Consider the area between the graphs x+1y=12 and x+8=y2 . This area can be computed in two different ways using integrals

Question

First of all it can be computed as a sum of two integrals
Interval a to b
f(x)= i put 12-x but it's wrong
Interval c to b
f(x)= i put (x-8)^(1/2) but it's wrong as well
I graphed this problem and got a=-8 and b=8. But the number my graph shows for c is not right.

Part two is
Alternatively this area can be computed as a single integral
Interval a to b
h(y)= ?

I assumed that the endpoints would be the same as above but it's not right. Can someone explain the steps i can take to get it. I've been working on this for a couple of hours.

Answer 1

You have a parabola opening to the right, crossed by a line. The two curves intersect at (8,4) and (17,-5).

It's much easier to integrate on y, so you end up with

∫[-5,4] (12-y)-(y^2-8) dy = 243/2