Answer
$A(x)=(x+2)^2$
Work Step by Step
Let $f(x) =2x+4$
The area function with a lower limit $a=-2$ is $A(x)$.
$A(x)=\int_{-2}^x(2t+4)dt$
$A(x)=[t^2+4t]_{-2}^x$
$A(x)=(x^2+4x)-((-2)^2-4(2))$
$A(x)=x^2+4x+4$
$A(x)=(x+2)^2$
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: 1
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