Calculus (3rd Edition)

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



Work Step by Step

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