Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 12


a)-$\sqrt{2}$ b)$\sqrt{2}$ c)-$\sqrt{2}$ d)1

Work Step by Step

a)$\int_0^{-3} g(t) dt$=-$\int_0^{3} g(t) dt$=-$\sqrt{2}$ b)$\int^0_{-3} g(t) dt$=$\int^0_{-3} g(t) dt$=$\sqrt{2}$ c)$\int^0_{-3} g(t) dt$[-g(x)]dx=-$\int^0_{-3} g(x) dx$=-$\sqrt{2}$ d)$\int^0_{-3} \frac{g(t)}{\sqrt{2}} dr$=$\frac{1}{\sqrt{2}}$$\int^0_{-3} g(t) dt$=($\frac{1}{\sqrt{2}})(\sqrt{2})=1$
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.