Answer
$\displaystyle \int_{0}^{1} (3x+1)^5dx = \dfrac{455}{2} = 227.5$
Work Step by Step
$\displaystyle \int_{0}^{1} (3x+1)^5dx$
To evaluate this integral let $u = 3x+1 \longrightarrow du = 3dx \longrightarrow \dfrac{du}{3} = dx$
Lower limit: When $x = 0$, $u = 3\times0 + 1 = 1$
Upper limit: When $x = 1$, $u = 3\times1 + 1 = 4$
Now, we substitute to obtain
$\displaystyle \int_{0}^{1} (3x+1)^5dx = \displaystyle \int_{1}^{4} u^5\dfrac{du}{3} = \displaystyle \dfrac{1}{3}\int_{1}^{4} u^5du = \dfrac{1}{3} \bigg[\dfrac{u^6}{6}\bigg] \Bigg\rvert_{1}^{4} = \dfrac{1}{3} \bigg[\dfrac{4^6}{6} - \dfrac{1^6}{6}\bigg] = \dfrac{1}{3} \bigg[\dfrac{4096}{6} - \dfrac{1}{6}\bigg] = \dfrac{1}{3} \bigg[\dfrac{1365}{2}\bigg] = \dfrac{455}{2} = 227.5$
