Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 35


$y=\int_5^x (\frac{sin(u)}{u})du -3$

Work Step by Step

Step 1. Letting $f(x)=\frac{sin(x)}{x}$ and $F(x)=\int_a^x f(u)du+C$, we have $F'(x)=f(x)$ where $C$ is a constant. Step 2. In the case of the exercise, we can write $y=\int_a^x f(u)du +C$, where $C$ is a constant. Step 3. Given $y(5)=-3$, we have $f(5)=\int_a^5 f(u)du +C=-3$; thus we can let $a=5, C=-3$. Step 4. The solution can be written as $y=\int_5^x f(u)du -3$
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