22.3 The Denite Integral: Qualitative Analysis and Signed Area 731
(a) The lead the chicken has when the goat begins to move
(b) The distance between the goat and the chicken at time 1.5 (c) The distance between the goat and the chicken at time 3 (d) The time at which the goat is farthest ahead of the chicken
(e) Approximate the time(s) at which the chicken and the goat pass one another on the path.
(f) Approximate the time at which the distance between the goat and the chicken is increasing most rapidly. At this time, what is the relationship between and ?
6. Suppose we want to approximate 2 by partitioning the interval [0, 2] into # equal pieces and constructing left- and right-hand sums. Let 3.
(a) Put the following expressions in ascending order.
, "4, 4, "20, 20, "100, 100 0
(b) Find 4 "4.
(c) Find 100 "100.
(d) How large must # be to assure that # "#0.05?
(e) Write out 4, once using summation notation, once without. Evaluate 4.
7. TurnbacktoProblem1ofProblemsforSection22.1onpage723.Expressthefollowing quantities using a denite integral or the sum or difference of denite integrals.
(a) The length of the line at 11:00 a.m. (b) The length of the line at its longest
(c) The number of people who came to the clinic for u shots (d) The number of people actually served by the clinic
(e) The length of the line at 3:00 p.m.
(f) The amount of time a person arriving at noon has to wait in line
(g) The amount of time by which the clinic must extend its hours in order to serve everyone who is in line before 4:00 p.m.
8. Repeat parts (a) through (e) of Problem 6, letting 1.
22.3 THE DEFINITE INTEGRAL: QUALITATIVE ANALYSIS AND SIGNED AREA
InSection22.1weestablishedthatthesignedareaunderarategraphisofparticularinterest to us; this area corresponds to the net change in amount. In Section 22.2 we established that if is a reasonably well-behaved function,10 we can, in theory, nd the signed area under
the graph of on the interval [, ] as follows. Partition the interval [, ] into # equal pieces, each of length , and label as shown on the following page.
10 A continuous function on a closed interval is reasonably well-behaved.
732 CHAPTER 22 Net Change in Amount and Area: Introducing the Denite Integral
t0 t1 t2 tn1 tn
The signed area is dened to be lim# %1 %, provided this limit exists,11 and is denoted by . In this section we interpret as signed area (positive for 0 and negative for 0) and use our knowledge of functions and their graphs, along with a bit of basic geometry, to approximate some denite integrals and evaluate others.12
REMARK: The denite integrals , , ’’, and are all equivalent. Each gives the signed area between the graph of and the horizontal axis from to . The variables , , ’, and are all dummy variables.
EXAMPLE 22.7 The function ! gives the rate at which water is owing into (or out of) a backyard swimming pool. The graph of ! is given below. At time 0 there are 200 gallons
of water in the pool.
r (t) (gal/hr)
120 (5, 120)
5 9 14 20 t (hrs)
(If ’ is the number of gallons of water in the pool at time , where is given in hours, then ! ’.)
Express your answers to the questions below using denite integrals wherever appro-priate and evaluate these integrals.
(a) When is the amount of water in the pool greatest? At that time, how many gallons of water are in the pool?
(b) What is the net ow in or out of the pool between 4 and 12? (c) At what time is the pools water level back at 200 gallons?
(d) Sketch a graph of the amount of water in the pool at time .
SOLUTIONS (a) Theamountofwaterismaximumat 9becausewaterisenteringthepoolfrom 0 to 9 and leaving for greater than 9.
11 If is continuous we can guarantee that this limit exists. 12 We will rely on the following bits of geometry:
The area of a disk is !2.
The area of a trapezoid is 2 (1 (2, where 2 (1 (2 is the average of the heights.
We can think of a triangle as a trapezoid with one height of length zero and a rectangle as a trapezoid with both heights of equal length.
22.3 The Denite Integral: Qualitative Analysis and Signed Area 733
The amount of water entering the pool on the interval [0, 9] is 9 !, the area under the graph from 0 to 9.
! the area of the rectangle the area of the triangle 0
5120124120 600 240
5 9 t
840 gallons of water have entered the pool in those 9 hours. Adding this to the original 200 gallons gives a total of 1040 gallons of water in the pool.
(b) The net ow in or out of the pool between 4 and 12 is given by 12 !.
Compute this by summing the signed area under ! from 4 to 5, from 5 to 9, and from 9 to 12. In other words, compute 12 ! by expressing it as follows.
12 5 9 12
! ! ! ! 4 4 5 9
area of rectangle area of triangle ( area of triangle) 1201121204123height of triangle13 120 240 12390
360 135 225
The pool has a net gain of 225 gallons of water.
4 5 9 90
13 The slope of the line forming the triangles is 120430, so between 9 and 12 the value of ! drops by 90, making the height of this triangle 90.
734 CHAPTER 22 Net Change in Amount and Area: Introducing the Denite Integral
(c) There are many ways to approach this question. From part (a), we know that by time 9thepoolhasgained840gallonsofwater.After 9thepoolloseswater.Between 9and 14thepoolhaslost125150375gallonsofwater, soitnowhasa net gain of 840 gallons 375 gallons = 465 gallons. Subsequently water is owing out atarateof150gallonsperhour.In morehoursthepoolwillhavelostanadditional150 gallons of water. For what will 150 465? When 465150 3.1. Therefore, at time 14 3.1 17.1, (17 hours and 6 minutes) the pool will be back at its original level of 200 gallons. Using integral notation, we have solved the following equation for .
Or, equivalently, ! !. 0 9
areas must be equal
5 9 t
(d) Here is a graph of ’, where ’ is the amount of water in the pool at time .
(9, 1040) 1000
800 (5, 800) 600
5 9 14 17.1 t(hrs)
EXAMPLE 22.8 Evaluate the following denite integrals by interpreting each as a signed area.
(a) 2 cos
(c) 7 421
(d) 2 2
SOLUTIONS (a) 2 cos 0. The positive area and the negative area cancel.
22.3 The Denite Integral: Qualitative Analysis and Signed Area 735
3 2 x
2 2 1
(b) sin 0. Sine is an odd function (sin sin , so again the negative and positive areas cancel regardless of the value of .
(c) The integrand is an odd function: 421 421. Thus, the negative and positive areas cancel and we conclude that 7 421 0.
.4 y = x4 + x2 + 1
7 7 x
(d) Sketch the graph of 2. The area of the triangle lying below the -axis is 12424.Therefore, 2 2 4.Theareaofthetrianglelyingabove
the -axis between 2 and 3 is 121112. We conclude that
3 2 3 2 2 2
2 2 2
4 0.5 3.5.