## Calculus: Early Transcendentals (2nd Edition)

$$Area = \ln \left( 6 \right)$$
\eqalign{ & y = \frac{{10}}{{{x^2} - 2x - 24}},{\text{ }}x = - 2,{\text{ }}x = 2,{\text{ }}x{\text{ - axis}} \cr & x{\text{ - axis}} \to y = 0 \cr & 0 \geqslant \frac{{10}}{{{x^2} - 2x - 24}}{\text{ on }}\left[ { - 2,2} \right],{\text{ then}} \cr & {\text{the area of the region is}} \cr & Area = \int_{ - 2}^2 {\left( {0 - \frac{{10}}{{{x^2} - 2x - 24}}} \right)} dx \cr & Area = - \int_{ - 2}^2 {\frac{{10}}{{{x^2} - 2x - 24}}} dx \cr & {\text{integrand }} \cr & = \frac{{10}}{{{x^2} - 2x - 24}} \cr & = \frac{{10}}{{\left( {x - 6} \right)\left( {x + 4} \right)}} \cr & {\text{partial fractions}} \cr & \frac{{10}}{{\left( {x - 6} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 6}} + \frac{B}{{x + 4}} \cr & {\text{multiplying}} \cr & 10 = A\left( {x + 4} \right) + B\left( {x - 6} \right) \cr & {\text{Letting }}x = 6 \cr & 10 = A\left( {10} \right) + B\left( 0 \right) \cr & A = 1 \cr & {\text{Letting }}x = - 4 \cr & 10 = A\left( 0 \right) + B\left( { - 10} \right) \cr & B = - 1 \cr & {\text{substituting constants}} \cr & \frac{A}{{x - 6}} + \frac{B}{{x + 4}} = \frac{1}{{x - 6}} - \frac{1}{{x + 4}} \cr & Area = - \int_{ - 2}^2 {\frac{{10}}{{{x^2} - 2x - 24}}} dx = - \int_{ - 2}^2 {\left( {\frac{1}{{x - 6}} - \frac{1}{{x + 4}}} \right)} dx \cr & {\text{integrating}} \cr & Area = - \left[ {\ln \left| {x - 6} \right| - \ln \left| {x + 4} \right|} \right]_{ - 2}^2 \cr & Area = \left[ {\ln \left| {x + 4} \right| - \ln \left| {x - 6} \right|} \right]_{ - 2}^2 \cr & Area = \left[ {\ln \left| {\frac{{x + 4}}{{x - 6}}} \right|} \right]_{ - 2}^2 \cr & {\text{evaluate limits}} \cr & Area = \ln \left| {\frac{{2 + 4}}{{2 - 6}}} \right| - \ln \left| {\frac{{ - 2 + 4}}{{ - 2 - 6}}} \right| \cr & Area = \ln \left| { - \frac{3}{2}} \right| - \ln \left| { - \frac{1}{4}} \right| \cr & Area = \ln \left( 6 \right) \cr}