1. The problem statement, all variables and given/known data Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=5 sqrt(x), y=3 and 2y+4x=9 2. Relevant equations Integral of right function - left function if integrating with respect to y F(b)-F(a) 3. The attempt at a solution I decided to integrate with respect to y since after I got the graph, the area is kind of split up into two different parts if you decide to do it with respect x. So if I do it with respect to y the right function would be 2y=5 sqrt(x) and to make it with respect to y it would be (4/5)y^2. The left function is 2y+4x=9 or with respect to y it would be -(2y-9)/4. To find the bounds I found where the line y=3 intersected with 2y=5 sqrt(x) and where 2y=5 sqrt(x) intersected with 2y+4x=9. I found my "y bounds" to be 2 to 3. So I plug everything in. Integral from 2 to 3 of: (4/25)*y^2 - [(-2y-9)/4] So then I took the anti-derivative and got 4/75y^3 - 1/4y^2 + 9/4y Then I plugged 3 into the above equation to get my F(b) and then plugged in 2 into the equation to get my F(a). I did F(b)-F(a) (4/75*(3)^3+1/4*(3)^2+9/4*(3)) - (4/75*(2)^3+1/4*(2)^2+9/4*(2)) = 4.513333 ... which is not the answer :( Can someone please help me? I'm sorry if this is hard to read, I tried using the symbols but it kept messing everything up; I'm new to this forum and I don't really know how to use the symbols.