Answer
$\int_{0}^{4} (x^2-4x+4)~dx \geq 0$
Work Step by Step
On the interval $0 \leq x \leq 4$:
$x^2-4x+4 = (x-2)^2 \geq 0$
Therefore, by the comparison property:
$\int_{0}^{4} (x^2-4x+4)~dx \geq 0$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 390: 55
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