## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 4 - Integration - 4.9 Evaluating Definite Integrals By Substitution - Exercises Set 4.9: 22

#### Answer

$\dfrac{38}{15}$

#### Work Step by Step

In order to integrate this function, we have to use an «u» substitution. Choose an u, derivate it: $u=5x-1$ $du = 5dx$ $\dfrac{1}{5}du=dx$ Change limits: $x_1 =1$ $u_1=5(1)-1=4$ $x_2 =2$ $u_2=5(2)-1=9$ Then substitute: $\dfrac{1}{5} \int_{4}^{9} \sqrt{u} du = \dfrac{1}{5} \int_{4}^{9} u^{1/2}du$  Then integrate $\dfrac{1}{5} \bigg [\dfrac{2}{3}u^{2/3} \bigg ]_{4}^{9} = \dfrac{2}{15} \bigg [u^{2/3} \bigg ]_{4}^{9}$ Apply calculus 2nd theorem and reduce: $\dfrac{2}{15} [ 9^{3/2} -4^{3/2}] = \dfrac{2}{15}[27-8]=\dfrac{38}{15}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.