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Thread: cal 2

  1. #1
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    cal 2

    intergral x(x^2-9)^3 dx
    intergral with limits 3/2 and 0 xsin(x^2)dx give answer in decimal number.
    intergral with limits 4 and -2 x^2(x^3 + 8)^2dx. give answer in decimal number.
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  2. #2
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    Quote Originally Posted by lay137 View Post
    intergral x(x^2-9)^3 dx
    $\displaystyle \int x(x^2-9)^3 dx$
    Use $\displaystyle u=x^2-9$.
    Then you get,
    $\displaystyle \frac{1}{2} \int u^3 du$
    $\displaystyle \frac{1}{2} \cdot \frac{1}{4}u^4 +C$
    $\displaystyle \frac{1}{8}u^4+C$
    $\displaystyle \frac{1}{8}(x^2-9)^4+C$

    intergral with limits 3/2 and 0 xsin(x^2)dx give answer in decimal number.
    $\displaystyle \int_0^{3/2} x\sin x^2 dx$
    Let $\displaystyle u=x^2$ then,
    $\displaystyle \frac{1}{2}\int_0^{9/4} \sin u du$
    $\displaystyle \frac{1}{2} \left( -\cos u \big|_0^{9/4} \right)$
    You can evaluate that.
    intergral with limits 4 and -2 x^2(x^3 + 8)^2dx. give answer in decimal number.
    $\displaystyle \int_{-2}^4 -2 x^2(x^3+8)^2 dx$
    Let $\displaystyle u=x^3+8$ then,
    $\displaystyle \frac{1}{3}\int_0^72 -2 u^2 du$
    $\displaystyle \frac{1}{3} \left( \frac{-2}{3} u^3 \big|_0^{72} \right) $
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