Eka_Oke: Soal dan Pembahasan Integral

Soal-soal ini ditujukan untuk murid yang request sedang belajar menjelang UN, karena mendadak, jadi masih sedikit soal dan pembahasannya. Kalo ada soal yang masih belum jelas, boleh langsung di comment atau kirim email…okey!

1. $\int\frac{x^3-1}{\sqrt x}dx$ adalah…

a. $\frac 27.\sqrt x(x^3-7) +c$

b. $\frac 27.\sqrt x(x^3+7) +c$

c. $\frac 17.\sqrt x(x^3+7) +c$

d. $\frac 17.\sqrt x(x^3-7) +c$

e. $\frac 27.\sqrt x(x^3+1) +c$

jawab:

$\int\frac{x^3-1}{\sqrt x}dx$ karena penyebut satu suku,maka pisahkan fungsi pembilangnya

$\dpi{100} \bg_white \begin{align*}\int\frac{x^3-1}{\sqrt x}dx & = &\int(\frac{x^3}{\sqrt x} - \frac{1}{\sqrt x})dx\\ & = &\int \frac{x^3}{x^{\frac 12}}dx - \int\frac{1}{x^{\frac 12}}dx\\ & = &\int x^{\frac 52}dx - \int x^{-\frac 12}dx\\ & = &\frac{1}{\frac 52+1}.x^{\frac 52+1} - \frac{1}{-\frac 12+1}.x^{-\frac 12+1}+c\\ & = &\frac{1}{\frac 72}.x^{\frac 72} - \frac{1}{\frac 12}.x^{\frac 12}+c\\ & = &\frac 27.x^{\frac 72} - 2x^{\frac 12}+c\\ & = &\frac 27.x^3.\sqrt x-2.\sqrt x+c\\ & = &\frac{2}{7}\sqrt x (x^3-7)+c\end{align*}$

2. $\intop_{1}^{2}(4x^3+3x^2+2x+1)dx$ = …

a. 10

b. 16

c. 20

d. 26

e. 35

jawab:

$\dpi{100} \bg_white \begin{align*}\intop_{1}^{2}(4x^3+3x^2+2x+1)dx & = &\left[\frac 44.x^4+\frac 33.x^3+\frac 22.x^2+x\right]_{1}^{2}\\ & = &\left[x^4+x^3+x^2+x\right]_{1}^{2}\\ & = &(2^4+2^3+2^2+2)-(1^4+1^3+1^2+1)\\ & = &30-4\\ & = &26\end{align*}$

3. $\intop_{\frac{\pi}{3}}^{\frac{\pi}{2}}(cosx-sinx)dx$ = …

a. $\frac 12(3-\sqrt 2)$

b. $\frac 12(3+\sqrt 2)$

c. $\frac 12(3-\sqrt 3)$

d. $\frac 12(1+\sqrt 3)$

e. $\frac 12(1-\sqrt 3)$

jawab :

$\dpi{100} \bg_white \begin{align*}\intop_{\frac{\pi}{3}}^{\frac{\pi}{2}}(cos\;x-sin\;x)dx & = &\left[sin\;x-(-cos\;x)\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}\\ & = &\left[sin\;x+cos\;x\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}\\ & = &(sin(\frac{\pi}{2})+cos(\frac{\pi}{2}))-(sin(\frac{\pi}{3}+cos(\frac{\pi}{3}))\\ & = &(1+0)-(\frac 12\sqrt 3+\frac 12)\\ & = &\frac 12-\frac 12\sqrt 3\\ & = &\frac 12(1-\sqrt 3)\end{align*}$

4. Volume benda putar yang terjadi jika daerah pada kuadran pertama yang dibatasi kurva $y=1-\frac{x^2}{4}$ , sumbu x, sumbu y, dan diputar mengelilingi sumbu x adalah…

a. $\frac{52}{15}\pi$

b. $\frac{16}{12}\pi$

c. $\frac{16}{15}\pi$

d. $\frac{32}{15}\pi$

e. $\frac{12}{15}\pi$

jawab :
* mencari batas kurva
* untuk kurva $y_1=1-\frac{x^2}{4}$
* sumbu $x$ maka $y_2=0$
$\begin{array}{rcl} y_1 & = &y_2\\1-\frac{x^2}{4} & = &0\:kalikan\:4\\4-x^2 & = &0\\(2-x)(2+x) & = &0\\x_1=2 & V&x_2=-2\end{array}$

$Volume=\pi\intop_{-2}^{2}(1-\frac{x^2}{4})^2dx$ ingat $(a+b)^2=(a^2-2ab+b^2)$

$\dpi{100} \bg_white \begin{align*} Volume & = &\pi\intop_{-2}^{2}(1-\frac 14.x^2)^2dx\\ & = &\pi\intop_{-2}^{2}(1-\frac 12.x^2+\frac{1}{16}.x^4)dx\\ & = &\pi\left[x-\frac{\frac 12}{3}.x^3+\frac{\frac {1}{16}}{5}.x^5\right]_{-2}^{2}\\ & = &\pi\left[x-\frac 16.x^3+\frac{1}{80}.x^5\right]_{-2}^{2}\\ & = &\pi[(2-\frac 16(2^3)+\frac{1}{80}(2^5)-((-2)-\frac 16(-2)^3+\frac{1}{80}(-2)^5)]\\ & = &\pi[(2-\frac 86+\frac{32}{80})-(-2+\frac 86-\frac{32}{80})]\\ & = &\frac{32}{15}\pi\:\:satuan\:volume\end{align*}$

Eka_Oke

Minggu, 15 April 2012

Soal dan Pembahasan Integral

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