I have a question here that is asking for the area of two curves to be expressed in terms of one or more integrals with respect to x, and then the same for y. The two curves are \(\displaystyle y = 9x\) and \(\displaystyle y = x^2-10\), and when I set them equal to each other and solve for x I get upper and lower bounds that sort of look on the right track but are "flipped". All of the documentation I've been able to find regarding area between two curves tells me that the x-values (roots of the quadratic) are the upper/lower bounds. For that question I found that \(\displaystyle x = -1\) and \(\displaystyle x = 10\) satisfy both curves yet they do not seem to be right:

For x = -1

\(\displaystyle 9x = x^2-10\)

\(\displaystyle 9(-1) = (-1)^2-10\)

\(\displaystyle -9 = -9\)

For x = 10

\(\displaystyle 9x = x^2-10\)

\(\displaystyle 9(10) = (10)^2-10\)

\(\displaystyle 90 = 90\)

The answer I end up with is:

\(\displaystyle \int_{-1}^{10}(x^2-9x-10)dx\)

I guess I could separate my answer into two parts, but the signs are flipped for the integrand as well so it wouldn't be the same.

However the answer they are looking for in terms of x is:

\(\displaystyle \int_{-\sqrt{10}}^{-1}(-x^2+10)dx+\int_{-1}^{0}-9xdx\)

Why do the bounds somewhat resemble the x-values that I found? They have -1 as the upper instead of the lower, and \(\displaystyle -\sqrt{10}\) as the lower instead of 10 in the upper. They also have a separate set of bounds for the other integral, but how? I know they are supposed to have separate ones, but the only way I can see to get two x-values is when I "connect" the curves by equating them and then factoring.