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
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: ?
2 answers
how did you get the -4 in your first step? i don't understand where that came from
1 answer