Soal Integral

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  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

    \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:

    \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 :

    \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 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 $(a+b)^2=(a^2-2ab+b^2)$

    \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*}

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