1. ## Bounds of integration

What would be my bounds of integration if I'm trying to integrate a function f(x,y) over a region bounded by y = x and y = -x2 ?

Giest

2. ## Re: Bounds of integration

see the graph and find them your self

3. ## Re: Bounds of integration

It's not quite simple as that. The multivariate function that I'm integrating requires two iterated integrals, and the bounds for the inner integrand, I'm pretty sure, need to expressed as a function of one of the two variables.

4. ## Re: Bounds of integration

If you have a double integral like this
$\int^a_b (\int^c_d f(x,y) dy)dx$

The limits c and d on the y integral are c=x and d=-x2
The limits a and b on the x integral are a=0 and b=-1 as minoanman shows in his graph

You may wonder why I said c=x and d=-x2 not d=x and c=-x2. x is always higher than -x2 and you want to integrate in the positive direction of y so the integral limits are FROM the lower value TO the higher value.

5. ## Re: Bounds of integration

Of course, using a double integral here is overkill when you can you can just do \displaystyle \begin{align*} A = \int_0^1{x - \left( -x^2 \right) \, dx} = \int_0^1{x + x^2 \, dx} \end{align*}

6. ## Re: Bounds of integration

Oh, I wish I could go back to those good days of single variable calculus. However, my problem is actually a line integral.

and C is the boundary of the region in the third quadrant bounded by y=x and y=-x^2 (traversed counterclockwise).

But, regardless, I appreciate the gesture.

7. ## Re: Bounds of integration

Originally Posted by Giestforlife
Oh, I wish I could go back to those good days of single variable calculus. However, my problem is actually a line integral.

and C is the boundary of the region in the third quadrant bounded by y=x and y=-x^2 (traversed counterclockwise).

But, regardless, I appreciate the gesture.
And this is why one should post the ENTIRE problem they are working on, so as not to waste the helpers' time...

8. ## Re: Bounds of integration

Originally Posted by Prove It
Of course, using a double integral here is overkill when you can you can just do \displaystyle \begin{align*} A = \int_0^1{x - \left( -x^2 \right) \, dx} = \int_0^1{x + x^2 \, dx} \end{align*}
That doesn't make sense because you get the same answer no matter what f(x,y) is

9. ## Re: Bounds of integration

Originally Posted by Shakarri
That doesn't make sense because you get the same answer no matter what f(x,y) is
This answer WOULD have been sufficient if the OP had posted the entire problem in the first place.

10. ## Re: Bounds of integration

I didn't mean to waste anyone's time. I just thought I'd keep things simple by giving the part of the problem that was putting me off--that and the fact it takes me a long time to do anything in LaTex. But I'll keep that in mind next time.