Help with Double Integral?

1. Mar 14, 2015

RJLiberator

1. The problem statement, all variables and given/known data

Double integral of y*e^(x^4-1)
with bounds
0=<y=<1
y^(2/3)=<x=<1

2. Relevant equations

3. The attempt at a solution

Well, the first key thing to recognize is that we need the correct order for the bounds to compute this double integral.
So I switch it from x=y^(2/3) and x=1 TO y=x^(3/2) and y=1
and x=0 to x=1 becomes the x boundaries.

So now I integrate with respects to the y boundary first as that is the only way to solve this problem.
I get y^2/2*e^(x^4-1) from y=1 to y=x^(3/2)

This then becomes the integral from x=0 to x=1 of 1/2(e^(x^4-1)-x^3*e^(x^4-1))dx

And I am clueless on how to solve this. I've been trying to do u-substitution for a while now knowing that letting u = x^4-1 and du=x^3 I can work with something.

Did I do something wrong in my previous steps?

2. Mar 14, 2015

SammyS

Staff Emeritus
Graph the region over which the integration takes place.

If $\displaystyle\ x > y^{2/3}\,,\$ then $\displaystyle\ y < x^{3/2}\$ .

So your limits of integration are not correct for integral with the order of integration reversed.

3. Mar 14, 2015

RJLiberator

I believe I did, where it is 'close' to a x^2 parabola and the region is beneath x^(3/2) from x=0 to x=1 and from y=0 to y=x^3/2

No?

4. Mar 14, 2015

SammyS

Staff Emeritus
Right.

y goes from 0 to x3/2..

What is $\displaystyle\ \int_0^{x^{3/2}} y\, dy\$ ?

5. Mar 14, 2015

RJLiberator

Well, that is just y^2/2 evaluated at the bounds so
x^(3) is what it is.

However, ah........
So you are saying my bounds for y in the initial setup were messed up.

I had them going from y=x^(3/2) to y=1 and you are claiming that they go from y=0 to y=x^(3/2)

Hm. Let me check to see if this makes sense. I'm not sure how this makes sense.
Original bounds have y=0 to y=1. And x=y^(2/3) to x=1.
Switching them results in x=0 to x=1 and y=x^(3/2) to y=1, no?

Last edited: Mar 14, 2015
6. Mar 14, 2015

RUber

You should integrate from y=0. Then that -1 will go away.

7. Mar 14, 2015

RJLiberator

Why can I integrate from y=0 to y=x^(3/2) instead of y=x^(3/2) to y=1.

How can I do this?

8. Mar 14, 2015

SammyS

Staff Emeritus
Look at the following.

Let me repeat, graph the region of integration.

9. Mar 14, 2015

RJLiberator

BINGO.
That makes sense.
In my original interpretation I was taking the incorrect region. This means I need to use y=0 to y=x^(3/2) as the correct region. BINGO. Now let's see how I handle this.

10. Mar 14, 2015