Answer
$H(-2)$ = $0$
$H'(-2)$ = $\frac{1}{5}$
Work Step by Step
$H(x)$ = $\int_{-2}^x\frac{du}{u^{2}+1}$
$H(-2)$ = $\int_{-2}^{-2}\frac{du}{u^{2}+1}$ = $0$
$H'(x)$ = $\frac{1}{x^{2}+1}$
$H'(-2)$ = $\frac{1}{(-2)^{2}+1}$ = $\frac{1}{5}$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 5 - The Integral - 5.5 The Fundamental Theorem of Calculus, Part II - Exercises - Page 262: 6
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