Answer
$A(x)=x^2+4x$
Work Step by Step
Let $f(x)=2x+4$
The area function with the lower limit $a+0$ is A(x).
$A(x)=\int_0^x(2t+4)dt$
$A(x)=[t^2+4t]_0^x$
$A(x)=x^2+4x$
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: 2
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