Answer
$$A = \frac{{22}}{3}$$
Work Step by Step
$$\eqalign{
& y = {x^2} - 1;\,\,\,\left[ {0,3} \right] \cr
& {\text{From the graph we can see that the area is given by}} \cr
& A = \int_0^1 {\left( {1 - {x^2}} \right)} dx + \int_1^3 {\left( {{x^2} - 1} \right)dx} \cr
& A = \left[ {x - \frac{1}{3}{x^3}} \right]_0^1 + \left[ {\frac{1}{3}{x^3} - x} \right]_1^3 \cr
& {\text{Evaluating}} \cr
& A = \left[ {1 - \frac{1}{3}{{\left( 1 \right)}^3}} \right] + \left[ {\frac{1}{3}{{\left( 3 \right)}^3} - \left( 3 \right)} \right] - \left[ {\frac{1}{3}{{\left( 1 \right)}^3} - \left( 1 \right)} \right] \cr
& A = \frac{2}{3} + 6 + \frac{2}{3} \cr
& A = \frac{{22}}{3} \cr} $$
