Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.2 Exercises - Page 264: 46

Answer

$${\text{Area}} = \frac{{51}}{2}$$

Work Step by Step

$$\eqalign{ & y = 3x - 2,{\text{ }}\left[ {2,5} \right] \cr & f\left( x \right) = 3x - 2 \cr & f\left( 2 \right) = 4{\text{ and }}f\left( 5 \right) = 13,{\text{ }} \cr & f\left( x \right){\text{ is continuous and there are no negatives on the interval }} \\ & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta x,{\text{ }}\Delta x = \frac{{5 - 2}}{n} = \frac{3}{n},{\text{ }} \cr & {c_i} = a + i\Delta x \to 2 + \frac{{3i}}{n} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {3\left( {2 + \frac{{3i}}{n}} \right) - 2} \right]} \left( {\frac{3}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {4 + \frac{{9i}}{n}} \right]} \left( {\frac{3}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{27i}}{{{n^2}}}} \right)} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{12}}{n}} \right)} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{27}}{{{n^2}}}\sum\limits_{i = 1}^n {\left( i \right)} + \mathop {\lim }\limits_{n \to \infty } \frac{{12}}{n}\sum\limits_{i = 1}^n {\left( 1 \right)} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{27}}{{{n^2}}}\frac{{n\left( {n + 1} \right)}}{2} + \mathop {\lim }\limits_{n \to \infty } \frac{{12}}{n}\left( n \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{{27n + 27}}{{2n}} + \mathop {\lim }\limits_{n \to \infty } 12 \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{27}}{2} + \frac{{27}}{{2n}}} \right) + \mathop {\lim }\limits_{n \to \infty } 12 \cr & {\text{Evaluate the limit when }}n \to \infty \cr & {\text{Area}} = \frac{{27}}{2} + 12 \cr & {\text{Area}} = \frac{{51}}{2} \cr} $$
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