Question

Please help me with this question

Answer 1

Explanation:

Given:
`f'(x) = x^2e^(2x^3)` and
`f(0) = 0`

We can solve for f(x) by writing

`\displaystyle f(x) = \int f'(x)dx=\int x^2e^(2x^3)dx`

Let
`u = 2x^3`

`\:\:\:\:du=6x^2dx`

Then

`\displaystyle f(x) = \int x^2e^(2x^3)dx = (1)/(6)\int e^u du`

`\displaystyle \:\:\:\:\:\:\:=(1)/(6)e^(2x^3) + k`

We know that f(0) = 0 so we can find the value for k:

`f(0) = (1)/(6)(1) + k \Rightarrow k = -(1)/(6)`

Therefore,

`\displaystyle f(x) = (1)/(6) \left(e^(2x^3) - 1 \right)`