Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 70

$$1$$

Given $$ \int_{0}^{1}(3 t-1)^{2} d t $$ Let $$u=3 t-1 \ \ \ \ \Rightarrow \ \ \ du =3dt $$ At $t=0 \to u= -1$ and at $t=1 \to u=2 $ Then \begin{aligned} \int_{0}^{1}(3 t-1)^{2} d t &=\frac{1}{3} \int_{-1}^{2} u^{2} d u \\ &=\left.\frac{1}{3}\left(\frac{1}{3} u^{3}\right)\right|_{-1} ^{2} \\ &=\frac{1}{9}\left(2^{3}-(-1)^{3}\right) \\ &=1 \end{aligned}