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: 56

Answer

$\displaystyle \frac{12}{5}$

Work Step by Step

Rewrite the equations as functions of y: $\left[\begin{array}{lll} x-y^{2/3}=0 & ... & x+y^{4}=2\\ x=y^{2/3} & & x=2-y^{4}\\ & & \end{array}\right]$ Graphing the given equations, we find the intersections or we solve the equation $f(y)=g(y), $ where $ f(y)=2-y^{4} \quad$ (the curve on the right side) and $ g(y)=y^{2/3}\quad$ (the curve on the left). When $y\in[c,d]=[-1,1]$, the area between the graphs is $A=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=\displaystyle \int_{-1}^{1}[2-y^{4}-y^{2/3}]dy $ $=\displaystyle \left[2y-\frac{y^{5}}{5}-\frac{y^{5/3}}{5/3} \right]_{-1}^{1}$ $=2(1+1)-\displaystyle \frac{(1+1)}{5}-\frac{3(1+1)}{5}$ $=4-\displaystyle \frac{8}{5}$ $=\displaystyle \frac{12}{5}$
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