Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 304: 35


$=\displaystyle \frac{5}{6}$

Work Step by Step

On the interval $x\in[0,1]$, the graph of $y= x$ is above the graph of $y=\displaystyle \frac{x^{2}}{4}$ On the interval $x\in[1,2]$, the graph of $y= 1$ is above the graph of $y=\displaystyle \frac{x^{2}}{4}$ The area between the two graphs over [a,b], where one graph is above the other, is given with $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dt$. Thus, $A=\displaystyle \int_{0}^{1}[x-\frac{x^{2}}{4}]dx+\int_{1}^{2}[1-\frac{x^{2}}{4}]dx$ $=[\displaystyle \frac{x^{2}}{2}]_{0}^{1}-\frac{1}{4}[\frac{x^{3}}{3}]_{0}^{1}+[x]_{1}^{2}-\frac{1}{4}[\frac{x^{3}}{3}]_{1}^{2}$ $=(\displaystyle \frac{1}{2}-0)-\frac{1}{12}(1-0)+(2-1)-\frac{1}{12}(2^{3}-1)$ $=\displaystyle \frac{1}{2}-\frac{1}{12}+1-\frac{7}{12}$ $=\displaystyle \frac{6-1-7+12}{12}$ $=\displaystyle \frac{10}{12}$ $=\displaystyle \frac{5}{6}$
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