Question

Question #0e2de

Answer 1

`104/15`

Explanation:

The two graphs intersect at (1,0) and (-1,0)

104/151

The red area is the area under the curve of `y=-4x^2+4`, or

`int_-1^1(-4x^2+4)dx`
`=int_-1^1(-4x^2)dx+int_-1^1(4)dx`
`=[4x-4/3x^3],(1,-1)`
`=(4(1)-4/3(1)^3)-(4(-1)-4/3(-1)^3)`
`=8/3-(-8/3)`
`=16/3`

The blue area is the area under the curve `x^4-1=y` or

`int_-1^1(x^4-1)dx`
`=int_-1^1(x^4)-int_-1^1(1)`
`=[x^5/5-x],(1,-1)`
`=((1)^5/5-(1))-((-1)^5/5-(-1))`
`=(-4/5)-(4/5)`
`=-8/5`
However, since we are taking the area, the integral must be positive.

Area under `y=-4x^2+4`: `16/3`
Area under `y=x^4-1`: `8/5`

Add them together to get: `16/3+8/5=104/15`