University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 73



Work Step by Step

$$A=\int^1_{-1}(3x^2-4x+7)dx$$ $$A=\Big(\frac{3x^3}{3}-\frac{4x^2}{2}+7x\Big)\Big]^1_{-1}$$ $$A=\Big(x^3-2x^2+7x\Big)\Big]^1_{-1}$$ $$A=(1^3-2\times1^2+7\times1)-\Big((-1)^3-2\times(-1)^2+7\times(-1)\Big)$$ $$A=(1-2+7)-(-1-2-7)$$ $$A=6+10=16$$
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.