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Math Help - integrate e^(2+y) 0<= y <= ln2

  1. #1
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    integrate e^(2+y) 0<= y <= ln2

    how to integrate e^(2+y) dy where 0<= y <= ln2?

    Thanks
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  2. #2
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    integrate e^(2+y) dy 0<= y <= ln2

    Note: the real problem is integrating e^(x + y) dydx where 0<= y <= ln2 and 0 <= x <= ln3. My solution manual says that the inner integral is e^x. I just can't see how you get e^x from integral e^(x + y) dy = e^x with the given limits (0 <= y <= ln2). I replaced x with 2 since x is constant in the inner integral on my original post.
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  3. #3
    Super Member Deadstar's Avatar
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    e^{2+y} = e^2 e^y.

    So you have...

    \int_0^{\ln(2)} e^{2+y} dy = e^2 \int_0^{\ln(2)} e^y dy = e^2 \left [e^y \right]_0^{\ln(2)} dy

    = e^2 (e^{\ln(2)} - e^0) = e^2 (2 - 1) = e^2.
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  4. #4
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    integrate e^(2+y) dy 0<= y <= ln2

    Thanks. I get it now. The mistake I was making was this:

    e^(2 + y) = e^2 + e^y instead of e^2e^y. Now it works out.

    Many thanks.
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