Question

int_{0}^{1}int_{8y}^{8}e ^{x ^{2}}dxdy

Answer 1

\[\int\limits_{0}^{1}\int\limits_{8y}^{8}e ^{x ^{2}}dxdy\]

Answer 2

The region is the triangle in the first quadrant indicated above.

It's tough to integrate \[e ^{x ^{2}}\] with respect to x (in fact, you can't really do it).

So, your best bet is to change the order of integration from dxdy to dydx.

If you look at the region I drew, the limits on x are 8y and 8 (as given), so x is bound by x=8y from the left and x=8 from above. The point of intersection of the two lines is (8,1), so the new limits on x would be from 0 to 8. Since x=8y, we know that y=(1/8)x, so the limits on y are 0 to (1/8)x. So our new integral is

\[\int\limits_{0}^{8}\int\limits_{0}^{(1/8)x}e ^{x ^{2}}dydx\]

Integrating the first part yields

\[\int\limits_{0}^{8}(1/8)xe ^{x ^{2}}dx\]

If we let
\[u=x ^{2}\]

Then du=2xdx, xdx=(1/2)du, and (1/8)xdx=(1/16)du

Also note that for x=0, u=0 and for x=8, u=64.

So our integral becomes

\[\int\limits\limits_{0}^{64}(1/16)e ^{u}du\]

So we get

\[(1/16)e ^{u}\]

and evaluate from 0 to 64, yielding

\[(1/16)(e ^{64}-1)\]

Hope this makes sense.