Definite integral problem?

Question

I'm having trouble with this one...I don't understand why f(x) isn't 3 - (x/4) - sqrt(x+9)? That's the right answer, right?

Consider the area between the graphs x + 4y = 12 and x + 9 = y^2.

integral from 0 to -9 of f(x)dx + integral from 40 to 0 of g(x)dx


f(x) = ?
g(x) = ?

Answer 1

Please take a look at the following image:
http://img85.imageshack.us/img85/1726/graphku3.png

The red and blue curves are the upper and lower part of x+9=y², the green line is the graph x+4y=12.

It should be obvious from the image that the area between these two graphs is given by [-9, 0]∫√(x+9) - (-√(x+9)) + [0, 40]∫3-x/4 - (-√(x+9)). The lower bound of the region is in every case the lower part of x+9=y², which is y=-√(x+9). Which curve forms the upper bound of the region however changes - from -9 to 0 it is the upper half of x+9=y², which is y=√(x+9), and it is only from 0 to 40 that the upper bound is given by the line y=3-x/4.

Thus, for this problem:

f(x) = 2√(x+9)
g(x) = 3-x/4+√(x+9)