Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 327: 22

Answer

$\pi t+C$

Work Step by Step

$\int \pi dt = \pi \int dt$ (constant is taken out) = $\pi\times t+C$ (As $\int dt= t+constant$) = $\pi t+C$ Now, $\frac{d}{dx}(\pi t+C)= \frac{d}{dx}\pi t + \frac{d}{dx}C = \pi +0 = \pi$ Here, derivative of the integral is the integrand. So the answer is correct.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.