Answer
$\displaystyle \int_{0}^{1}x^2(x^3-2)^3dx= -\dfrac{5}{4}$
Work Step by Step
To evaluate the integral $\displaystyle \int_{0}^{1}x^2(x^3-2)^3dx$, let $u = x^3-2$.
Then we obtain $u = x^3-2 \longrightarrow du = 3x^2dx \longrightarrow \dfrac{du}{3} = x^2dx$
Before substituting, determine the new upper and lower limits of integration
Lower limit: When $x = 0$, $u = 0^3-2 = -2$
Upper limit: When $x = 1$, $u = 1^3-2 = -1$
$\displaystyle \int_{0}^{1}x^2(x^3-2)^3dx = \displaystyle \int_{0}^{1}(x^3-2)^3(x^2dx) = \displaystyle \int_{-2}^{-1}u^3\dfrac{du}{3} = \displaystyle \dfrac{1}{3}\int_{-2}^{-1}u^3du = \dfrac{1}{3}\dfrac{u^4}{4} \Bigg\vert_{-2}^{-1} = \dfrac{1}{3} \bigg[ \dfrac{(-1)^4}{4} - \dfrac{(-2)^4}{4}\bigg] = \dfrac{1}{3} \bigg[ \dfrac{1}{4} - \dfrac{16}{4}\bigg] = \dfrac{1}{3} \dfrac{-15}{4} = -\dfrac{5}{4}$
