Chapter 4 - Integration - 4.2 Exercises - Page 264: 58

$${\text{Area}} = \frac{{11}}{3}$$

$$\eqalign{ & f\left( y \right) = 4y - {y^2},{\text{ }}1 \leqslant y \leqslant 2 \cr & f\left( 1 \right) = 3{\text{ and }}f\left( 2 \right) = 4, \cr & f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\ & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta y,{\text{ }}\Delta x = \frac{{2 - 1}}{n} = \frac{1}{n} \cr & {c_i} = a + i\Delta x \to 1 + \frac{i}{n} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {4\left( {1 + \frac{i}{n}} \right) - {{\left( {1 + \frac{i}{n}} \right)}^2}} \right]} \left( {\frac{1}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {4 + \frac{{4i}}{n} - 1 - \frac{{2i}}{n} - \frac{{{i^2}}}{{{n^2}}}} \right)} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {3 + \frac{{2i}}{n} - \frac{{{i^2}}}{{{n^2}}}} \right)} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{3}{n} + \frac{2}{{{n^2}}}i - \frac{1}{{{n^3}}}{i^2}} \right)} \cr & {\text{Using the Summation Formulas from THEOREM 4}}{\text{.2}} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{3}{n}\left( n \right) + \frac{2}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) - \frac{1}{{{n^3}}}\left( {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right)} \right] \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {3 + \frac{{{n^2} + n}}{{{n^2}}} - \frac{{2{n^2} + 3n + 1}}{{6{n^2}}}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {3 + 1 + \frac{1}{n} - \frac{1}{3} - \frac{1}{{2n}} - \frac{1}{{6{n^2}}}} \right) \cr & {\text{Evaluate the limit when }}n \to \infty \cr & {\text{Area}} = 3 + 1 + 0 - \frac{1}{3} - 0 - 0 \cr & {\text{Area}} = \frac{{11}}{3} \cr} $$