Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.2 Exercises - Page 265: 69

Answer

(a) 15.33 (b) 21.73 (c) 18.44

Work Step by Step

(a) For part A we’ve been instructed to find the lower sum of the figure when n=4. Wondering where the “n” came from? it’s not necessarily a variable, its simply letting you know that they want you to solve the sum using 4 rectangles. Start off by redrawing the graph they’ve given us. Limited to $0 \leq x \leq$ 4 we are given easy separation for our four future rectangles. Now, lower sum means we’ll be underestimating the area of the graph. The LEFT endpoint of each rectangle will determine the height. start at x=0, move to x=1. Not a rectangle, so thats an easy height, and therefore area, of 0. Now move to x=1. We need to find the height of our rectangle! Thats where our formula of f(x) = 8x/(x+1) comes in handy. Plug 1 into the equation and you get a height of y=4. Multiply it by the width (which is 1 in this case) and this will make a rectangle with an area of 4. Repeat this process for x=2 and x=3. You should get an area of 16/3 (or 5.33) for x=2 and an area of 6 for x=3. We don’t need the height of x=4 in this case since we’re underestimating the area, so the height of x=3 determines the area of the last rectangle. Now add all four areas together to get a sum of 15.33, and that’s the answer to part A. (b) For part B we’re going to pretty much the same thing as we did in part A, only this time the height of each rectangle will be determined by the RIGHT endpoint. So this time we start at x=4. Find the height by plugging 4 into our equation and you should get 32/5 (or 6.4). Multiply by our width of 1 and the area is 6.4. The rest is quiet simple since we did the math in part A. Use the height x=3 to get a rectangle with an area of 6, from x=2 to get a rectangle with an area of 16/3 (5.33), and from x=1 to get a rectangle with an an area of 4. Add them together and you should get 21.73, which is the upper sum, or an overestimate, of our area. (c) Part C is no different than A or B, so don’t let the Midpoint Rule intimidate you. All we are doing is taking the height from the midpoint of each rectangle rather than an endpoint. All we need to do is find our y-value when x=0.5, 1.5, 2.5, and 3.5. Do this the same way we found the height last time - by plugging each x-value into the formula. You should get 4/1.5 (2.67), 9.6/2.5 (3.84), 20/3.5 (5.71), and 28/4.5 (6.22). Make your rectangles with a width of 1 and multiply by the height for your areas. Now add all of them up for a midpoint sum of 18.44.
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