Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 60

Answer

$(-1,1)$

Work Step by Step

Given \begin{aligned}F(x)&=\int_1^xf(t)dt \end{aligned} To find the interval of concave downward for $F''(x)$, since \begin{aligned} F'(x)& = \frac{d}{dx}\int_1^xf(t)dt \\ &=f(x) \end{aligned} Then \begin{aligned} F''(x)&= f'(x) \end{aligned} Hence the interval of concave downward for $F''(x)$ is the decreasing interval for $f(x)$, from the given graph, we know that f(x) is decreasing on $(-1,1)$. It follows that $F(x)$ is concave downward on $(-1,1)$
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