Double Intergral with Substitution

[BenMcC]

I have a problem with Double Integral that I can not seem to get correct.

4 2
∫ ∫ e^(y^2)dydx
0 (x/2)

The answer is (e^4)-1, but I can't seem to get the Substitution at all right. I have literally spent hours on this problem. Any help would be greatly appreciated, its also due by 3:30, so I am kind of limited by time. Thanks

 

[mfb]

3:30 in which time zone? ;)
Here at me, you asked your question at 5:09 PM.

Converted in a format that is easier to read:
$$\int_0^4 \left(\int_{x/2}^2 e^{y^2} dy\right)dx$$
Did you draw a sketch of the integration area (in the x,y-plane)? The integral is equivalent to
$$\int_0^2 \left(\int_{0}^{2y} e^{y^2} dx\right)dy$$
There, both integrals are easy to evaluate in that order.

 

[BenMcC]

I have 4 hours, so I have time. I'm just really confused how to do the initial integral. I tried u substitution of u*dv=uv-∫v*du, and I can't get it to come out quite right

 

[mfb]

You cannot find an antiderivative of $e^{y^2}$. It is nice to try, but you won't get a result unless you just define it (but that does not help here).
The change of the integration order (or something equivalent) is the key point here. It allows to evaluate one integral and simplifies the other one.

 

[BenMcC]

It's possible. It says to change the order of integration, and I have no idea how to set that up

 

[Vorde]

As you may or may not see, evaluating it with regards to x first will let you solve the whole thing. The trick is just the figure out the new bounds on the integrals - it's not as simple as just switching the order.

What I would do is the draw a graph and draw the lines x=4, x=0, y=2 and y=x/2.
Then think about what the bounds would be if you were looking at it 'sideways'.

EDIT: Just realized mfb gave you the integral already, so you don't need what I said.