Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.4 The Fundamental Theorem of Calculus - 7.4 Exercises - Page 395: 27

Answer

$$49$$

Work Step by Step

$$\eqalign{ & \int_0^8 {{x^{1/3}}\sqrt {{x^{4/3}} + 9} } dx \cr & = \int_0^8 {\sqrt {{x^{4/3}} + 9} \left( {{x^{1/3}}} \right)} dx \cr & {\text{use substitution}}{\text{. Let }}u = {x^{4/3}} + 9,{\text{ so that }}du = \frac{4}{3}{x^{1/3}}dx,\,\,\,\,\,\,\,\frac{3}{4}du = {x^{1/3}}du \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 8,{\text{ then }}u = {\left( 8 \right)^{4/3}} + 9 = 25 \cr & \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = {\left( 0 \right)^{4/3}} + 9 = 9 \cr & {\text{Then}} \cr & \int_0^8 {\sqrt {{x^{4/3}} + 9} \left( {{x^{1/3}}} \right)} dx = \int_9^{25} {\sqrt u \left( {\frac{3}{4}du} \right)} \cr & = \frac{3}{4}\int_9^{25} {\sqrt u du} \cr & = \frac{3}{4}\int_9^{25} {{u^{1/2}}du} \cr & {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr & = \frac{3}{4}\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_9^{25} \cr & = \frac{3}{4}\left( {\frac{{2{u^{3/2}}}}{3}} \right)_9^{25} \cr & = \frac{1}{2}\left( {{u^{3/2}}} \right)_9^{25} \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = \frac{1}{2}\left( {{{\left( {25} \right)}^{3/2}} - {{\left( 9 \right)}^{3/2}}} \right) \cr & = \frac{1}{2}\left( {125 - 27} \right) \cr & = \frac{{98}}{2} \cr & = 49 \cr} $$
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