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Math Help - Integral

  1. #1
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    Integral

    I can not seem to get the correct answer on the following integral.

    Integral from (1-x) up to 2 of (2x+2-y)/(4) dy

    Please show the steps so i can see what I am doing wrong. Thanks
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  2. #2
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    Quote Originally Posted by PensFan10 View Post
    I can not seem to get the correct answer on the following integral.

    Integral from (1-x) up to 2 of (2x+2-y)/(4) dy

    Please show the steps so i can see what I am doing wrong. Thanks
    It's interesting that your integrand has both x and y. Is this what you really want

    \frac{1}{4} \int_{1-x}^2 2 x + 2 - y\; dy\;\;?
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  3. #3
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    Now that's what I call steps (must be a dance teacher)
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    Yes, that is what I want. The answer the book gives is (5x^2+6x-y)/8

    It contains both x and y because its a double integral. I cant get the first part.
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  5. #5
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    Quote Originally Posted by PensFan10 View Post
    Yes, that is what I want. The answer the book gives is (5x^2+6x-y)/8

    It contains both x and y because its a double integral. I cant get the first part.
    Can you give the whole problem (i.e as in the double integral)?
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  6. #6
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    Yeah here it is:

    i will use S to denote integral

    S(from 0 to 1)S(from (1-x) to 2) of (2x+2-y)/(4) dydx
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    Quote Originally Posted by PensFan10 View Post
    Yeah here it is:

    i will use S to denote integral

    S(from 0 to 1)S(from (1-x) to 2) of (2x+2-y)/(4) dy dx
    \frac{1}{4}\int_0^1 \int 2x+2-y \,dy\,dx = \frac{1}{4} \int_0^1 2(x+1)y - \left. \frac{y^2}{2} \right|_{1-x}^2 dx
    = \frac{1}{4} \int_0^1 \left( 4(x+1) - 2\right) - \left( 2(1+x)(1-x) - \frac{(1-x)^2}{2}\right)dx = \frac{1}{8}\int_0^1 5x^2 + 6x + 1\,dx
    Did you get this?
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  8. #8
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    Thats the correct answer, but not what i got. Not sure where I was making my mistake. Thanks
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