Double Integral of Exponential Function with Changing Bounds

[RJLiberator]

Homework Statement

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

Homework Equations

The Attempt at a Solution

[/B]
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?

 

[SammyS]

Homework Statement

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

Homework Equations

The Attempt at a Solution

[/B]
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?

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.

 

[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?

 

[SammyS]

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?

Right.

y goes from 0 to x^3/2..

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

 

[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?

 

[RUber]

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

 

[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?

 

[SammyS]

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?

Look at the following.

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.

Let me repeat, graph the region of integration.

 

[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.

 

[RJLiberator]

My answer becomes 1/8-1/8e
and I am fairly confident that this is correct thanks to you guys showing me my wrong interpretation of the region. Thank you.