Answer
$$\frac{1}{153}(2^{51}+1)$$
Work Step by Step
Using the substitution method:
Let $u=3t-1$. Hence, $du=3dt$.
Substituting $u$ for $3t-1$ and $\frac{1}{3}du$ for $dt$:
$\int^{1}_{0}(3t-1)^{50}dt$
$=\int^2_{-1}(u)^{50}(\frac{1}{3}du)$
$=\frac{1}{3}\int^2_{-1}u^{50}du$
$=\frac{1}{3}[\frac{1}{51}u^{51}]|^2_{-1}$
$=\frac{1}{3}[(\frac{1}{51}2^{51})-(\frac{1}{51}(-1)^{51})]$
$=\frac{1}{153}(2^{51}+1)$
