Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.1 Approximating Areas under Curves - 5.1 Exercises - Page 343: 14

Answer

$((0^2/2 + 4) * 2 + (2^2/2 + 4) * 2 + (4^2/2 + 4) * 2 + (6^2/2 + 4) * 2 + (8^2/2 + 4) * 2 + (10^2/2 + 4) * 2) ft= 268 ft$

Work Step by Step

We begin by computing the length of the sub-intervals. To do this, we take the length of our interval, and divide by the number of sub intervals. (12−0)/6=2. Next, we compute the height of each rectangle. Since this is a left-hand sum, we take the lower bound of each sub interval. (0,2,4,6,8,10). Finally, we evaluate the function at each bound, then multiply by the height of the sub interval, and sum. $(0^2/2 + 4) * 2 + (2^2/2 + 4) * 2 + (4^2/2 + 4) * 2 + (6^2/2 + 4) * 2 + (8^2/2 + 4) * 2 + (10^2/2 + 4) * 2 = 268$ since we are multiplying velocity (ft/sec) by time (sec), our units must be ft. This is logical as we are solving for displacement, which is measured in terms of distance.
Small 1557726176
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.