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Math Help - Evaluating a definite integral

  1. #1
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    Evaluating a definite integral

    Hello, I am having trouble with a calculus problem which requires us to evaluate this sum of two integrals:

    \int^0_{-5} x + 2 + \sqrt{4-x}\,dx  +  \int^4_0 2\sqrt{4-x}\,dx

    In the answer key the next step is written as:

    \left[ \frac{x^2}{2} + 2x - \frac{(4-x)^{3/2}}{3/2} \right]_{-5}^0 + 2\left[ \frac{(4-x)^{3/2}}{-3/2} \right]^4_0

    However, I don't understand why the third term in the first integral is being subtracted, and why the denominator in the second integral is negative. If it helps, the original problem was to find the area of the region bounded by x = 4-y^2 and x = y-2 by integrating with respect to x. I set up the integral properly but I don't understand why it is being evaluted like this.
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  2. #2
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    Re: Evaluating a definite integral

    Quote Originally Posted by Ragnarok View Post
    Hello, I am having trouble with a calculus problem which requires us to evaluate this sum of two integrals:

    \int^0_{-5} x + 2 + \sqrt{4-x}\,dx  +  \int^4_0 2\sqrt{4-x}\,dx

    In the answer key the next step is written as:

    \left[ \frac{x^2}{2} + 2x - \frac{(4-x)^{3/2}}{3/2} \right]_{-5}^0 + 2\left[ \frac{(4-x)^{3/2}}{-3/2} \right]^4_0

    However, I don't understand why the third term in the first integral is being subtracted, and why the denominator in the second integral is negative. If it helps, the original problem was to find the area of the region bounded by x = 4-y^2 and x = y-2 by integrating with respect to x. I set up the integral properly but I don't understand why it is being evaluted like this.
    Take the derivative of each. That should show you why.
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  3. #3
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    Re: Evaluating a definite integral

    Ah, okay. I forgot to use u-substitution. Thank you!
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