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 303: 25


$\displaystyle \frac{16}{3}$

Work Step by Step

We treat the x-axis as the graph of $y=f_{1}(x)=0$, and the given function as $y=f(x)=x\sqrt{4-x^{2}}$. The area between the two graphs is over [a,b], where one graph is above the other; it is given as: $\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx$. On $[-2,0] \quad y=0$ is above $y=x\sqrt{4-x^{2}}$. On $[0,2] \quad y=x\sqrt{4-x^{2}}$is above $ y=0$ . so $A=\displaystyle \int_{-2}^{0}[0-x\sqrt{4-x^{2}}]dx++\int_{0}^{2}[x\sqrt{4-x^{2}}-0]dx$ $A=-\displaystyle \int_{-2}^{0}x\sqrt{4-x^{2}}dx+\int_{0}^{2}x\sqrt{4-x^{2}}dx$ $A=-I_{1}+I_{2}$. Applying (to both integrals): $\quad \left[\begin{array}{ll} u=4-x^{2} & du=g'(x)dx\\ & du=-2xdx \\ & xdx=\frac{du}{-2} \end{array}\right]$, The borders change to integral $I_{1}:\quad x=-2 \ \rightarrow\ u=0$ and $\quad x=0 \ \rightarrow\ u=4$ integral $I_{2}:\quad x=0 \ \rightarrow\ u=4$ and $\quad x=2 \ \rightarrow\ u=0$ $I_{1}=$ (after substituing) $=\displaystyle \int_{0}^{4}u^{1/2}(-\frac{du}{2})$ $=-\displaystyle \int_{0}^{4}u^{1/2}du=$ $=-\displaystyle \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{4}$ $=-\displaystyle \frac{2}{3}(\frac{4^{3/2}}{2}-0)$ $=-\displaystyle \frac{2}{3}\cdot\frac{(2^{2})^{3/2}}{2}$ $=-\displaystyle \frac{1}{3}\cdot 8$ $=-\displaystyle \frac{8}{3}$ $I_{2}=$ (after substituing) $=-\displaystyle \int_{4}^{0}u^{1/2}du=$ Reversing the bounds, we find that $I_{2}=-I_{1}=\displaystyle \frac{8}{3}$ $A=-(-\displaystyle \frac{8}{3})+\frac{8}{3}=\frac{16}{3}$
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