Exercise Zone : Integral Tentu

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

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No. 1

Jika \displaystyle\intop_0^3f(x)\ dx=-5 dan \displaystyle\intop_0^7f(x)\ dx=8 maka \displaystyle\intop_3^7f(x)\ dx= ....
  1. 3
  2. 13
  3. -8
  1. 5
  2. 12
\(\begin{aligned} \displaystyle\intop_0^7f(x)\ dx&=\displaystyle\intop_0^3f(x)\ dx+\displaystyle\intop_3^7f(x)\ dx\\ 8&=-5+\displaystyle\intop_3^7f(x)\ dx\\ \displaystyle\intop_3^7f(x)\ dx&=8+5\\ &=\boxed{\boxed{13}} \end{aligned}\)

No. 2

Jika \displaystyle\intop_0^4f(x)\ dx=5 dan \displaystyle\intop_4^{10}f(x)\ dx=10 maka 2\displaystyle\intop_0^{10}f(x)\ dx= ....
  1. 30
  2. 32
  3. 34
  1. 36
  2. 38
\(\begin{aligned} \displaystyle\intop_0^{10}f(x)\ dx&=\displaystyle\intop_0^4f(x)\ dx+\displaystyle\intop_4^{10}f(x)\ dx\\ &=5+10\\ &=15\\ 2\displaystyle\intop_0^{10}f(x)\ dx&=2(15)\\ &=\boxed{\boxed{30}} \end{aligned}\)

No. 3

Jika \displaystyle\intop_0^22f(x)\ dx=4 dan \displaystyle\intop_4^23f(x)\ dx=12 maka nilai \displaystyle\intop_0^4f(x)\ dx=
  1. -2
  2. -3
  3. -4
  1. -5
  2. -6
\(\begin{aligned} \displaystyle\intop_0^22f(x)\ dx&=4\\ 2\displaystyle\intop_0^2f(x)\ dx&=4\\ \displaystyle\intop_0^2f(x)\ dx&=2 \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_4^23f(x)\ dx&=12\\ 3\displaystyle\intop_4^2f(x)\ dx&=12\\ \displaystyle\intop_4^2f(x)\ dx&=4\\ \displaystyle\intop_2^4f(x)\ dx&=-4 \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_0^4f(x)\ dx&=\displaystyle\intop_0^2f(x)\ dx+\displaystyle\intop_2^4f(x)\ dx\\ &=2+(-4)\\ &=\boxed{\boxed{-2}} \end{aligned}\)

No. 4

Nilai dari \displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx adalah
\(\begin{aligned} \displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx&=\displaystyle\intop_1^4\left(2x^{\frac12}-\dfrac3{x^{\frac12}}-1\right)dx\\ &=\displaystyle\intop_1^4\left(2x^{\frac12}-\dfrac3x^{-\frac12}-1\right)dx\\ &=\left[2\cdot\dfrac23x^{\frac32}-3\cdot2x^{\frac12}-x\right]_1^4\\ &=\left[\dfrac43x\sqrt{x}-6\sqrt{x}-x\right]_1^4\\ &=\left[\dfrac43(4)\sqrt4-6\sqrt4-4\right]-\left[\dfrac43(1)\sqrt1-6\sqrt1-1\right]\\ &=\left[\dfrac{16}3(2)-6(2)-4\right]-\left[\dfrac43(1)-6(1)-1\right]\\ &=\left[\dfrac{32}3-12-4\right]-\left[\dfrac43-6-1\right]\\ &=-\dfrac{16}3-\left(-\dfrac{17}3\right)\\ &=-\dfrac{16}3+\dfrac{17}3\\ &=\boxed{\boxed{\dfrac13}} \end{aligned}\)

No. 5

Nilai dari \displaystyle\intop_{-1}^1\left(x^2+2x-3\right)\ dx-\displaystyle\intop_{-1}^1\left(x^2-3x+4\right)\ dx=
  1. 0
  2. 7
  3. 14
  1. -7
  2. -14
\(\begin{aligned} \displaystyle\intop_{-1}^1\left(x^2+2x-3\right)\ dx-\displaystyle\intop_{-1}^1\left(x^2-3x+4\right)\ dx&=\displaystyle\intop_{-1}^1\left(x^2+2x-3-\left(x^2-3x+4\right)\right)\ dx\\ &=\displaystyle\intop_{-1}^1\left(x^2+2x-3-x^2+3x-4\right)\ dx\\ &=\displaystyle\intop_{-1}^1\left(5x-7\right)\ dx\\ &=\left[\dfrac52x^2-7x\right]_{-1}^1\\[10pt] &=\left[\dfrac52(1)^2-7(1)\right]-\left[\dfrac52(-1)^2-7(-1)\right]\\[10pt] &=\left[\dfrac52-7\right]-\left[\dfrac52+7\right]\\[10pt] &=\dfrac52-7-\dfrac52-7\\ &=\boxed{\boxed{-14}} \end{aligned}\)

No. 6

\displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx=
Misal u=x^2+1
\(\begin{aligned} du&=2x\ dx\\ x\ dx&=\dfrac12du \end{aligned}\)

x=-1\rightarrow u=(-1)^2+1=2,
x=2\rightarrow u=2^2+1=5,

\(\begin{aligned} \displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx&=\dfrac12\displaystyle\intop_2^5u\ du\\ &=\dfrac12\left[\dfrac12u^2\right]_2^5\\ &=\dfrac12\left[\left(\dfrac12(5)^2\right)-\left(\dfrac12(2)^2\right)\right]\\ &=\dfrac12\left[\dfrac{25}2-\dfrac42\right]\\ &=\dfrac12\left[\dfrac{21}2\right]\\ &=\boxed{\boxed{\dfrac{21}4}} \end{aligned}\)

No. 7

Jika \displaystyle\intop_1^p3x\left(x+\dfrac23\right)\ dx=78 maka nilai 3p=
  1. 18
  2. 12
  3. 15
  1. 9
  2. 6
\(\eqalign{ \displaystyle\intop_1^p3x\left(x+\dfrac23\right)\ dx&=78\\ \displaystyle\intop_1^p\left(3x^2+2x\right)\ dx&=78\\ \left[x^3+x^2\right]_1^p&=78\\ \left[p^3+p^2\right]-\left[1^3+1^2\right]&=78\\ p^3+p^2-2&=78\\ p^3+p^2-80&=0\\ (p-4)\left(p^2+5p+20\right)&=0\\ p&=4\\ 3p&=3(4)\\ &=\boxed{\boxed{12}}}\)

No. 8

Nilai dari \displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx adalah
\(\begin{aligned} \displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx&=\left[\dfrac13x^3+5x^2+25x\right]_2^3\\ &=\left[\dfrac13(3)^3+5(3)^2+25(3)\right]-\left[\dfrac13(2)^3+5(2)^2+25(2)\right]\\ &=\left[\dfrac13(27)+5(9)+75\right]-\left[\dfrac13(8)+5(4)+50\right]\\ &=\left[9+45+75\right]-\left[\dfrac83+20+50\right]\\ &=129-\dfrac83-70\\ &=59-2\dfrac23\\ &=\boxed{\boxed{56\dfrac13}} \end{aligned}\)

No. 9

\displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx=
  1. \dfrac{3\pi+4}{4\pi}
  2. \dfrac{3\pi+8}{8\pi}
  3. \dfrac{3\pi-4}{4\pi}
  1. \dfrac{3\pi-8}{8\pi}
  2. \dfrac34+\pi
\(\begin{aligned} \displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx&=\displaystyle\intop_{\frac12}^1\left(\sin\pi x+(2x-1)^{\frac13}\right)\ dx\\ &=\left[-\dfrac1{\pi}\cos\pi x+\dfrac12\cdot\dfrac34(2x-1)^{\frac43}\right]_{\frac12}^1\\ &=\left[-\dfrac1{\pi}\cos\pi x+\dfrac38(2x-1)\sqrt[3]{2x-1}\right]_{\frac12}^1\\ &=\left[-\dfrac1{\pi}\cos\pi(1)+\dfrac38(2(1)-1)\sqrt[3]{2(1)-1}\right]-\left[-\dfrac1{\pi}\cos\pi\left(\dfrac12\right)+\dfrac38\left(2\left(\dfrac12\right)-1\right)\sqrt[3]{2\left(\dfrac12\right)-1}\right]\\ &=\left[-\dfrac1{\pi}\cos\pi+\dfrac38(1)\sqrt[3]{1}\right]-\left[-\dfrac1{\pi}\cos\dfrac{\pi}2+\dfrac38\left(0\right)\sqrt[3]{0}\right]\\ &=\left[-\dfrac1{\pi}(-1)+\dfrac38\right]-\left[-\dfrac1{\pi}(0)+0\right]\\ &=\left[\dfrac1{\pi}+\dfrac38\right]-0\\ &=\dfrac{8+3\pi}{8\pi}\\ &=\boxed{\boxed{\dfrac{3\pi+8}{8\pi}}} \end{aligned}\)

No. 10

\displaystyle\intop_2^6\sqrt{x}\ dx= ....
\(\begin{aligned} \displaystyle\intop_9^16\sqrt{x}\ dx&=\displaystyle\intop_9^16x^{\frac12}\ dx\\ &=\left.\dfrac23x^{\frac32}\right|_9^{16}\\ &=\left.\dfrac23x\sqrt{x}\right|_9^{16}\\ &=\dfrac23(16)\sqrt{16}-\dfrac23(9)\sqrt{9}\\[8pt] &=\dfrac{32}3(4)-6(3)\\[8pt] &=\dfrac{128}3-18\\ &=\boxed{\boxed{\dfrac{74}3}} \end{aligned}\)

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