Posted by Donna on Thursday, July 11, 2013 at 9:15pm.
odd that one function is written as
2y = 4√x
why not just y = 2√x ?
Anyway, assuming the functions are correctly written, the area is roughly triangular, with vertices at (-1,4),(1,2),(4,4)
Integrating along x, we need to divide it up into two regions, over [-1,1] and [1,4]
a = ∫[-1,1] 4 - (3-x) dx + ∫[1,4] 4 - 2√x dx
= (x + 1/2 x^2)[-1,1] + (4x - 4/3 x^(3/2))[1,4]
= 2 + 8/3
= 14/3
Integrating along y is easier, since we just subtract the left from the right:
a = ∫[2,4] y^2/4 - (3-y) dy
= y^3/12 + y^2/2 - 3y [2,4]
= 14/3