\(\frac{56}{12}\)
\(\frac{65}{12}\)
12
65
Correct answer is B
\({1_0^∫} x^2(x^3+2)^3\)dx
let \( u = x^3 + 2, du = 3x^2dx\)
when x = 1, u = 3
when x = 0, u = 2
dx = \(\frac{du}{3x^2}\)
\({3_2^∫}\) \(\frac{x^2[u]^3}{3x^2}\)
\({3_2^∫}\) \(\frac{u^3}{3}\) du
= \(\frac{u^4}{3*4}\)\(_2\)3
\(\frac{1}{12} [u^4]\)\(_2\)3
\(\frac{1}{12} [3^4 - 2^4]\)
\(\frac{1}{12}[81 - 16]\)
\(\frac{65}{12}\)