Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 434: 28

Answer

$\begin{equation}\frac{3}{2}\end{equation}$

Work Step by Step

The graph as depicted in the answer can be divided into the parts with $\begin{equation}0\leq x\leq1\end{equation}$ and $\begin{equation}1\leq x\leq2\end{equation}$. The left part is bounded on the top by $\begin{equation}2x^{2}\end{equation}$ and on the bottom by $\begin{equation}\frac{1}{4}x^{2}\end{equation}$. The right part is bounded by $\begin{equation}-x+3\end{equation}$ on the top and $\begin{equation}\frac{1}{4}x^{2}\end{equation}$ on the bottom. So we have two integrals. $\begin{equation} \int_{0}^{1}\left[\left(2x^{2}\right)-\left(\frac{1}{4}x^{2}\right)\right]dx+\int_{1}^{2}\left[\left(-x+3\right)-\left(\frac{1}{4}x^{2}\right)\right]dx\\ =\int_{0}^{1}\frac{7}{4}x^2dx+\int_{1}^{2}\left(-\frac{1}{4}x^2-x+3\right)dx\\ =\frac{7}{4}\int_0^1 x^2 dx-\frac{1}{4}\int_1^2x^2dx-\int_1^2xdx+3\int_1^21dx\\ =\frac{7}{4}\left[\frac{1}{3}x^3\right]_0^1-\frac{1}{4}\left[\frac{1}{3}x^3\right]_1^2-\left[\frac{1}{2}x^2\right]_1^2+3\left[x\right]_1^2\\ =\frac{7}{4}\left[\frac{1}{3}(1)^3-\frac{1}{3}(0)^3\right]-\frac{1}{4}\left[\frac{1}{3}(2)^3-\frac{1}{3}(1)^3\right]-\left[\frac{1}{2}(2)^2-\frac{1}{2}(1)^2\right]+3\left[(2)-(1)\right]\\ =\frac{7}{4}\left(\frac{1}{3}\right)-\frac{1}{4}\left(\frac{7}{3}\right)-\frac{3}{2}+3\\ =\frac{7}{12}-\frac{7}{12}-\frac{3}{2}+3\\ =\frac{3}{2} \end{equation}$
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