## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 279: 9

#### Answer

$$\left( {\bf{a}} \right)\frac{{{x^9}}}{9} + C,\left( {\bf{b}} \right)\frac{{7{x^{12/7}}}}{{12}} + C,\left( {\bf{c}} \right)\frac{{2{x^{9/2}}}}{9} + C$$

#### Work Step by Step

\eqalign{ & \left( {\bf{a}} \right)\int {{x^8}} dx \cr & {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^8}} dx = \frac{{{x^{8 + 1}}}}{{8 + 1}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^8}} dx = \frac{{{x^9}}}{9} + C \cr & \cr & \left( {\bf{b}} \right)\int {{x^{5/7}}} dx \cr & {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^{5/7}}} dx = \frac{{{x^{5/7 + 1}}}}{{5/7 + 1}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^{5/7}}} dx = \frac{{{x^{12/7}}}}{{12/7}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^{5/7}}} dx = \frac{{7{x^{12/7}}}}{{12}} + C \cr & \cr & \left( {\bf{c}} \right)\int {{x^3}\sqrt x } dx \cr & {\text{rewriting the integrand}} \cr & \,\,\,\,\,\,\,\, = \int {{x^3}{x^{1/2}}} dx \cr & \,\,\,\,\,\,\,\, = \int {{x^{7/2}}} dx \cr & {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^{7/2}}} dx = \frac{{{x^{7/2 + 1}}}}{{7/2 + 1}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^{7/2}}} dx = \frac{{{x^{9/2}}}}{{9/2}} + C \cr & \,\,\,\,\,\,\,\,\int {{x^{7/2}}} dx = \frac{{2{x^{9/2}}}}{9} + C \cr}

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