∫0^1 8x^(2)e^(2x^(3))dx

Question

Wasn't exactly sure how to put it but it was an integral with 1 at the top and 0 at the bottom. Starting with 8x^2 is what's inside the integral.

Answer 1

The good news is that the derivative of x^3 is 3x^2.

Thus, using I[a, b] for the integral from a to b and E[a, b] for the evaluation from a to b,

I[0, 1] 8x^2 e^(2x^3) dx = 4/3 e^(2x^3) E[0, 1] = 4/3 e^2 - 4/3