1. ## Double Integral

Evaluate the following integral:

$\int^{1}_{0} \int^{\frac{1}{2}(1-y)}_{0} e^{x-x^2}dxdy$

I'm thinking the only way I can get through this one is if I switch the order in which we integrate. To do this, I'm going to need to fix the bounds on the integrals.

Firstly,

$x = \frac{1}{2}(1-y)$

Then,

$y = 1-2x$

and I'm not sure what I should do about the bottom bound of the inner integral, i.e.

$x=0$

The bounds on the outter integral should stay the same I think.

Can somebody clarify my confusions before I continue?

Thanks again!

2. Draw a picture! Your region of integration is a triangle. The inner integral you should think of as going from one function to another, and the outer integral you should think of as going from one number to another. What do you get?

3. Originally Posted by Ackbeet
Draw a picture! Your region of integration is a triangle. The inner integral you should think of as going from one function to another, and the outer integral you should think of as going from one number to another. What do you get?
Okay I see the triangle now, and for this integral we are creating horizontal strips of area where each horizontal strip starts at x=0 and ends at x = 1/2 - y/2. We then sum up all these horizontal strips across the range of y values from 0 to 1.

I still don't see how I'm suppose to integrate e^(x-x^2).

4. Oops. I see I didn't make myself clear. Your idea of switching the order of integration is essential to solving this problem. I'm merely trying to help you see what the new limits should be. Before you switch, you're integrating first from the function x = 0 to x = (1-y)/2, and then from the number y = 0 to y = 1.

After you switch, you're integrating first from the function y = ? to y = ?, and then from the number x = ? to x = ?

Fill in the blanks using your triangle.

5. Originally Posted by Ackbeet
Oops. I see I didn't make myself clear. Your idea of switching the order of integration is essential to solving this problem. I'm merely trying to help you see what the new limits should be. Before you switch, you're integrating first from the function x = 0 to x = (1-y)/2, and then from the number y = 0 to y = 1.

After you switch, you're integrating first from the function y = ? to y = ?, and then from the number x = ? to x = ?

Fill in the blanks using your triangle.
Okay so then I'd be using vertical strips of area where each strip starts at y = 0 and ends at y = 1 - 2x. We will then add these strips of area up across the range of x values 0 to 1/2.

That should be the whole problem.

Thanks, I didn't think of actually drawing the region and thinking.

6. On these sorts of problems, I would highly recommend that you draw the region. I always have to do that, or I get it wrong.

Everything looks good. And you'll see, I think, that something cool happens after you perform the first integration.

7. Funnily enough, I end up confusing myself if I draw the function - it used to infuriate my lecturers hahaha. I manipulate inequalities instead.

At the moment you have $\displaystyle 0 \leq x \leq \frac{1}{2}(1 - y)$ and $\displaystyle 0 \leq y \leq 1$.

From the first inequality:

$\displaystyle 2x \leq 1 - y$

$\displaystyle y \leq 1 - 2x$.

So that means one of your new regions of integration is $\displaystyle 0 \leq y \leq 1 - 2x$.

You also know

$\displaystyle 0 \leq y$

$\displaystyle -y \leq 0$

$\displaystyle 1-y \leq 1$

$\displaystyle \frac{1}{2}(1-y) \leq \frac{1}{2}$.

From the second inequality, we know $\displaystyle 0 \leq x \leq \frac{1}{2}(1 - y)$, so therefore your second region of integration is

$\displaystyle 0 \leq x \leq \frac{1}{2}$.

Therefore, reversing the order of integration gives:

$\displaystyle \int_{0}^{1}{\int_{0}^{\frac{1}{2}(1 - y)}{e^{x-x^2}\,dx}\,dy}=\int_{0}^{\frac{1}{2}}{\int_{0}^{1-2x}{e^{x-x^2}\,dy}\,dx}$.