## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# 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.

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