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: 58


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

Work Step by Step

With respect to $y$, the formulas of the curves are - Curve $x+y=2$: $x=2-y$ - Curve $y=x^2$: $x=\pm\sqrt y=\pm y^{1/2}$ However, from the graph, we can see that we work only with the positive values of $x$, so there is no need for the negative sign, meaning that $x=y^{1/2}$. 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 $x=2-y$ and on the left by the curve $x=y^{1/2}$ and runs from $y=0$ to $y=1$. Therefore, the area of the shaded region is $$A=\int^{1}_{0}(2-y-y^{1/2})dy$$ $$A=\Big(2y-\frac{y^2}{2}-\frac{2y^{3/2}}{3}\Big)\Big]^{1}_{0}$$ $$A=2-\frac{1}{2}-\frac{2}{3}$$ $$A=\frac{5}{6}$$
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