Chapter 4 - Integration - 4.2 Exercises - Page 264: 52

$${\text{Area}} = \frac{3}{4}$$

$$\eqalign{ & y = 2x - {x^3},{\text{ }}\left[ {0,1} \right] \cr & f\left( x \right) = 2x - {x^3} \cr & f\left( 0 \right) = 0{\text{ and }}f\left( 1 \right) = 1,{\text{ }} \cr & f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\ & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta x,{\text{ }}\Delta x = \frac{{1 - 0}}{n} = \frac{1}{n} \cr & {c_i} = a + i\Delta x \to \frac{i}{n} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {2\left( {\frac{i}{n}} \right) - {{\left( {\frac{i}{n}} \right)}^3}} \right]} \left( {\frac{1}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{2i}}{{{n^2}}} - \frac{{{i^3}}}{{{n^4}}}} \right)} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{2}{{{n^2}}}\sum\limits_{i = 1}^n i - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^4}}}\sum\limits_{i = 1}^n {{i^3}} \cr & {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{2}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^4}}}\left( {\frac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{n} - \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^2}}}{{4{n^2}}} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right) - \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{4} + \frac{1}{n} + \frac{1}{{4{n^2}}}} \right) \cr & {\text{Evaluate the limit when }}n \to \infty \cr & {\text{Area}} = 1 - \left( {\frac{1}{4} + 0 + 0} \right) \cr & {\text{Area}} = \frac{3}{4} \cr} $$