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Double Integral with base e |
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| Nov20-09, 12:08 PM | #1 |
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Double Integral with base e
[tex]
\int_0^1\int_0^y e^{x^2} dx dy [/tex] 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 [/tex] integrating gives ex2y then substituting the values for y gives [tex] \int_0^1 e^{x^2} - e^{x^2}x dx [/tex] Am I integrating over the wrong bounds? I know if 0 < y < x, it would work. |
| Nov20-09, 12:13 PM | #2 |
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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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