## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 6 - Review - Exercises: 1

#### Answer

$\frac{8}{3}$

#### Work Step by Step

Step 1 of 2: Consider the function $y=x^{2},y=4x-x^{2}.$ The region enclosed by the curves $y=x^{2},y=4x-x^{2}$ is shown below: Step 2 of 2: From the above graph the point of intersection of the curves $y=x^{2},y=4x-x^{2}$ are $\left( 0,0 \right) and$ $\left( 2,4 \right)$ . Consider $y=4x-x^{2}=f \left( x \right)$ $y=x^{2}=g \left( x \right)$ The area between the curves $y=f \left( X \right) ,y=g \left( x \right)$ and between x=a and x=b is $A= \int _{0}^{2} \vert f \left( x \right) -g \left( x \right) \vert dx$ . Here the area of the region bounded by the curves $y=x^{2},y=4x-x^{2}$ and between $x=0$ and $x=2$ is $A= \int _{0}^{2} \vert 4x-x^{2}-x^{2} \vert dx$ $= \vert 2x^{2}-\frac{2x^{3}}{3} \vert _{0}^{2}$ $= \vert \left( 2 \left( 2 \right) ^{2}-\frac{2 \left( 2 \right) ^{3}}{3} \right) - \left( 2 \left( 0 \right) ^{2}-\frac{2 \left( 0 \right) ^{3}}{3} \right) \vert$ $=\frac{8}{3}$ Hence the required area of the given region is $\frac{8}{3}$.

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