Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 238: 17



Work Step by Step

$\int(x+1)dx= \int xdx +\int1dx$ $\int xdx = \frac{x^{2}}{2}+ C_{1}$ (As $\int x^{n}dx= \frac{x^{n+1}}{n+1}+ C_{1})$. $\int1dx = x+C_{2}$ (As $\frac{d}{dx}(x+C_{2})=1$). Now, $\int(x+1)dx= \frac{x^{2}}{2}+x+(C_{1}+C_{2}) $= $\frac{x^{2}}{2}+x+C$ (Note that $C_{1}$ and $C_{2}$ are arbitrary constants and C is the sum of these two constants which is again a constant). $\frac{d}{dx}(\frac{x^{2}}{2}+x+C)= \frac{2x}{2}+1+0$= x+1. Here derivative of the integral is the integrand. Hence the answer is correct.
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