`y = (1/4)x^2, y = 2x^2, x + y = 3, x>0` Sketch the region enclosed by the given curves and find its area.

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.`