Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.1 Area Between Two Curves - Exercises Set 5.1 - Page 353: 2

Answer

$$A = \frac{{22}}{3}$$

Work Step by Step

$$\eqalign{ & {\text{From the graph we can note that the area is given by}} \cr & A = \int_0^4 {\left( {\sqrt x - \left( { - \frac{1}{4}x} \right)} \right)} dx \cr & A = \int_0^4 {\left( {{x^{1/2}} + \frac{1}{4}x} \right)} dx \cr & {\text{Integrate}} \cr & A = \left[ {\frac{2}{3}{x^{3/2}} + \frac{1}{8}{x^2}} \right]_0^4 \cr & {\text{Evaluate }} \cr & A = \left[ {\frac{2}{3}{{\left( 4 \right)}^{3/2}} + \frac{1}{8}{{\left( 4 \right)}^2}} \right] - \left[ {\frac{2}{3}{{\left( 0 \right)}^{3/2}} + \frac{1}{8}{{\left( 0 \right)}^2}} \right] \cr & {\text{Simplify}} \cr & A = \frac{{22}}{3} + 0 \cr & A = \frac{{22}}{3} \cr} $$
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