Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises: 35

Answer

$$\int^1_0(x^{10}+10^x)dx=\frac{1}{11}+\frac{9}{\ln10}$$

Work Step by Step

$$A=\int^1_0(x^{10}+10^x)dx$$ According to Table 1, we have $$\int x^ndx=\frac{x^{n+1}}{n+1}+C (n\ne-1)$$ $$\int b^xdx=\frac{b^x}{\ln|b|}+C$$ Therefore, $$A=(\frac{x^{11}}{11}+\frac{10^x}{\ln10})\Bigg]^1_0$$ $$A=(\frac{1^{11}}{11}+\frac{10^1}{\ln10})-(\frac{0^{11}}{11}+\frac{10^0}{\ln10})$$ $$A=(\frac{1}{11}+\frac{10}{\ln10})-(0+\frac{1}{\ln10})$$ $$A=\frac{1}{11}+\frac{9}{\ln10}$$
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