Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.1 Area Between Two Curves - Exercises Set 5.1 - Page 353: 8

$$A = 4$$

$$\eqalign{ & {\text{From the graph we can note that the area is given by}} \cr & A = \int_0^2 {\left( {0 - \left( {{x^3} - 4x} \right)} \right)} dx \cr & A = \int_0^2 {\left( {4x - {x^3}} \right)} dx \cr & {\text{Integrate}} \cr & A = \left[ {2{x^2} - \frac{1}{4}{x^4}} \right]_0^2 \cr & {\text{Evaluate}} \cr & A = \left[ {2{{\left( 2 \right)}^2} - \frac{1}{4}{{\left( 2 \right)}^4}} \right] - \left[ {2{{\left( 0 \right)}^2} - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr & {\text{Simplify}} \cr & A = 4 - 0 \cr & A = 4 \cr} $$