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Math Help - Area under the shaded region answer check please?

  1. #1
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    Area under the shaded region answer check please?

    I solved that the integral for this is Integral -1 to 1 (y^2 -3 -e^y) dy

    antiderivative of that is y^3/3 -3y -e^y from -1 to 1


    plug in those intervals, you get

    (1^3/3-3(1)-e^1) - (-1^3 -3(-1)-e^-1)

    What would be the final answer of that? I'm terrible with calculating the end result, but I understand the problem as a whole though. Thanks!
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  2. #2
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    Re: Area under the shaded region answer check please?

    Hello, Steelers72!

    I solved: . \displaystyle \int^1_{\text{-}1}\left(y^2 - 3 - e^y)\,dy

    and got: . \tfrac{1}{3}y^3 -3y - e^y\,\bigg]^1_{\text{-}1}

    Plug in those limits and get: . \left[\tfrac{1}{3}(1^3) - 3(1)- e^1\right] - \left[\tfrac{1}{3}(\text{-}1)^3 -3(\text{-}1)-e^{\text{-}1}\right]

    What would be the final answer of that? . . . . . . . . . . LOL!
    I'm terrible with calculating the end result, . . . . . . . . Your calculus is excellent,
    but I understand the problem as a whole though. . . . but you suck at arithmetic ??

    We have: . \left(\tfrac{1}{3} - 3 - e\right) - \left(\text{-}\tfrac{1}{3} + 3 - \tfrac{1}{e}\right)

    . . . . . . =\;\tfrac{1}{3} - 3 - e  + \tfrac{1}{3} - 3 + \tfrac{1}{e}\right)

    . . . . . . =\;\tfrac{2}{3} - 6 + \tfrac{1}{e} - e

    . . . . . . =\;-\frac{16}{3} + \frac{1-e^2}{e}
    Thanks from Steelers72
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  3. #3
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    Re: Area under the shaded region answer check please?

    Haha yes I absolutely suck at doing that type of stuff without a calculator; no lies. I appreciate your help so much!!
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