Question

Please help me with this question

Answer 1

Step-by-step 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

$\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du$

$\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k$

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

$f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}$

Therefore,

$\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)$