## Calculus: Early Transcendentals 8th Edition

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