# Thread: Calculating an integral between two curves.

1. ## Calculating an integral between two curves.

This is multi-part problem, starting with:

Consider the region in the first quadrant that is bounded above by $y=Ax-Ax^2$, and below by $y=3x$. A is the unknown parameter, which allows the above to be true, and should be considered a constant.

We are to write a definite integral for this, leaving the final answer in terms of A.

So, when graphing this, you can see that $y=Ax-Ax^2$ only goes from 0 to 1. I figured that this would mean that our definite integral would go from 0 to 1, but apparently this is wrong. The integral I got was
${\int_{{0}}^{{1}}}{\left[{A}{x}-{A}{{x}}^{{2}}-{3}{x}\right]}{\left.{d}{x}\right.}$
but, I guess the bounds should be from 0 to some number, say z. I understand that the bounds of integration are dependent on A, but I'm not sure how, exactly.

Any help would be great, I've been wracking my brain for a couple days on this problem.

2. Originally Posted by momopeaches
This is multi-part problem, starting with:

Consider the region in the first quadrant that is bounded above by $y=Ax-Ax^2$, and below by $y=3x$. A is the unknown parameter, which allows the above to be true, and should be considered a constant.

We are to write a definite integral for this, leaving the final answer in terms of A.

So, when graphing this, you can see that $y=Ax-Ax^2$ only goes from 0 to 1. I figured that this would mean that our definite integral would go from 0 to 1, but apparently this is wrong. The integral I got was
${\int_{{0}}^{{1}}}{\left[{A}{x}-{A}{{x}}^{{2}}-{3}{x}\right]}{\left.{d}{x}\right.}$
but, I guess the bounds should be from 0 to some number, say z. I understand that the bounds of integration are dependent on A, but I'm not sure how, exactly.

Any help would be great, I've been wracking my brain for a couple days on this problem.
The limits of the integral is given by the intersection points of the two curves. To find that, simply set them equal to each other. So solve $Ax - Ax^2 = 3x$ for $x$ and that will give you your limits.

3. Originally Posted by momopeaches

So, when graphing this, you can see that $y=Ax-Ax^2$ only goes from 0 to 1. I figured that this would mean that our definite integral would go from 0 to 1, but apparently this is wrong.
It is wrong, you need the intergral to go from 0 to a where a is the intersection of $Ax-Ax^2=3x$ other than that you are on the right track!

4. So, solving for x, I get $x=(A-3)/A$. The next part in the question asks to evaluate the integral, leaving it in terms on A. So I would have $((Ax^2)/2)-((Ax^3)/3)-((3x^2)/2)$, just subbing in $(A-3)/A$ for x?

5. Yep, looks good!

6. Awesome, thanks a lot! Been stumped on this one for a while. :]