## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# 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.

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