# Double Integral with Strange Limits of Integration

## Homework Statement

For the double integral ∫[0,1]∫[0,x^3] e^(y/x) dxdy
(a) sketch the region of integration
(b) evaluate the integral and
(c) re-express the integral with the order of integration reversed

None

## The Attempt at a Solution

The problem is that I've never seen a double integral problem with the limits of integration with respect to x in terms of a function of x, not y. I couldn't find any examples online or in my book where this is the case. The problem is written such that your boundaries should be 0≤x≤x^3 and 0≤y≤1... but x=x^3 doesn't make any sense to me as an upper boundary for x.

I know how to do most double integral problems, reverse the order of integration, etc. but this has me stumped. Am I right in thinking this might be a typo? or is there some way to make sense of the region that I'm just not seeing?

## Answers and Replies

DryRun
Gold Member
Hi bossman27 and welcome to PF!

Using LaTeX, here is how your double integral appears:
$$\int^1_0 \int^{x^3}_0 e^{(y/x)}\,.dxdy$$
But it should have been:
$$\int^{x=1}_{x=0} \int^{y=x^3}_{y=0} e^{(y/x)}\,.dydx$$

Mark44
Mentor

## Homework Statement

For the double integral ∫[0,1]∫[0,x^3] e^(y/x) dxdy
(a) sketch the region of integration
(b) evaluate the integral and
(c) re-express the integral with the order of integration reversed

None

## The Attempt at a Solution

The problem is that I've never seen a double integral problem with the limits of integration with respect to x in terms of a function of x, not y. I couldn't find any examples online or in my book where this is the case. The problem is written such that your boundaries should be 0≤x≤x^3 and 0≤y≤1... but x=x^3 doesn't make any sense to me as an upper boundary for x.

I know how to do most double integral problems, reverse the order of integration, etc. but this has me stumped. Am I right in thinking this might be a typo? or is there some way to make sense of the region that I'm just not seeing?

0 <= x <= x3 doesn't make sense to me either. That should be 0 <= y <= x3, and 0 <=x <= 1.

Are you sure you don't have dx and dy switched?
This would make more sense.
$$\int_{x = 0}^1\int_{y = 0}^{x^3}e^{y/x}dy~dx$$

0 <= x <= x3

Are you sure you don't have dx and dy switched?
This would make more sense.
$$\int_{x = 0}^1\int_{y = 0}^{x^3}e^{y/x}dy~dx$$

I'm sure I did not switch them, that was exactly how the problem appears on this practice test... I was thinking that it was probably a typo as well, but as the final is tomorrow morning I wanted to make sure I wasn't missing something.

Thanks to both of you for the quick responses!

It's entirely possible that you are essentially being asked for the anti-derivative, so you should think of the bound for the integral as a parameter, and the internal integral as being taken with respect to some dummy variable $x^\prime$.

Mark44
Mentor
It's entirely possible that you are essentially being asked for the anti-derivative, so you should think of the bound for the integral as a parameter, and the internal integral as being taken with respect to some dummy variable $x^\prime$.
That's an interesting idea, but I don't think it's applicable here. bossman27 said that the integral was exactly as it appeared on the practice test. I believe that the instructor switched the order of dx and dy by mistake.