Double Integral with base e

by r_swayze
Tags: base, double, integral
r_swayze is offline
Nov20-09, 12:08 PM
P: 66
\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.
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LCKurtz is offline
Nov20-09, 12:13 PM
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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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