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Thread: Definite Integral Problem-Work needs checked

  1. #1
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    Definite Integral Problem-Work needs checked

    $\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

    $\displaystyle \int (x^2+e^x-6)dx$

    $\displaystyle \frac{x^3}{3}+e^x-6x+c$

    $\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

    $\displaystyle \frac{x^3}{3}+e^x-6x\int_{x=0}^{x=1}$

    $\displaystyle \frac{1^3}{3}+e^1-6(1)$

    $\displaystyle =-5\frac{2}{3}+e$

    $\displaystyle \frac{0^3}{3}+e^0-6(0)$

    $\displaystyle =e$

    $\displaystyle -5\frac{2}{3}+e-e$

    $\displaystyle =-5\frac{2}{3}$
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Jim Marnell View Post
    $\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

    $\displaystyle \int (x^2+e^x-6)dx$

    $\displaystyle \frac{x^3}{3}+e^x-6x+c$

    $\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

    $\displaystyle \frac{x^3}{3}+e^x-6x\int_{x=0}^{x=1}$

    $\displaystyle \frac{1^3}{3}+e^1-6(1)$

    $\displaystyle =-5\frac{2}{3}+e$

    $\displaystyle \frac{0^3}{3}+e^0-6(0)$

    $\displaystyle =e$

    $\displaystyle -5\frac{2}{3}+e-e$

    $\displaystyle =-5\frac{2}{3}$

    $\displaystyle e^0 \neq e$
    $\displaystyle
    e^0 = 1$
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