## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 6 - Review - Exercises: 6

#### Answer

$\frac{10}{3}-\frac{4\sqrt[]{2}}{3}$

#### Work Step by Step

Step 1 of 3: First, find the points of intersection of the curves $y=\sqrt[]{x}$ and $y=x^{2}$ before $x=2$ $x^{2}=\sqrt[]{x}$ $x^{4}-x=0$ or $x=0$ or $x=1$ The points of intersection are (0, 0) and (1, 1). Step 2 of 3: Sketch the curves. Find the area of the shaded region. Step 3 of 3: Area of the shaded region $A= \int _{0}^{1} \left( \sqrt[]{x}-x^{2} \right) dx+ \int _{1}^{2} \left( x^{2}-\sqrt[]{x} \right) dx$ $A= \int _{0}^{1} \left( x^{\frac{1}{2}}-x^{2} \right) dx+ \int _{1}^{2} \left( x^{2}-x^{\frac{1}{2}} \right) dx$ $A= \left[ \frac{2}{3}x^{\frac{3}{2}}-\frac{1}{3}x^{3} \right] _{0}^{1}+ \left[ \frac{1}{3}x^{3}-\frac{2}{3}x^{\frac{3}{2}} \right] _{1}^{2}$ By the fundamental Theorem of calculus $A= \left[ \frac{2}{3}-\frac{1}{3} \right] - \left[ \frac{8}{3}-\frac{4}{3}\sqrt[]{2}-\frac{1}{3}+\frac{2}{3} \right]$ $A=\frac{10}{3}-\frac{4\sqrt[]{2}}{3}$

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