Answer
$$\frac{15}{4}$$
Work Step by Step
Given
$$\int_{-1}^2x^3dx$$
Since
\begin{aligned}
x^3\lt 0 \ \ \ \ \text{for }\ \ -1\leq x\leq 0,\\
x^3\geq 0 \ \ \ \ \ \ \ \ \text{for }\ \ 0\leq x\leq 2
\end{aligned}
Then
\begin{aligned}
\int_{-1}^2x^3dx&= \int_{-1}^0x^3dx+\int_{0}^2x^3dx\\
&= \frac{1}{4}x^4 \bigg|_{-1}^0+\frac{1}{4}x^4 \bigg|_{0}^2\\
&= \left(0-\frac{1}{4}\right)+ \left(4-0\right)\\
&= \frac{15}{4}
\end{aligned}
