Question

$\underset{1}{\overset{3}{\int }}\left|x^2-4\right|dx$

Answer 1

$$\begin{gathered} {\int }_1^3\left|x^2-4\right|dx \\ ={\int }_1^2-\left(x^2-4\right)dx+{\int }_2^3\left(x^2-4\right)dx \\ ={\left[-\frac{x^3}{3}+4x\right]}_1^2+{\left[\frac{x^3}{3}-4x\right]}_2^3 \\ =\frac{-8}{3}+8+\frac{1}{3}-4+9-12-\frac{8}{3}+8 \\ =4 \end{gathered}$$