Step-by-step explanation:
Given: [tex]f'(x) = x^2e^{2x^3}[/tex] and [tex]f(0) = 0[/tex]
We can solve for f(x) by writing
[tex]\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx[/tex]
Let [tex]u = 2x^3[/tex]
[tex]\:\:\:\:du=6x^2dx[/tex]
Then
[tex]\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du[/tex]
[tex]\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k[/tex]
We know that f(0) = 0 so we can find the value for k:
[tex]f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}[/tex]
Therefore,
[tex]\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)[/tex]
