Chapter 7 - Integration - 7.5 The Area Between Two Curves - 7.5 Exercises - Page 405: 8

$$A = \frac{{343}}{6}$$

$$\eqalign{ & y = {x^2} - 18,\,\,\,\,\,y = x - 6 \cr & {\text{Find the points of intersection of the graphs}} \cr & {x^2} - 18 = x - 6 \cr & {x^2} - 18 - x + 6 = 0 \cr & {x^2} - x - 12 = 0 \cr & {\text{factor}} \cr & \left( {x - 4} \right)\left( {x + 3} \right) = 0 \cr & x = - 3{\text{ and }}x = 4 \cr & {\text{We have the interval }}\left( { - 3,4} \right) \cr & {\text{Evaluate both functions at }}x = 0 \cr & y\left( 0 \right) = {\left( 0 \right)^2} - 18 = - 18 \cr & y\left( 0 \right) = 0 - 6 = - 6 \cr & {\text{For the interval }}\left( { - 3,4} \right){\text{ we have that }}x - 6 \geqslant {x^2} - 18 \cr & \cr & {\text{The area is given by}}: \cr & A = \int_{ - 3}^4 {\left[ {\left( {x - 6} \right) - \left( {{x^2} - 18} \right)} \right]} dx \cr & A = \int_{ - 3}^4 {\left( {x - 6 - {x^2} + 18} \right)} dx \cr & A = \int_{ - 3}^4 {\left( {x - {x^2} + 12} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {\frac{1}{2}{x^2} - \frac{1}{3}{x^3} + 12x} \right)_{ - 3}^4 \cr & A = \left( {\frac{1}{2}{{\left( 4 \right)}^2} - \frac{1}{3}{{\left( 4 \right)}^3} + 12\left( 4 \right)} \right) - \left( {\frac{1}{2}{{\left( { - 3} \right)}^2} - \frac{1}{3}{{\left( { - 3} \right)}^3} + 12\left( { - 3} \right)} \right) \cr & A = \left( {\frac{{104}}{3}} \right) - \left( { - \frac{{45}}{2}} \right) \cr & A = \frac{{343}}{6} \cr & A = 57.16 \cr} $$