Calculus: Early Transcendentals 8th Edition

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: 40



Work Step by Step

On the interval $0 \leq x \leq 1$, the graph of $\vert 2x-1 \vert$ forms two triangles above the x-axis. We can find the area of these two triangles: $A = \frac{1}{2}(\frac{1}{2})(1)+ \frac{1}{2}(\frac{1}{2})(1)$ $A = \frac{1}{4}+\frac{1}{4}$ $A = \frac{1}{2}$ Therefore: $\int_{0}^{1} \vert 2x-1 \vert~dx = \frac{1}{2}$
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