# Double Integral with base e

by r_swayze
Tags: base, double, integral
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 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 $$\int_0^1\int_x^1 e^{x^2} dy dx$$ integrating gives ex2y then substituting the values for y gives $$\int_0^1 e^{x^2} - e^{x^2}x dx$$ Am I integrating over the wrong bounds? I know if 0 < y < x, it would work. Attached Thumbnails
 HW Helper Thanks PF Gold P: 7,659 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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