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: 49

Answer

$${\text{Area}} = 54$$

Work Step by Step

$$\eqalign{ & y = 25 - {x^2},{\text{ }}\left[ {1,4} \right] \cr & f\left( x \right) = 25 - {x^2} \cr & f\left( 1 \right) = 24{\text{ and }}f\left( 4 \right) = 9,{\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{{4 - 1}}{n} = \frac{3}{n} \cr & {c_i} = a + i\Delta x \to 1 + \frac{{3i}}{n} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {25 - {{\left( {1 + \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[ {25 - 1 - \frac{{6i}}{n} - \frac{{9{i^2}}}{{{n^2}}}} \right]} \left( {\frac{3}{n}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{n}\sum\limits_{i = 1}^n {\left( {24 - \frac{{6i}}{n} - \frac{{9{i^2}}}{{{n^2}}}} \right)} \cr & {\text{Where }}\sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6},{\text{ and }}\sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}}{2} \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{n}\left( {24n - \frac{6}{n} \cdot \frac{{n\left( {n + 1} \right)}}{2} - \frac{9}{{{n^2}}} \cdot \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{n}\left( {24n - 3\left( {n + 1} \right) - \frac{{3n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{2n}}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {72 - 9\left( {1 + \frac{1}{n}} \right) - \frac{{9\left( {n + 1} \right)\left( {2n + 1} \right)}}{{2{n^2}}}} \right) \cr & {\text{Area}} = \mathop {\lim }\limits_{n \to \infty } \left( {72 - 9\left( {1 + \frac{1}{n}} \right) - 9\left( {1 + \frac{3}{n} + \frac{1}{{2n}}} \right)} \right) \cr & {\text{Evaluate the limit when }}n \to \infty \cr & {\text{Area}} = 72 - 9\left( {1 + 0} \right) - 9\left( {1 + 0 + 0} \right) \cr & {\text{Area}} = 72 - 9 - 9 \cr & {\text{Area}} = 54 \cr} $$
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