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 396: 60

Answer

$\int_{0}^{5}f(x)~dx = 17$

Work Step by Step

We can evaluate the integral: $\int_{0}^{5}f(x)~dx$ $=\int_{0}^{3}f(x)~dx+\int_{3}^{5}f(x)~dx$ $=\int_{0}^{3}3~dx+\int_{3}^{5}x~dx$ $=3(3-0)+(\frac{5^2-3^2}{2})$ $= 9+8$ $= 17$
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.