Home Integral Integral Tentu Kalkulus Soal dan Pembahasan Tingkat Dasar

Exercise Zone : Integral Tentu

Image
Idham Muqoddas
Updated: • 0 • min read
Table of Contents
Berikut ini adalah kumpulan soal mengenai Integral Tentu. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

StandarSNBTHOTS


No.

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
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}
Jadi, \(\displaystyle\intop_3^7f(x)\ dx=13\).
JAWAB: B

No.

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
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}
Jadi, \(2\displaystyle\intop_0^{10}f(x)\ dx=30\).
JAWAB: A

No.

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
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}
Jadi, \(\displaystyle\intop_0^4f(x)\ dx=-2\).
JAWAB: A

No.

Nilai dari \(\displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx\) adalah
  1. \(\dfrac13\)
  2. \(\dfrac23\)
  3. 1
  1. \(\dfrac12\)
  2. 3
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}
Jadi, \(\displaystyle\intop_1^4\left(2\sqrt{x}-\dfrac3{\sqrt{x}}-1\right)dx=\dfrac13\).
JAWAB: A

No.

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
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}
Jadi, \(\displaystyle\intop_{-1}^1\left(x^2+2x-3\right)\ dx-\displaystyle\intop_{-1}^1\left(x^2-3x+4\right)\ dx=-14\).
JAWAB: E

No.

\(\displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx=\)
  1. 5
  2. 6
  3. \(\dfrac{21}4\)
  1. \(\dfrac{27}4\)
  2. \(\dfrac{25}4\)
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
Misal u = x2 + 1 \begin{aligned} du&=2x\ dx\\ x\ dx&=\dfrac12du \end{aligned} x = −1 ⟶ u = (−1)2 + 1 = 2,
x = 2 ⟶ u = 22 + 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}
Jadi, \(\displaystyle\intop_{-1}^2x\left(x^2+1\right)\ dx=\dfrac{21}4\).
JAWAB: C

No.

Jika \(\displaystyle\intop_1^p3x\left(x+\dfrac23\right)\ dx=78\) maka nilai 3p =
  1. 18
  2. 12
  3. 15
  1. 9
  2. 6
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\begin{aligned} \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}} \end{aligned}
Jadi, 3p = 12.
JAWAB: B

No.

Nilai dari \(\displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx\) adalah
  1. \(56\dfrac13\)
  2. \(56\dfrac23\)
  3. \(57\dfrac13\)
  1. \(57\dfrac23\)
  2. \(58\dfrac13\)
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}
Jadi, \(\displaystyle\intop_2^3\left(x^2+10x+25\right)\ dx=56\dfrac13\).
JAWAB: A

No.

\[\displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx=\]
  1. \(\dfrac{3\pi+4}{4\pi}\)
  2. \(\dfrac{3\pi+8}{8\pi}\)
  3. \(\dfrac{3\pi-4}{4\pi}\)
  1. \(\dfrac34+\pi\)
  2. \(\dfrac{3\pi-8}{8\pi}\)
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}
Jadi, \(\displaystyle\intop_{\frac12}^1\left(\sin\pi x+\sqrt[3]{2x-1}\right)\ dx=\dfrac{3\pi+8}{8\pi}\).
JAWAB: B

No.

\[\displaystyle\intop_2^6\sqrt{x}\ dx= ....\]
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\begin{aligned} \displaystyle\intop_9^16\sqrt{x}\ dx&=\displaystyle\intop_9^16x^{\frac12}\ dx\\ &=\left.\dfrac23x^{\frac32}\right|_9^{16}\\[3.5pt] &=\left.\dfrac23x\sqrt{x}\right|_9^{16}\\[3.5pt] &=\dfrac23(16)\sqrt{16}-\dfrac23(9)\sqrt{9}\\[3.5pt] &=\dfrac{32}3(4)-6(3)\\[3.5pt] &=\dfrac{128}3-18\\ &=\boxed{\boxed{\dfrac{74}3}} \end{aligned}
Jadi, \(\displaystyle\intop_2^6\sqrt{x}\ dx=\dfrac{74}3\).


Post a Comment