Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 28

Given y=1/4x^2, y=2x^2, x+y=3, x>0
Find the intersection point of y=2x^2 and y=-x+3.
2x^2=-x+3
2x^2+x-3=0
(2x+3)(x-1)=0
x=-3/2, x=1
Ignore the x=-3/2. The original problem states that x>0.
When x=1, y=2. The intersection point is (1,2).
Find the intersection point of y=-x+3 and y=1/4x^2 .
1/4x^2=-x+3
x^2+4x-12=0
(x+6)(x-2)=0
x=-6, x=2
Ignore the x=-6. The original problem states that x^0.
When x=2, y=1. The intersection point is (2, 1).
A=int_0^1(2x^2-1/4x^2)dx+int_1^2(-x+3-1/4x^2)dx
=int_0^1(7/4x^2)dx-int_1^2(1/4x^2+x-3)dx
=[7/4*x^3/3]_0^1-[1/4*x^3/3+x^2/2-3x]_1^2
=[7/12x^3]_0^1-[1/12x^3+x^2/2-3x]_1^2
=[7/12(1)^3-0]-[(1/12(2)^3+(2)^2/2-3(2))-(1/12(1)^3+(1)^2/2-3(1))]
=[7/12]-[8/12+2-6-1/12-1/2+3]
=[7/12]-[7/12-1/2-1]
=[1/2+1]
=3/2
The area enclosed by the given curves is 3/2 units squared.
The black graph is y=1/4x^2.
The red graph is y=2x^2.
The green graph is x+y=3=>y=-x+3.