## Calculus with Applications (10th Edition)

$$A = 366.2$$
\eqalign{ & y = {x^2} - 30,\,\,\,\,\,y = 10 - 3x \cr & {\text{Find the points of intersection of the graphs}} \cr & {x^2} - 30 = 10 - 3x \cr & {x^2} + 3x - 40 = 0 \cr & {\text{factor}} \cr & \left( {x + 8} \right)\left( {x - 5} \right) = 0 \cr & {x_1} = - 8{\text{ and }}{x_2} = 5 \cr & {\text{We have the interval }}\left( { - 8,5} \right) \cr & \left( {0,5} \right){\text{ and }}\left( {5,6} \right) \cr & {\text{Evaluate both functions at }}x = 0 \cr & y\left( 0 \right) = {\left( 0 \right)^2} - 30 = - 30 \cr & y\left( 0 \right) = 10 - 3\left( 0 \right) = 10 \cr & {\text{For the interval }}\left( { - 8,5} \right){\text{ we have that }}10 - 3x \geqslant {x^2} - 30{\text{ }} \cr & \cr & {\text{The area is given by}}: \cr & A = \int_{ - 8}^5 {\left[ {\left( {10 - 3x} \right) - \left( {{x^2} - 30} \right)} \right]} dx \cr & A = \int_{ - 8}^5 {\left( {10 - 3x - {x^2} + 30} \right)} dx \cr & A = \int_{ - 8}^5 {\left( {40 - 3x - {x^2}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {40x - \frac{3}{2}{x^2} - \frac{{{x^3}}}{3}} \right)_{ - 8}^5 \cr & A = \left( {40\left( 5 \right) - \frac{3}{2}{{\left( 5 \right)}^2} - \frac{{{{\left( 5 \right)}^3}}}{3}} \right) - \left( {40\left( { - 8} \right) - \frac{3}{2}{{\left( { - 8} \right)}^2} - \frac{{{{\left( { - 8} \right)}^3}}}{3}} \right) \cr & A = \frac{{725}}{6} - \left( { - \frac{{736}}{3}} \right) \cr & A = \frac{{2197}}{6} \cr & A = 366.2 \cr}