Double Integral with base e

  1. [tex]
    \int_0^1\int_0^y e^{x^2} dx dy

    The region I am integrating over should look like this graph, right?

    I tried switching the bounds but I am left where what I started.

    since 0 < x < y, and 0 < y < 1

    I can switch to 0 < x < 1 , and x < y < 1

    leaving me with the integral [tex]
    \int_0^1\int_x^1 e^{x^2} dy dx

    integrating gives ex2y

    then substituting the values for y gives [tex]
    \int_0^1 e^{x^2} - e^{x^2}x dx

    Am I integrating over the wrong bounds? I know if 0 < y < x, it would work.

    Attached Files:

  2. jcsd
  3. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    Everything looks correct and you reversed the limits nicely. I suspect a typo in your textbook. (Of course you can do the second integral but that doesn't help).
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