## Calculus 8th Edition

$$\int_0^1(3t-1)^{50}dt=\frac{2^{51}+1}{153}$$
To evaluate the integral $$\int_0^1(3t-1)^{50}dt$$ we will use substitution $3t-1=z$ which gives us $3dt=dz\Rightarrow dt=\frac{dz}{3}$ and the integration bounds would be: for $t=0$ we have $z=-1$ and for $t=1$ we have $z=2.$ Now putting all this into this into the integral we get: $$\int_0^1(3t-1)^{50}dt=\int_{-1}^2z^{50}\frac{dz}{3}=\frac{1}{3}\left(\left.\frac{z^{51}}{51}\right|_{-1}^2\right)=\frac{2^{51}-(-1)^{51}}{3\cdot51}=\frac{2^{51}+1}{153}$$