## Calculus with Applications (10th Edition)

Published by Pearson

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

#### Answer

$$13$$

#### Work Step by Step

\eqalign{ & \int_0^2 {3\sqrt {4u + 1} } du \cr & {\text{use substitution}}{\text{. Let }}t = 4u + 1,{\text{ so that }}dt = 4du \cr & {\text{the new limits on }}t{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}u = 2,{\text{ then }}t = 4\left( 2 \right) + 1 = 9 \cr & \,\,\,\,\,\,{\text{If }}u = 0,{\text{ then }}t = 4\left( 0 \right) + 1 = 1 \cr & {\text{Then}} \cr & \int_0^2 {3\sqrt {4u + 1} } du = \int_1^9 {3\sqrt t } \left( {\frac{1}{4}dt} \right) \cr & = \frac{3}{4}\int_1^9 {\sqrt t } dt \cr & = \frac{3}{4}\int_1^9 {{t^{1/2}}} dt \cr & {\text{integrate}} \cr & = \frac{3}{4}\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right)_1^9 \cr & = \frac{1}{2}\left( {{t^{3/2}}} \right)_1^9 \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( 9 \right)}^{3/2}} - {{\left( 1 \right)}^{3/2}}} \right) \cr & = \frac{1}{2}\left( {27 - 1} \right) \cr & = 13 \cr}

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