# Calculus

Consider the area between the graphs x+2y=4 and x+4=y2. This area can be computed
in two different ways using integrals

1) Compute as a sum of two integrals
2) Compute as a single integral
3) Either way, what is the area?

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So for 1) I rewrote the two to be:
x = 4-2y
x = y^2-4
and setting them equal:
y^2-4 = 4-2y gives:
y^2+2y-8 or (y+4)(y-2)
Therefore the roots are y = -4 and 2

So if its the integral of f(x) from a to b plus the integral of g(x) from b to c:
a: -4, b: 0, c: -12
But what would be the functions f(x) and g(x)? Nothing I tried has worked, and I have no idea how to approach it as a single integral.

Thanks!

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1. p.s. I wrote y2 for the second equation but it is y^2

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2. let's look at the graph
http://www.wolframalpha.com/input/?i=plot+x%2B2y%3D4+and+x%2B4%3Dy%5E2

You found the correct y values of the intersection points, the points are (0,2) and (12, -4)

So why not take horizontal slices?
the effective width = rightmost x - left x
= (4 - 2y) - (y^2 - 4)
= -y^2 - 2y + 8

area = ∫ (-y^2 - 2y + 8) dy from -4 to 2
= [ (-1/3)y^3 - y^2 + 8y] from -4 to 2
= ( (-1/3)(8) - 4 + 16) - ( (-1/3)(-64) - 16 - 32)
= -8/3 + 12 - 64/3 + 48
= 36

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