Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 395: 25

Answer

$\int_1^3\sqrt{4+x^2}dx=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\sqrt{\frac{5n^2+4in+4i^2}{n^2}}$

Work Step by Step

Step 1: Find $\Delta x$ and $x_i=a+i\Delta x$. The lower bound: $a=1$ The upper bound: $b=3$ $\Delta x=\frac{b-a}{n}=\frac{3-1}{n}=\frac{2}{n}$ $x_i=1+i\cdot \frac{2}{n}$ $x_i=1+\frac{2i}{n}$ Step 2: Find $R_n$. $R_n=\sum_{i=1}^n\sqrt{4+x_i^2}\cdot \frac{2}{n}$ $R_n=\sum_{i=1}^n\sqrt{4+(1+\frac{2i}{n})^2}\cdot \frac{2}{n}$ $R_n=\sum_{i=1}^n\sqrt{4+1+\frac{4i}{n}+\frac{4i^2}{n^2}}\cdot \frac{2}{n}$ $R_n=\sum_{i=1}^n\frac{2}{n}\sqrt{\frac{5n^2+4in+4i^2}{n^2}}$ Step 3: Write the define as a limit of Riemann sum using right endpoints. $\int_1^3\sqrt{4+x^2}dx=\lim\limits_{n \to \infty}R_n$ $\int_1^3\sqrt{4+x^2}dx=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\sqrt{\frac{5n^2+4in+4i^2}{n^2}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.