b^2+36&=4b^4+36b^2\\ 36&=12b^2\\ b^2&=\boxed{\boxed{3}} \end{aligned}\)

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
   

\(\begin{aligned} \displaystyle\intop_0^1\dfrac1{1-x}dx&=\displaystyle\intop_0^1\dfrac{1-x+x}{1-x}dx\\ &=\displaystyle\intop_0^1\left(1+\dfrac{x}{1-x}\right)dx \end{aligned}\)

CARA BIASA

\(\begin{aligned} \displaystyle\intop_0^1\left(1+\dfrac{x}{1-x}\right)dx&=\displaystyle\intop_0^11\ dx+\displaystyle\intop_0^1\dfrac{x}{1-x}dx\\ &=[x]_0^1+b\\ &=1-0+b\\ &=\boxed{\boxed{1+b}} \end{aligned}\)

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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