Question

What is the net area between `f(x) = (x-2)^3` and the x-axis over `x in [1, 5 ]`?

Answer 1

`20.5`

Explanation:

The integrand `f(x) = (x-2)^3` crosses 0 at `x=2`. The area enclosed between `x=1` and `x=2` is given by

`A_1 = |int_1^2 (x-2)^3dx|`
`qquadquad = |(x-2)^4/4]_1^2|`
`qquadqquad = |-1/4| = 1/4`

while, the area enclosed between `x=2` and `x=5` is given by

`A_2 = |int_2^5 (x-2)^3dx|`
`qquadquad = |(x-2)^4/4]_2^5|`
`qquadqquad = |81/4| = 81/4`

Thus, the net area is `81/4+1/4=82/4=20.5`