University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 57


The area of the shaded region is $5/6$.

Work Step by Step

With respect to $y$, the formulas of the curves are - Curve $y=x$: $x=y$ - Curve $y=\frac{x^2}{4}$: $x=\pm\sqrt{4y}=\pm2\sqrt y=\pm2y^{1/2}$ However, from the graph, we can see that we work only with the positive values of $x$, so $x=2y^{1/2}$. In this exercise, it might be easier if we carry out integration with respect to $y$ by the formula $$A=\int^d_c[f(y)-g(y)]dy$$ to calculate the area, in which $f$ denotes the right-hand curve and $g$ the left-hand curve. Here, the shaded region is bounded on the right by the curve $x=2y^{1/2}$ and on the left by the curve $x=y$ and runs from $y=0$ to $y=1$. Therefore, the area of the shaded region is $$A=\int^{1}_{0}(2y^{1/2}-y)dy$$ $$A=\Big(\frac{2y^{3/2}}{\frac{3}{2}}-\frac{y^2}{2}\Big)\Big]^{1}_{0}=\Big(\frac{4y^{3/2}}{3}-\frac{y^2}{2}\Big)\Big]^{1}_{0}$$ $$A=\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$$
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