Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises - Page 422: 3

Answer

$\int_{0}^{1} (x+\sqrt{1-x^2})~dx = \frac{\pi+2}{4}$

Work Step by Step

$\int_{0}^{1} (x+\sqrt{1-x^2})~dx = \int_{0}^{1} x~dx+\int_{0}^{1} \sqrt{1-x^2}~dx$ On the interval $0 \leq x \leq 1$, the graph of $f(x) = x$ forms a triangle above the x-axis. We can find the area of this triangle: $A = \frac{1}{2}(1)(1) = \frac{1}{2}$ On the interval $0 \leq x \leq 1$, the graph of $\sqrt{1-x^2}$ forms a quarter of a circle above the x-axis. We can find the area: $A = \frac{\pi~r^2}{4} = \frac{\pi(1)^2}{4} = \frac{\pi}{4} $ Therefore: $\int_{0}^{1} (x+\sqrt{1-x^2})~dx = \frac{1}{2}+ \frac{\pi}{4} = \frac{\pi+2}{4}$
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