Posted by **bobby** on Wednesday, September 12, 2012 at 8:03am.

Consider the area between the graphs x+y=16 and x+4= (y^2). This area can be computed in two different ways using integrals.

First of all it can be computed as a sum of two integrals integrate from a to b of f(x)dx + integrate from b to c of g(x)dx

What is the value of a, b, c and what are f(x) and g(x) equal to?

Alternatively this area can be computed as a single integral

integrate from alpha to beta of h(y)dy

Alpha=?, Beta=?, h(y)=?

Either way we find that the area is: ?

- math, calculus 2 -
**Steve**, Wednesday, September 12, 2012 at 11:42am
the graphs intersect at (12,4) and (21,-5)

a = ∫[-4,12] 2√(x+4) dx + ∫[12,21] 16-x+√(x+4) dx

a = ∫[-5,4] (16-y) - (y^2-4) dy

a = 243/2

- math, calculus 2 -
**bobby**, Wednesday, September 12, 2012 at 6:26pm
how did you get the -4 in your first step? i dont understand where that came from

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