## Calculus with Applications (10th Edition)

$$A = 8$$
\eqalign{ & x = - 3,\,\,\,\,\,x = 0,\,\,\,\,\,y = 1 - {x^2},\,\,\,\,\,\,y = 0 \cr & {\text{Check if the graph crosses the }}x{\text{ - axis}} \cr & \,\,\,\,\,\,\,\,1 - {x^2} = 0 \cr & \,\,\,\,\,\,\,x = \pm 1 \cr & {\text{The critical points of the graph is }}x = \pm 1.{\text{ Evaluaing the graph at }}x = - 3 \cr & \,\,\,\,\,\,\,y\left( { - 3} \right) = 1 - {\left( { - 3} \right)^2} \cr & \,\,\,\,\,\,\,y\left( { - 3} \right) = - 8 \cr & {\text{Then}}{\text{, the region of the graph lies below for the interval }}\left( { - 3, - 1} \right). \cr & {\text{The area is given by}}: \cr & A = \left| {\int_{ - 3}^{ - 1} {\left( {1 - {x^2}} \right)dx} } \right| + \int_{ - 1}^0 {\left( {1 - {x^2}} \right)} dx \cr & A = \left| {\left( {x - \frac{{{x^3}}}{3}} \right)_{ - 3}^{ - 1}} \right| + \left( {x - \frac{{{x^3}}}{3}} \right)_{ - 1}^1 \cr & {\text{Evaluating}} \cr & A = \left| {\left( {\left( { - 1} \right) - \frac{{{{\left( { - 1} \right)}^3}}}{3}} \right) - \left( {\left( { - 3} \right) - \frac{{{{\left( { - 3} \right)}^3}}}{3}} \right)} \right| + \left( {\left( 1 \right) - \frac{{{{\left( 1 \right)}^3}}}{3}} \right) - \left( {\left( { - 1} \right) - \frac{{{{\left( { - 1} \right)}^3}}}{3}} \right) \cr & {\text{Simplifying}} \cr & A = \left| { - \frac{2}{3} - 6} \right| + \frac{2}{3} + \frac{2}{3} \cr & A = \frac{{20}}{3} + \frac{4}{3} \cr & A = 8 \cr}