# Calculate the double integral : int int xye^((x^2)(y))

1. Feb 7, 2013

### Rancy

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

Calculate the double integral

int int xye^((x^2)(y)) , 0<= x <= 1 , 0<= y <= 2

2. Relevant equations

Integral by parts

uv - int vdu

3. The attempt at a solution

The answer in the back of the book is (1/2)((e^2) -3) , but I get (1/2)((e^2) -1) .

I think I made a positive/negative sign error, but I can't find it. I've had similar encounters where I would get close to the answer for questions involving integration by parts. I might of made a consistent error in one of my lines for each other question involving integration by parts, but I don't know where.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 7, 2013

### clamtrox

Your error is in the first step

3. Feb 7, 2013

### Karnage1993

You have a $dx$ on the first line. That means you integrate with respect to x and hold y constant.

$\int\limits_{0}^1 xye^{x^2y} \ dx = y\int\limits_{0}^1 xe^{x^2y}\ dx$

Does this make it clearer?

4. Feb 7, 2013

### Rancy

Not really, I already knew that I should think of y as a constant when integrating with respect to x, but I rarely ever factor it out and just imagine that y is a number, an integrate the function.

But I think I know where I made my mistake:
The derivative of e^((x^2)(y)) = (2xy)(e^((x^2)(y))) , and I forgot to take out the y in my answer.

Thanks!

5. Feb 7, 2013

### HallsofIvy

That is, by the way, the hard way to do this problem. Change the order of integration:
$$\int_{y=0}^2\int_{x=0}^1 xye^{x^2y}dx dy$$
Let $u=x^2y$ so that $du= 2xydy$ and so $(1/2)du= xydy$. That simplifies the problem a lot!