## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 5 - Integration - 5.4 Working with Integrals - 5.4 Exercises: 13

#### Answer

$$0$$

#### Work Step by Step

\eqalign{ & \int_{ - 10}^{10} {\frac{x}{{\sqrt {200 - {x^2}} }}dx} \cr & {\text{testing for symmetry}} \cr & f\left( x \right) = \frac{x}{{\sqrt {200 - {x^2}} }} \cr & f\left( { - x} \right) = \frac{{\left( { - x} \right)}}{{\sqrt {200 - {{\left( { - x} \right)}^2}} }} \cr & f\left( { - x} \right) = - \frac{x}{{\sqrt {200 - {x^2}} }} \cr & f\left( { - x} \right) = - f\left( x \right){\text{ so the function }}\frac{x}{{\sqrt {200 - {x^2}} }}{\text{ is odd}} \cr & {\text{Use the theorem 5}}{\text{.4}} \cr & {\text{If }}f\left( x \right){\text{ is odd}}{\text{, }}\int_{ - a}^a {f\left( x \right)dx} = 0 \cr & then \cr & \int_{ - 10}^{10} {\frac{x}{{\sqrt {200 - {x^2}} }}dx} = 0 \cr}

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