Question

How do you solve `F(x) = int_(x^4)^(x^3) (2t-1)^3` `dt`?

Answer 1

`F(x) = 1/8{(2x^3-1)^4-(2x^4-1)^4}`

Explanation:

We have:

`F(x) = int_(x^4)^(x^3) (2t-1)^3 dt`
`" " = [(2t-1)^4/8]_(x^4)^(x^3)`

`" " = 1/8{(2x^3-1)^4-(2x^4-1)^4}`