Chapter 7 - Applications of Integration - 7.1 Exercises - Page 443: 62

$A = \frac{{27}}{4}$

$$\eqalign{ & y = {x^3} - 2x,{\text{ }}\left( { - 1,1} \right) \cr & {\text{Differentiate }} \cr & y' = 3{x^2} - 2 \cr & {\text{Find the slope at }}\left( { - 1,1} \right) \cr & m = y'\left( { - 1} \right) = 3{\left( { - 1} \right)^2} - 2 \cr & m = 1 \cr & {\text{The equation of the tangent line at }}\left( { - 1,1} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 1 = x - \left( { - 1} \right) \cr & y = x + 2 \cr & \cr & {\text{The intersection points of the functions }}y = {x^3} - 2x,{\text{ }}y = x + 2 \cr & {\text{is}} \cr & {x^3} - 2x = x + 2 \cr & {x^3} - 2x - x - 2 = 0 \cr & {x^3} - 3x - 2 = 0 \cr & {\text{Solving we obtain }}x = 2 \cr & {\text{The points are }}\left( { - 1,1} \right){\text{ and }}\left( {2,f\left( 2 \right)} \right) \cr & {\text{The area is given by }}\left( {{\text{See graph below}}} \right) \cr & A = \int_{ - 1}^2 {\left[ {\left( {x + 2} \right) - \left( {{x^3} - 2x} \right)} \right]} dx \cr & A = \int_{ - 1}^2 {\left( {x + 2 - {x^3} + 2x} \right)} dx \cr & A = \int_{ - 1}^2 {\left( {3x + 2 - {x^3}} \right)} dx \cr & {\text{Integrate and evaluate}} \cr & A = \left[ {\frac{3}{2}{x^2} + 2x - \frac{1}{4}{x^4}} \right]_{ - 1}^2 \cr & A = \left[ {\frac{3}{2}{{\left( 2 \right)}^2} + 2\left( 2 \right) - \frac{1}{4}{{\left( 2 \right)}^4}} \right] - \left[ {\frac{3}{2}{{\left( { - 1} \right)}^2} + 2\left( { - 1} \right) - \frac{1}{4}{{\left( { - 1} \right)}^4}} \right] \cr & A = 6 + \frac{3}{4} \cr & A = \frac{{27}}{4} \cr} $$