Calculus, 10th Edition (Anton)

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

Chapter 4 - Integration - 4.9 Evaluating Definite Integrals By Substitution - Exercises Set 4.9 - Page 340: 7

Answer

$$\frac{{1192}}{{15}}$$

Work Step by Step

$$\eqalign{ & \int_0^8 {x\sqrt {1 + x} } dx \cr & u = 1 + x,{\text{ then }}du = dx \cr & {\text{With this substitution}}{\text{,}} \cr & {\text{if }}x = 0,{\text{ }}u = 1 + 0 = 1 \cr & {\text{if }}x = 8,{\text{ }}u = 1 + 8 = 9 \cr & {\text{so}} \cr & \int_0^8 {x\sqrt {1 + x} } dx = \int_1^9 {\left( {u - 1} \right)} {u^{1/2}}du \cr & = \int_1^9 {\left( {{u^{3/2}} - {u^{1/2}}} \right)} du \cr & {\text{find the antiderivative by the power rule}} \cr & = \left( {\frac{{{u^{5/2}}}}{{5/2}} - \frac{{{u^{3/2}}}}{{3/2}}} \right)_1^9 \cr & = \left( {\frac{2}{5}{u^{5/2}} - \frac{2}{3}{u^{3/2}}} \right)_1^9 \cr & {\text{part 1 of fundamental theorem of calculus}} \cr & = \left( {\frac{2}{5}{{\left( 9 \right)}^{5/2}} - \frac{2}{3}{{\left( 9 \right)}^{3/2}}} \right) - \left( {\frac{2}{5}{{\left( 1 \right)}^{5/2}} - \frac{2}{3}{{\left( 1 \right)}^{3/2}}} \right) \cr & = \frac{{396}}{5} - \frac{2}{5} + \frac{2}{3} \cr & = \frac{{1192}}{{15}} \cr} $$
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