Chapter 4 Overview: Definite Integrals In this chapter, we ...
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225 Chapter 4 Overview: Definite Integrals In this chapter, we will study the Fundamental Theorem of Calculus, which establishes the link between the algebra and the geometry, with an emphasis on the mechanics of how to find the definite integral. We will consider the differences implied between the context of the definite integral as an operation and as an area accumulator. We will learn some approximation techniques for definite integrals and see how they provide theoretical foundation for the integral. We will revisit graphical analysis in terms of the definite integral and view another typical AP context for it. Finally, we will consider what happens when trying to integrate at or near an asymptote.
Transcript of Chapter 4 Overview: Definite Integrals In this chapter, we ...
225
Chapter 4 Overview: Definite Integrals In this chapter, we will study the Fundamental Theorem of Calculus, which establishes the link between the algebra and the geometry, with an emphasis on the mechanics of how to find the definite integral. We will consider the differences implied between the context of the definite integral as an operation and as an area accumulator. We will learn some approximation techniques for definite integrals and see how they provide theoretical foundation for the integral. We will revisit graphical analysis in terms of the definite integral and view another typical AP context for it. Finally, we will consider what happens when trying to integrate at or near an asymptote.
226
4.1 The Fundamental Theorem of Calculus As noted in the overview, Anti-derivatives are known as Indefinite Integrals and this is because the answer is a function, not a definite number. But there is a time when the integral represents a number. That is when the integral is used in an Analytic Geometry context of area. Though it is not necessary to know the theory behind this to be able to do it, the theory is a major subject of Integral Calculus, so we will explore it briefly. We know, from Geometry, how to find the exact area of various polygons, but we never considered figures where one side is not made of a line segment. Here we want to consider a figure where one side is the curve y = f (x) and the other sides are the x-axis and the lines x = a and x = b.
1 2 3 4
1
2
3
x
y
As we can see above, the area can be approximated by rectangles whose height is the y-value of the equation and whose width we will call xΔ . The more rectangles we make, the better the approximation. The area of each rectangle would be
f x( ) ⋅ Δx and the total area of n rectangles would be A= f xi( )⋅Δx
i=1
n
∑ . If we could
make an infinite number of rectangles (which would be infinitely thin), we would have the exact area. The rectangles can be drawn several ways--with the left side at the height of the curve (as drawn above), with the right side at the curve, with
y = f (x)
a b
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the rectangle straddling the curve, or even with rectangles of different widths. But once they become infinitely thin, it will not matter how they were drawn--they will have virtually no width (represented by dx instead of the xΔ ) and a height equal to the y-value of the curve. We can make an infinite number of rectangles mathematically by taking the Limit as n approaches infinity, or
Limn→∞ f xi( )⋅Δx
i=1
n
∑ .
This limit is rewritten as the Definite Integral:
! f x( ) !dxa
b
∫ b is the "upper bound" and a is the "lower bound," and would not mean much if it were not for the following rule: The Fundamental Theorem of Calculus If f x( ) is a continuous function on
a,b⎡⎣ ⎤⎦ , then
1) d dx
f t( ) dtc
x
∫ = f x( ) or d dx
f t( ) dtc
u
∫ = f u( )iDu
2) If F ' x( ) = f x( ) , then f x( ) dx
a
b
∫ = F b( )− F a( ) . The first part of the Fundamental Theorem of Calculus simply says what we already know--that an integral is an anti-derivative. The second part of the Fundamental Theorem says the answer to a definite integral is the difference between the anti-derivative at the upper bound and the anti-derivative at the lower bound. This idea of the integral meaning the area may not make sense initially, mainly because we are used to Geometry, where area is always measured in square units. But that is only because the length and width are always measured in the same kind of units, so multiplying length and with must yield square units. We are expanding
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our vision beyond that narrow view of things here. Consider a graph where the x-axis is time in seconds and the y-axis is velocity in feet per second. The area under the curve would be measured as seconds multiplied by feet/sec--that is, feet. So the area under the curve equals the distance traveled in feet. In other words, the integral of velocity is distance. Objectives Evaluate Definite Integrals Find the average value of a continuous function over a given interval Differentiate integral expressions with the variable in the boundary Let us first consider Part 2 of the Fundamental Theorem, since it has a very practical application. This part of the Fundamental Theorem gives us a method for evaluating definite integrals. Ex 1 Evaluate
4x+3( )dx
2
8
∫
4x + 3( )dx
2
8
∫ = 2x2 + 3x2
8 The antiderivative of 4x + 3 is 2x2 + 3x. We
use this notation when we apply the Fundamental Theorem
=
128+ 24( )− 8+ 6( )⎡⎣
⎤⎦ Plug the upper limit of integration into the
antiderivative and subtract it from the lower limit when plugged into the antiderivative
= 138 Check with Math 9.
Ex 2 Evaluate
1x
dx1
4
∫
1x
dx1
4
∫ = 2 x1
4
= 4( )− 2( )⎡⎣
⎤⎦
= 2
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Ex 3 Evaluate
sin xdx0
π2∫
sin xdx0
π2∫ = −cos x
0
π2
= − 0−1⎡⎣ ⎤⎦=1
Ex 4 Evaluate
4+u2
u3 du1
2
∫
4+u2
u3 du1
2
∫ = 4u−3 +u−1( )du1
2
∫= −2u−2 + ln u
1
2
= −2 ⋅ 122 + ln2⎛
⎝⎜⎞⎠⎟− −2+ ln1( )⎡
⎣⎢⎢
⎤
⎦⎥⎥
= 32 + ln2
Ex 5 Evaluate
1x3 dx
−5
5
∫
This is a trick! We use the Fundamental Theorem on this integral because
the curve is not continuous on ,x a b⎡ ⎤⎣ ⎦∈ . When x = 0, 31x
does not exist, so
the Fundamental Theorem of Calculus does not apply. One simple application of the definite integral is the Average Value Theorem. We all recall how to find the average of a finite set of numbers, namely, the total of the numbers divided by how many numbers there are. But what does it mean to take the average value of a continuous function? Let’s say you drive from home to school – what was your average velocity (velocity is continuous)? What was the average temperature today (temperature is continuous)? What was your average height for the first 15 years of your life (height is continuous)? Since the integral is the sum of infinite number of function values, the formula below answers those questions.
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Average Value Formula: The average value of a function f on a closed interval
a,b⎡⎣ ⎤⎦ is defined by
favg =
1b− a
f x( )dxa
b
∫
If we look at this formula in the context of the Fundamental Theorem of Calculus, it can start to make a little more sense.
f x( ) dxa
b
∫ = F b( )− F a( ) The Fundamental Theorem of Calculus
( ) ( ) ( )1 1 b
af x dx F b F a
b a b a⎡ ⎤⎣ ⎦−
− −=∫
( ) ( )F b F ab a−−
=
Notice that this is just the average slope for ( )F x on ,x a b⎡ ⎤⎣ ⎦∈ . The average slope of ( )F x would be the average value of ( )'F x . But since the definition in the Fundamental Theorem of Calculus says that ( ) ( )'F x f x= , this is actually just the average value of ( )f x .
Ex 6 Find the average value of f x( ) = x2 +1 on
0,5⎡⎣ ⎤⎦ .
favg =
1b− a
f x( )dxa
b
∫
favg =
15− 0
x2 +1( )dx0
5
∫ Substitute in the function and interval
favg = 28
3 Use Math 9 (or integrate analytically) to calculate answer
231
Ex 7 Find the average value of h θ( ) = secθ tanθ on
0,π
4⎡
⎣⎢
⎤
⎦⎥ .
havg =1
b− a h θ( )dθa
b
∫= 1π4 − 0
secθ tanθdθ0
π4∫
= .524
Now let’s take a look at the First Part of the Fundamental Theorem of Calculus:
ddx
f t( )a
x
∫ dt = f x( ) or ddx
f t( )a
u
∫ dt = f u( ) ⋅ dudx
This part of the Fundamental Theorem of Calculus says that derivatives and integrals are inverses operations of each other. Confusion usually pops up with the seemingly “extra” variable t showing up. It is a dummy variable, somewhat like the parameter in parametric mode. It would disappear in the process of integrating. In terms of the AP Test, the symbology is what is important and usually only comes into a process with L’Hopital’s Rule. Therefore, we will discuss this further in a later chapter. For now, the key idea is that the First Part of the Fundamental Theorem of Calculus simply tells us that an integral is an anti-derivative.
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4.1 Homework Set A Use Part II of the Fundamental Theorem of Calculus to evaluate the integral, or explain why it does not exist. 1.
x5dx
−1
3
∫ 2.
5x −1( )dx2
7
∫
3.
2x3 dx
−5
5
∫ 4. x7 − 4x3 − e
x⎛
⎝⎜
⎞
⎠⎟
−3
−1⌠
⌡
⎮⎮⎮
dx
5.
3t4 dt
1
2⌠⌡⎮
6. 3
4
4csc cot y ydy
π
π∫
7. 240
sec ydyπ∫ 8.
32z dz
1
9⌠⌡⎮
9. If F(x)= f (t)dt
1
x
∫ , where f (t)= 1+ u4
u du1
t2
⌠
⌡⎮ , find F ''(2) .
Find the Average Value of each of the following functions. 10. F x( ) = x − 3( )2 on x∈ 3, 7⎡⎣ ⎤⎦ 11. H x( ) = x on x∈ 0, 3⎡⎣ ⎤⎦
12. F x( ) = sec2 x on x∈ 0, π4
⎡⎣⎢
⎤⎦⎥
13. F x( ) = 1x on x∈ 1, 3⎡⎣ ⎤⎦
14. If a cookie taken out of a 450˚F oven cools in a 60˚F room, then according to Newton’s Law of Cooling, the temperature of the cookie t minutes after it has been taken out of the oven is given by
T t( ) = 60+ 390e−.205t .
What is the average value of the cookie during its first 10 minutes out of the oven?
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15. We know as the seasons change so do the length of the days. Suppose the length of the day varies sinusoidally with time by the given equation
L t( ) =10− 3cos πt
182⎛
⎝⎜⎞
⎠⎟,
where t the number of days after the winter solstice (December 22, 2007). What was the average day length from January 1, 2008 to March 31, 2008? 16. During one summer in the Sunset, the temperature is modeled by the function T t( ) = 50 +15sin π12 t , where T is measured in F° and t is measured in
hours after 7 am. What is the average temperature in the Sunset during the six-hour Chemistry class that runs from 9 am to 3 pm? Use Part I of the Fundamental Theorem of Calculus to find the derivative of the function. 17.
g( y) = t2 sint
2
y
∫ dt 18. g(x) = 1+ 2tdt
0
x
∫
19. F(x) = cos(t2)dt
x
2
∫ 20. h(x) = arctantdt
2
1x∫
21. y = cost
tdt
3
x
∫
4.1 Homework Set B
1.
1x +1
dx0
e2 −1⌠⌡⎮
2.
sin y dyπ
5π4∫
3. Find the average value of
f t( ) = t2 − t +5 on
t ∈ 1,4⎡⎣ ⎤⎦
4.
x2 +5x + 6( )dx3
5⌠⌡
5.
1x− 2 dx
3
e2+2⌠⌡⎮
, show your work, you may use the calculator to verify your
answer.
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6.
cos y dyπ
3π4∫ , show your work, you may use the calculator to verify your
answer.
7.
1x − 4
dx5
e3+4⌠⌡⎮
, show your work, you may use the calculator to verify your
answer. 8. Find the average value of
f t( ) = t2 − t +5 on
t ∈ 4,9⎡⎣ ⎤⎦
9.
sin y dyπ4
3π4∫ , show your work, you may use the calculator to verify your
answer.
10.
x2 − 4x + 7x
⎛
⎝⎜
⎞
⎠⎟ dx
1
2⌠
⌡⎮⎮ 11.
ddx
ln t2 +1( )dte
x2
∫⎡
⎣⎢⎢
⎤
⎦⎥⎥
12. If h x( ) = e5t dt
π
x
∫ , find h ' x( ) 13.
ddx
t ln t( )dt10
x2
∫
14. h m( ) = t2 cos−1 t( )dt
5
cos m
∫ , find h ' m( )
15. h y( ) = et
t4 dt5
ln y
∫ , find h ' y( ) 16.
ddx
t3 + t +1( )dtex
5
∫
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4.2 Definite Integrals and The Substitution Rule Let’s revisit u – subs with definite integrals and pick up a couple of more properties for the definite integral. Objectives Evaluate definite integrals using the Fundamental Theorem of Calculus. Evaluate definite integrals applying the Substitution Rule, when appropriate. Use proper notation when evaluating these integrals. Ex 1 Evaluate
t2 t3 +1( )dt
0
2
∫ .
t2 t3 +1( )dt
0
2
∫ =13
3t2 t3 +1( )dt0
2
∫
Let u = t3 +1 du = 3t2dt
u 0( ) = 1
u 2( ) = 9
= 13
u( )du1
9
∫
= 13
u3
2
32
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥1
9
= 29
93
2 −13
2⎡⎣⎢
⎤⎦⎥
= 529
Ex 2 Evaluate
xcos x2( )( )dx0
π
∫ .
xcos x2( )( )dx
0
π
∫ =
12
cosu dux=0
x= π
∫
Let u = x2
du = 2xdx
u 0( ) = 0
u π( ) = π
=
12
sinu0
π
=0
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Ex 3
e1
x
x2 dx1
2⌠
⌡⎮⎮
e1
x
x2 dx1
2⌠
⌡⎮⎮ =
− eudu
1
12∫
Let u = 1x
du = − 1x2 dx
u 1( ) = 1
u 2( ) = 12
= −eu
1
12
= −e1
2 + e Ex 4
x x2 + a2 dx
2
0
∫ a > 0
x x2 + a2 dx
2
0
∫ = − x x2 + a2 dx
0
2
∫
Let u = x2 + a2
du = 2xdx
=
= − x x2 + a2 dx0
2
∫= − 1
2u du
a2
4+a2
∫
= − 12
u3
2
32
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥a2
4+a2
= −13
4+ a2( )32 − a3⎡
⎣⎢
⎤
⎦⎥
Ex 5 Find the derivatives of the following functions. (a)
F x( ) = tanθdθ
x
10
∫
F x( ) = tanθdθ
x
10
∫ ⇒ ′F x( ) = − tan x (b)
y = sin3 tdt
ex
0
∫
y = sin3 tdt
ex
0
∫ ⇒ ′y = −sin3 ex ⋅ex
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(c) F x( ) = tanθdθ
10
10
∫
F x( ) = tanθdθ
10
10
∫ =0
238
4.2 Homework Set A Evaluate the definite integral, if it exists. 1.
x2(1+ 2x3)5dx
0
1
∫ 2.
sec2(t / 4)dt0
π
∫
3.
e1
x
x2 dx1
2
∫ 4.
sinθcos2θ
dθ0
π3∫
5.
dxx ln xe
e4
∫ 6. dx2x + 5−1
2⌠
⌡⎮⎮
7. sin x2 − cos x0
π⌠
⌡⎮⎮
dx 8.
dxx Lnx2
4⌠
⌡⎮⎮
9.
ex
1+ e2x dx0
Ln2⌠
⌡
⎮⎮ 10.
cos5 x sin x dx
π 6
π 2∫
11.
cos 12θ dθ
π
2π⌠⌡⎮
12.
f (x) dx0
2
∫ where f (x)= x4, if 0 ≤ x <1
x5, if 1≤ x ≤ 2⎧⎨⎪
⎩⎪
Find the Average Value of each of the following functions. 13. f x( ) = cosxsin4 x on x∈ 0, π⎡⎣ ⎤⎦ 14. g x( ) = xe−x2 on x∈ 1, 5⎡⎣ ⎤⎦ 15. G x( ) = x
1+ x2( )3 on x∈ 0, 2⎡⎣ ⎤⎦ 16. h x( ) = x
1+ x2( )2 on x∈ 0, 4⎡⎣ ⎤⎦
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4.2 Homework Set B
1.
sec2 2x( )dx0
π8
∫ 2.
cos x1+ sin x0
π
∫ dx
3.
x2 −16x2 +5x + 6( ) dx
3
5⌠
⌡⎮⎮
4.
cos x2+ sin x0
π
∫ dx
5.
sec2 4x( )dx0
π16
∫ 6.
sin y2+ cos y0
π
∫ dy
7.
csc2 ln y( )y dy
eπ4
eπ2
⌠
⌡⎮⎮ 8.
msec m2( ) tan m2( )dm
0
π4
∫
9.
cos9 x( )sin x( )dxπ2
π
∫ 10.
sec2 x tan3 x dx0
π4
∫
11.
yey2 −3
3
4
∫ dy
12. If a t( ) = cos 2t( ) , find
v t( ) and
x t( ) if
v π( ) = 6 and
x π
2⎛
⎝⎜⎞
⎠⎟= 0 .
13. If
a t( ) = e3t + cos 2t( ) , find
v t( ) when
v 0( ) = 3 . Then find
x t( ) when
x 0( ) = 0 .
14. If f ' θ( ) = tan6 2θ( )sec2 2θ( ) , find
f θ( ) when
f 3π
8⎛
⎝⎜⎞
⎠⎟= 2 .
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15. If v t( ) = sec 3t( ) tan 3t( ) , find
a t( ) and
x t( ) when
x π
12⎛
⎝⎜⎞
⎠⎟= 3
16. If f ' x( ) = cos x
x, find
f x( ) when
f π 2
4⎛
⎝⎜
⎞
⎠⎟ = 2
17.
sec2 2x dxπ8
π6
∫ 18. Find the exact value of
1x +1
dx0
e2 −1
∫
19.
sin y dyπ
5π4
∫ 20.
5m2 + 3m+14
0
e⌠
⌡⎮ dm
21.
ln x −1( )( )4x −1
dx2
e3+1⌠
⌡⎮⎮
22.
cos6 x2
⎛
⎝⎜⎞
⎠⎟sin x
2⎛
⎝⎜⎞
⎠⎟dx
0
π
∫
23.
csc2 ln y( )y dy
eπ4
eπ2
⌠
⌡⎮⎮ 24.
msec m2( ) tan m2( )dm
0
π4
∫
25.
cos9 x( )sin x( )dxπ2
π
∫
26. Find the average value of f x( ) = csc2xcot2x on
x ∈ π
8,π4
⎡
⎣⎢
⎤
⎦⎥
241
4.3 Context for Definite Integrals: Area, Displacement, and Net Change Here are a couple of more properties for the definite integral. Properties of Definite Integrals 1.
f x( )dx = −
a
b
∫ f x( )dxb
a
∫ 2.
f x( )dx
a
a
∫ = 0 3. f x( )
a
b
∫ dx = f x( )a
c
∫ dx + f x( )c
b
∫ dx , where a < c < b Since we originally defined the definite integrals in terms of “area under a curve,” we need to consider what this idea of “area” really means in relation to the definite integral. Let’s say that we have a function,
y = xcos x2( )( ) on
x∈ 0, π⎡
⎣⎢⎤⎦⎥. The graph looks
like this:
In the last section, we saw that
xcos x2( )( )0
π⌠⌡⎮
dx = 0 . But we can see there is area
under the curve, so how can the integral equal the area and equal 0? Remember that the integral was created from rectangles with width dx and height f(x). So the area below the x-axis would be negative, because the f(x)-values are negative.
242
Ex 1 What is the area under y = xcos x2( )( ) on
x∈ 0, π⎡
⎣⎢⎤⎦⎥?
We already know that
xcos x2( )( )0
π⌠⌡⎮
dx = 0 , so this integral cannot
represent the area. As with example 1b, we really are looking for the positive number that represents the area (total distance), not the difference between the positive and negative “areas” (displacement). The commonly accepted context for area is a positive value. So,
Area = xcos x2( )0
π⌠⌡ dx
= xcos x2( )0
1.244∫ dx − xcos x2( )1.244
π∫ dx
=1
We could have used our calculator to find this answer:
When we use the phrase “area under the curve, we really mean the area between the curve and the x-axis. CONTEXT IS EVERYTHING. The area under the curve is only equal to the definite integral when the curve is completely above the x-axis. When the curve goes below the x-axis, the definite integral is negative, but the area, by definition, is positive. Objectives: Relate definite integrals to area under a curve. Understand the difference between displacement and total distance. Extend that idea to understanding the difference between the two concepts in other contexts.
243
Ex 2 Find the area under y = x3 − 2x2 − 5x + 6 on x∈ −1, 2⎡⎣ ⎤⎦ . A quick look at the graph reveals that the curve crosses the x-axis at x=1.
−1 1 2
−4
4
8
x
y
If we integrate y on x∈ −1, 2⎡⎣ ⎤⎦ , we will get the difference between the areas, not the sum. To get the total area, we need to set up two integrals:
Area = x3 − 2x2 −5x+6( )−1
1⌠⌡ dx+ − x3 − 2x2 −5x+6( )
1
2⌠⌡ dx⎛
⎝⎜⎞⎠⎟
= x4
4 − 2x3
3 − 5x2
2 +6x⎤
⎦⎥−1
1
− x4
4 − 2x3
3 − 5x2
2 +6x⎡
⎣⎢
⎤
⎦⎥1
2
= 323 + 29
12
= 15712
244
Ex 3 Find the area under y = x3 − 2x on x∈ −1, 2⎡⎣ ⎤⎦ .
4
2
–2
Note that there are three regions here, therefore there will be three integrals:
Area = x3 − 2x( )−1
0⌠⌡ dx+ − x3 − 2x( )
0
2⌠⌡ dx
⎛
⎝⎜⎞
⎠⎟+ x3 − 2x( )
2
2⌠⌡ dx
= x4
4 − x2 ⎤
⎦⎥−1
0
− x4
4 − x2⎡
⎣⎢
⎤
⎦⎥
0
2
+ x4
4 − x2⎡
⎣⎢
⎤
⎦⎥
2
2
= 34 − −1( )+1
= 114
F could be a function that describes anything – volume, weight, time, height, temperature. F ’ represents its rate of change. The left side of that equation accumulates the rate of change of F from a to b and the right side of the equation says that accumulation is difference in F from a to b. Imagine if you were leaving your house to go to school, and that school is 6 miles away. You leave your house and halfway to school you realize you have forgotten your calculus homework (gasp!). You head back home, pick up your assignment, and then head to school. There are two different questions that can be asked here. How far are you from where you started? And how far have you actually traveled? You are six miles
245
from where you started but you have traveled 12 miles. These are the two different ideas behind displacement and total distance. Vocabulary: Displacement – How far apart the starting position and ending position are. (It can be positive or negative.) Total Distance – how far you travel in total. (This can only be positive.)
Displacement = vdt
a
b
∫
Total Distance= v dta
b
∫
Ex 4 A particle moves along a line so that its velocity at any time t is
v t( ) = t2 + t − 6 (measured in meters per second). (a) Find the displacement of the particle during the time period 1≤ t ≤ 4 . (b) Find the distance traveled during the time period 1≤ t ≤ 4 .
a( ) vdta
b
∫ = t 2 + t − 6( )dt1
4
∫= t
3
3+ t
2
2− 6t
1
4
= −4.5
b( ) v dta
b
∫ = t 2 + t − 6( )dt1
4
∫= − t 2 + t − 6( )dt
1
2
∫ + t 2 + t − 6( )dt2
4
∫= − t
3
3− t
2
2+ 6t
1
2
+t3
3+ t
2
2− 6t
2
4
= 10 16
Note that we used the properties of integrals to split the integral into two integrals that represent the separate positive and negative distanced and then
246
made the negative one into a positive value by putting a – in front. We split the integral at t = 2because that would be where
v t( ) = 0 .
Ex 5 AB 1997 # 1
247
4.3 Homework Set A
Find the area under the curve of the given equation on the given interval.
1. y = x3 − 2x2 − 3x on x∈ −2, 2⎡⎣ ⎤⎦
2. y = x3 − 4x2 + 4x on x∈ −1, 2⎡⎣ ⎤⎦
3. y = x3 − 2x2 − x+ 2 on x∈ −3, 3⎡⎣ ⎤⎦
4. y = π2 cosx sin π +π sin x( )( ) on x∈ −π
2, π⎡
⎣⎢⎤⎦⎥
5. y = −xx2 + 4 on x∈ −2, 2⎡⎣ ⎤⎦
6. y = 4 − x2
x2 + 4 on x∈ −3, 3⎡⎣ ⎤⎦
7. y = sin xx
on x∈ .01, π 2⎡⎣ ⎤⎦
8. y = x 18 − 2x2 on x∈ −2, 1⎡⎣ ⎤⎦
9. y = 3sin x 1− cos x on x∈ −π2 , π3⎡⎣⎢
⎤⎦⎥
10. y = x2ex3 on x∈ 0, 1.5⎡⎣ ⎤⎦
11. The velocity function (in meters per second) for a particle moving along a line is v(t) = 3t −5 for 0 ≤ t ≤ 3 . Find (a) the displacement and (b) the distance traveled by the particle during the given time interval.
12. The velocity function (in meters per second) for a particle moving along a line is v(t)= t2 − 2t −8 for 1≤ t ≤ 6 . Find (a) the displacement and (b) the distance traveled by the particle during the given time interval. 13. Find the area under the curve
f x( ) = e−x2 − x on
x ∈ −1, 2⎡⎣ ⎤⎦ (do not use
absolute values in your setup, break it into multiple integrals).
248
14. Find the area under the curve
f x( ) = e−x2 − 2x on
x∈ −1, 2⎡⎣ ⎤⎦ (do not use
absolute values in your setup, break it into multiple integrals).
15. Find the area under the curve f x( ) = x
x2 +1+ cos x( ) on
x∈ 0,π⎡⎣ ⎤⎦
16. Find the area under the curve
g x( ) = −1− xsin x on
x∈ 0,2π⎡⎣ ⎤⎦
249
4.4 Graphical Analysis II An important part of calculus is being able to read information from graphs. Earlier, we looked at graphs of first derivatives and answered questions concerning the traits of the “original” function. This section will answer the same questions, even with similar graphs – but this time the “original” function in will be defined as an integral in the form of the FTC. This is a major AP topic. Objectives Use the graph of a function to answer questions concerning maximums, minimums, and intervals of increasing and decreasing Use the graph of a function to answer questions concerning points of inflection and intervals of concavity. Use the graph of a function to answer questions concerning the area under a curve. As we saw before, the graphs of a function and its derivatives are related. We summarized that relationship in a table that we will now expand.
G(x)= F t( )0
x∫ dt
The y-values of
G(x)
Increasing
EXTREME Decreasing
Concave up
POI Concave down
F(x)
Area under F(x)
Positive ZERO
Negative
Increasing
EXTREME Decreasing
F’(x)
Positive ZERO
Negative
NB. The AP Test will always require the statement that g x( ) = f t( )0
x∫ dt
means that g ' x( ) = f x( ) , and the arguments about extremes, concavity and increasing and decreasing be made from the point of view of the derivatives.
250
Ex 1 The graph of f is shown below. f consists of three line segments.
4
3
2
1
–1
–2
–3
–4
–4 –2 2 4
Let
g x( ) = f t( )dt
0
x
∫ .
a) Find g −2( ), g′ −2( ), and g″ −2( ) . b) For what values on x∈ −4, 4( ) is g x( ) increasing? Explain your reasoning. c) For what values on x∈ −4, 4( ) is g x( ) concave up? Explain your reasoning. d) Sketch a graph of g x( ) on x∈ −4, 4[ ] . a) g −2( ) = −3 . g −2( ) would equal the area of the triangle from -2 to 0, but the integral goes from 0 to -2, therefore the integral equals the negative of the area. g′ −2( ) = f −2( ) = 0 .
g″ −2( ) = f ′ −2( ) = 32 . This is the slope of f (x) at x=-2.
b) g x( ) is increasing when g′ x( ) is positive. g′ x( ) = f x( ) . Therefore, g x( ) is increasing on x∈ −2, 2[ ] c) g x( ) concave up when f x( ) is increasing. Therefore, g x( ) is concave up on x∈ −4, 0( ) d) Because of the equal triangle areas, g −4( ) = g 0( ) = g 4( ) = 0 g x( ) has a relative max at x = 2 and a relative min at x = −2 . g x( ) is increasing on x∈ −2, 2[ ] and decreasing on x∈ −4, − 2[ ]∪ 2, 4[ ]
251
g x( ) is concave up on x∈ −4, 0( ) and concave down on x∈ 0, 4( ) . The graph looks something like this: Ex 2 The graph of f is shown below. f consists of three line segments.
Let
g x( ) = f t( )dt
−3
x
∫ .
(a) Find g 3( ) .
(b) Find g −6( ) . (c) Find
g 1( ) , ′g 1( ) , and
′′g 1( ) . (d) On what intervals is g decreasing? (e) Where is g concave up? (f) Make a sketch of the original function. (g) How do the answers in (a) – (e) change if
m x( ) = 4+ f t( )dt
−3
x
∫ ? (a) Find
g 3( ) .
g 3( ) = f t( )dt
−3
3
∫ We must split this integral into four parts – the two triangles above the x – axis and the two triangles below the x- axis.
6
4
2
-2
- 4
- 6
-10 - 5 5 10
252
g 3( ) = f t( )dt
−3
3
∫ = f t( )dt−3
0
∫ + f t( )dt0
2
∫ + f t( )dt2
198∫ + f t( )dt19
8
3
∫
Where did
x =19
8 come from? That is the x – intercept of that line segment.
g 3( ) = 6− 3− 9
16+ 25
16= 4 Use the area formula for a triangle to
calculate the different areas. Area below the x – axis must be subtracted.
(b) Find g −6( ) .
g −6( ) = f t( )dt
−3
−6
∫ = − f t( )dt−6
−3
∫ Use the properties of integrals.
= − 1
2⋅2 ⋅4 − 1
2⋅1 ⋅2⎛
⎝⎜⎞⎠⎟
Area of triangles – triangle above the x –
axis is added to the integral, triangle below the x – axis is subtracted from the integral. =-3
(c) Find g 1( ) , ′g 1( ) , and
′′g 1( ) .
g 1( ) = f t( )dt
−3
1
∫ = f t( )dt−3
0
∫ + f t( )dt0
1
∫ Again, we must split the integral into two parts – one for each separate triangle.
g 1( ) = 6− 3
4= 21
4 How did we conclude that
3
4 as the area of that
second triangle? Find the equation of the line making up the hypotenuse of the triangle, plugged in x =1, and use that number as the height of the triangle. Subtract this number because that triangle was below the x – axis.
Now in order to find
′g 1( ) , we must find ′g x( ) . But since g was defined as an
integral, the FTC kicks in when we take the derivative of g.
253
′g x( ) = d
dxf t( )dt
−3
x
∫⎡⎣⎢⎤⎦⎥= f x( ) Derivatives and integrals are inverse
operations – the FTC
′g 1( ) = f 1( ) = − 3
2 You can read the value from the graph.
In order to find
′′g 1( ) , we must find ′′g x( ) . But this follows from what we just
did. Since ′g x( ) = f x( ) , we can see that
′′g x( ) = ′f x( ) .
′′g 1( ) = ′f 1( ) = − 3
2 The derivative is the slope of f – and the slope of
that line segment is − 3
2.
(d) On what intervals is g decreasing?
g is decreasing when ′g < 0 -- but as stated above
′g x( ) = f x( ) . And we can see
graphically that f is negative from x ∈ −7,−5( )∪ 0,19
8⎛
⎝⎜⎞
⎠⎟.
(e) Where is g concave up?
g is concave up when ′f is increasing. So g is concave up from
x ∈ −7,−3( )∪ 2,3( ) .
(f) Make a sketch of the original function.
254
6
4
2
-2
- 4
- 6
-10 - 5 5 10
(g) How do the answers in (a) – (e) change if m x( ) = 4+ f t( )dt
−3
x
∫ ?
The answers in (a) and the first part of (b) would both need a 4 added to them. The graph in part (e) would be shifted up 4 units. And all other answers would remain the same.
255
Ex 3 Let h x( ) = g t( )dt
0
x
∫ . The graph of g is shown below.
–5 5
4
2
–2
(a) Find h 4( ) .
(b) Find h 0( ) , ′h 0( ) , and
′′h 0( ) . (c) What is the instantaneous rate of change of h(x) at x = 1? (d) Find the absolute minimum value of h on
−6, 8⎡⎣ ⎤⎦ .
(e) Find the x – coordinate of all points of inflection of h on −6, 8( ) .
(f) Let F x( ) = g t( )dt
x
0
∫ . When is F increasing? (a)
h 4( ) = g t( )dt
0
4
∫ would equal the area from 0 to 4. This is one quarter of the
circle with radius 4. So, h 4( ) = g t( )dt
0
4
∫ = 4π . (b)
h 0( ) = g t( )dt
0
0
∫ = 0
′h 0( ) = d
dxg t( )dt
0
x
∫ = g 0( ) = 4
′′h 0( ) = g ' 0( ) = 0
256
(c) The instantaneous rate of change of h(x) at x = 1 would be g(1). To find g(1), we would need the equation of the circle. Since I t has its center at the origin and radius = 4, the equation is x
2 + y2 =16 or y = 16− x2 .
g 1( ) = 16−12 = 15
(d) Since
g x( ) = h ' x( ) , the critical values of h would be the zeros and endpoints
of g, namely, x = -6, -6, 4, 9. The graph of g(x) yields a sign pattern:
gx − 6 − 4 2 8
← →⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 0 + 0 + 0 − 0
From this, we can see that x = -4 and 2 cannot be minimums. x = 2 is a maximum because the sign of
g x( ) = h ' x( ) switches from + to - and x = -4 is not an extreme
because the sign does not change. Now all we need are h(-6) and h(-9). Whichever is smaller is the absolute minimum.
h −6( ) = g t( )dt
0
−6
∫ = − g t( )dt−6
0∫ = − 4 + 4π( )
h 9( ) = g t( )dt
0
9
∫ = 4π − 4
h(-6) is the smaller value so the absolute minimum is − 4 + 4π( )
(e)
g ' x( ) = h" x( ) so the points of inflection on h would be where the slope of g
changes sign, namely at x = -4, 0 and 6. (f)
F x( ) = g t( )dt
x
0
∫ → F ' x( ) = −g x( ) . So F is increasing when g is negative,
namely on x∈ 4, 8⎡⎣ ⎤⎦ .
257
4.4 Homework Handout of AP Questions 1. BC 2002B #4 2. BC 1999 #5 3. BC 2003 #5 4. BC 2002 #4 5. BC 2009B #5 6. Let
g x( ) = f t( )dt
2
x
∫ for 0 ≤ t ≤ 7 , where the graph of the differentiable function f is shown below.
4
2
5
a) Find
g 3( ) , g′ 3( ) , and
g″ 3( ) .
b) Find the average rate of change of g x( ) on 0 ≤ c ≤ 3?
c) For how many values of c, where 0 < c < 3 , is g′ c( ) equal to the average rate found in part (b)? Justify your answer.
258
d) Find the x-coordinate of each point of inflection of the graph of g on the interval 0 < t < 7 . Justify your answer. 7. Let
g x( ) = f t( )dt
−2
x
∫ for −7 ≤ t ≤ 7 , where the graph of the differentiable function f is shown below.
–5 5
4
2
–2
a) Find
g 4( ) , g′ 2( ) , and
g″ 4( ) .
b) Find the average rate of change of g x( ) on −7 ≤ t ≤ 0 ?
c) At what x-values is g x( ) decreasing and concave up? Justify your answer.
d) Find the x-coordinate of the absolute minimum of g x( ) . Justify your
answer. 8. Let
g x( ) = f t( )dt
−4
x
∫ for −7 ≤ t ≤ 7 , where the graph of the differentiable function f is shown below.
4
3
2
1
–1
–2
–3
–8 –6 –4 –2 2 4 6
259
a) Find g 0( ) , g′ 0( ) , and
g″ 0( ) .
b) Find the equation of the line tangent to g x( ) at x = 0 .
c) At what x-values is g x( ) decreasing and concave up? Justify your answer.
d) Find the x-coordinate of the absolute maximum of g x( ) . Justify your
answer. 9. Let
g x( ) =10+ f t( )dt
0
x
∫ for 0 ≤ t ≤8 and let f t( ) be the differentiable
function (shown below ) comprised of two horizontal line segments and one cycle
of the cosine wave y = 2cos π
4t − 2( )⎡
⎣⎢
⎤
⎦⎥ .
2
1
–1
–2
–3
2 4 6 8
a) Find
g 4( ) , g′ 4( ) , and
g″ 4( ) .
b) Find the average rate of change of g x( ) on 0 ≤ t ≤8?
c) Find the average value of f x( ) on 0 ≤ t ≤8 .
d) Find the absolute minimum of g x( ) . Justify your answer.
260
4.5 Approximate Integration – Riemann and Trapezoidal Sums We have been focusing on anti-derivatives of functions where the equation is known. But let us suppose we need to evaluate an integral where either the function is unknown of cannot be anti-differentiated—such as
ex2 dx
−2
3
∫ . If we had some exact y-values, we could approximate the area geometrically. Mathematicians came up with this technique of dividing the area in question into rectangles, and then finding the area of each rectangle. Why rectangles? Because their area formula was so simple. Objectives: Find approximations of integrals using different rectangles. Use proper notation when dealing with integral approximation. Ex 1 Use a left end Riemann sum with four equal subintervals to approximate
f x( )dx
1
5
∫ given the table of values below.
x 1 2 3 4 5
f x( ) 1 4 9 16 25
Let’s first figure out the width of our rectangles. If we need four rectangles in between x = 1 and x = 5, then our rectangles will be 1 unit wide. The height will be determined by how high the function 2x is for each rectangle.
−1 1 2 3 4 5 6
−1
1
2
3
4
5
x
y
261
x2dx
1
5
∫ ≈1⋅12 +1⋅22 +1⋅32 +1⋅42 = 30 **Notice the use of ≈ and = Ex 2 Use a right end Riemann sum with four rectangles to approximate
f x( )dx
1
5
∫ from Ex 1. Again, our rectangles would be 1 unit wide.
−1 1 2 3 4 5 6
−1
1
2
3
4
5
x
y
x2dx
1
5
∫ ≈1⋅22 +1⋅32 +1⋅42 +1⋅52 = 54 **Notice the use of ≈ and = These two approximations are very different, and both aren’t too close to the true value of the integral – you remembered how to find 5 2
1x dx∫ using the FTC …
x2dx
1
5
∫ =1243 ≈ 41.333
How could we find a better approximation? Let’s try using midpoints.
262
Ex 3 The function described in Ex 1is actually f x( ) = x2 . Use a midpoint
Riemann sum with four rectangles to approximate
x2 dx1
5
∫ .
−1 1 2 3 4 5 6
−1
1
2
3
4
5
x
y
Because we now have the function, we can determine the y-values that would be the midpoint values if they had been on the table. They would be:
y = 1.5( )2 ,
2.5( )2 ,
3.5( )2 , and
4.5( )2
Yet again, our rectangles would be 1 unit wide and the approximation would be:
x2dx
1
5
∫ ≈1⋅ 1.5( )2 +1⋅ 2.5( )2 +1⋅ 3.5( )2 +1⋅ 4.5( )2 = 41 This is a much better approximation than either left-hand or right-hand rectangles. Would our approximation get more accurate or less accurate if we added in more rectangles? Think about this graphically.
263
Steps to Approximating an Integral with Rectangles: 1. Determine the width of your rectangle. 2. Determine the height of your rectangle – whether that be using the left endpoint, right endpoint, midpoint, or any other height requested. 3. Calculate the areas of each rectangle. 4. Add the areas together for your approximation. 5. State answer using proper notation. Ex 4 Use the midpoint rule and the given data to approximate the value of
f x( )dx
0
2.6
∫ . x
�
f x( ) x
�
f x( ) 0 7.5 1.6 7.7
0.4 7.3 2.1 7.9 0.8 7.2 2.6 8.0 1.2 7.3
f x( )dx
0
2.6
∫ ≈ 0.8 7.3( ) + 0.8 7.3( ) +1.0 7.9( ) = 19.58 A geometric alternative to rectangles would be to use trapezoids. Ex 5 Using the table of data from Ex 4, approximate
f x( )dx
0
2.6
∫ with 6 trapezoids.
Keep in mind the area formula for a trapezoid is ATrap =
12
b1 + b2( ) ⋅h , where
b1and b2 are the parallel sides of the trapezoid.
Each trapezoid must have a height equal to the change in x and the bases must be the y-values (see the illustration below).
264
0.5 1.0 1.5 2.0 2.5
1
2
3
4
5
6
7
8
x
y
These look much more like trapezoids if you turn the page sideways – then it becomes obvious that the height is xΔ , and that the y-values are the bases:
f x( )dx0
2.6∫ ≈ 1
2 0.4( ) 7.5+7.3( )+ 12 0.4( ) 7.3+7.2( )+ 1
2 0.4( ) 7.2+7.3( )+ 1
2 0.4( ) 7.3+7.7( )+ 12 0.5( ) 7.7+7.9( )+ 1
2 0.5( ) 7.9+8.0( )
=19.635
This isn’t far off from the true answer for the integral – the FTC way. Notice what happened when we did our last calculation – do you see how each
trapezoid has a
12
and a 0.2 in its area formula, the outer endpoints showed up once
where the inner endpoints showed up twice.
0.4xΔ =
1 7.5b =
2 7.3b =
265
This can be summed up in the Trapezoidal Rule formula – feel free to memorize this formula, or just derive it when you come across these problems. The Trapezoidal Rule*
f x( )dx
a
b
∫ ≈ b− a2n
f x0( ) + 2 f x1( ) + 2 f x2( ) + ...+ 2 f xn−1( ) + f xn( )⎡⎣
⎤⎦
*The Trapezoidal Rule requires equal sub-intervals Ex 6 Estimate the area under the graph in the figure by first using Midpoint Rule and then by the Trapezoidal Rule with n = 5.
1 2 3 4 5 6 7
1
2
3
4
5
6
x
y
AMid ≈1⋅ .5( ) +1⋅ 2.5( ) +1⋅ 3( ) +1⋅ 1.5( ) +1⋅ 1.25( ) = 8.75
ATrap ≈
12
0+1( ) ⋅1+ 12
1+ 4( ) ⋅1+ 12
4+ 2( ) ⋅1+ 12
2+1.5( ) ⋅1+ 12
1.5+1( ) ⋅1= 9
266
Ex 7 The following table gives values of a continuous function. Approximate the average value of the function using the midpoint rule with 3 equal subintervals. Repeat the process using the Trapezoidal Rule.
x 10 20 30 40 50 60 70
f x( ) 3.649 4.718 6.482 9.389 14.182 22.086 35.115
Midpoint Rule:
favg =
170−10
f x( )dx10
70
∫ ≈ 170−10
20 4.718( ) + 20 9.389( ) + 20 22.086( )⎡⎣
⎤⎦ =12.064
Trapezoidal Rule:
favg =1
70−10f x( )dx
10
70
∫
≈ 170−10
70−1012
3.649+ 2 4.718( ) + 2 6.482( ) +2 9.389( ) + 2 14.182( ) + 2 22.086( ) + 35.115
⎛
⎝⎜⎜
⎞
⎠⎟⎟
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=12.707
Ex 8 Let us assume that the function that determined the values in the chart above is
f x( ) = 2+ e.05x . Calculate the average value of the function and compare it to
your approximations. (Can you figure out ahead of time whether our approximation will be an overestimate or an underestimate?)
favg =
170−10
2+ e.05x( )dx10
70
∫ =12.489
As noted in the Trapezoidal Rule footnote, the Trapezoidal Rule in the stated form only works for equal subintervals. This is because the Trapezoidal Rule as stated is a factored form of setting up the problem using the trapezoidal area formula from Geometry:
Trapezoid Area: A= 1
2h b1 + b2( )
267
Ex 8 Let us assume that the function that determined the values in the chart above is ( ) 0.052 xf x e= + . Calculate the average value of the function and compare it to your approximations. (Can you figure out ahead of time whether our approximation will be an overestimate or an underestimate?)
( )70 0.0510
1 2 12.48970 10
xavgf e dx= + =
− ∫
If we want to figure out if our approximations are overestimates or underestimates, we have to look at the graph of the function.
10 20 30 40 50 60 70
10
20
30
x
y
Since this is concave up throughout, the area of the trapezoid would be larger than the area under the curve (see the trapezoid above for an example, it is just slightly above the curve).
In looking at whether an approximation is an over- or under- estimate, we always look at the concavity and compare whether our rectangles (or trapezoids) are above the curve or below it. If they are above the curve, we have overestimates, below the curve, we have underestimates.
268
As noted in the Trapezoidal Rule footnote, the Trapezoidal Rule in the stated form only works for equal subintervals. This is because the Trapezoidal Rule as stated is a factored form of setting up the problem using the trapezoidal area formula from Geometry:
Trapezoid Area: A= 1
2h b1 + b2( )
Ex 9 Given the table of values below, find an approximation for
a) ( )6
0v t dt∫ using 3 midpoint rectangles.
b) ( )8
4v t dt∫ using 4 trapezoids.
c) ( )4
0v t dt∫ using 4 left Riemann rectangles.
Make sure you indicate the units for each.
t (in sec.)
0 1 2 3 4 5 6 7 8
v(t) (in m/sec.)
8 12 17 10 3 8 6 12 10
a) ( ) ( ) ( ) ( )6
02 12 2 10 2 8 60 metersv t dt ≈ + + =∫
b) ( ) ( ) ( ) ( ) ( )8
4
1 1 1 13 8 8 6 6 12 12 10 32.5 meters2 2 2 2
v t dt ≈ + + + + + + + =∫
c) ( ) ( ) ( ) ( ) ( )4
01 8 1 12 1 17 1 10 47 metersv t dt ≈ + + + =∫
269
The units were in meters for all of them by virtue of the multiplication: the widths were always in seconds, and the heights were in meters per second. When we multiply these quantities, the seconds cancel.
Ex 10 The function g is continuous on the closed interval [2,14] and has values
shown on the table below. Using the subintervals [2,5], [5,10], and [10,14],
what is the approximation of ( )14
2g x dx∫ found by using a right Riemann
sum?
x 2 5 10 14 g(x) 1.2 2.8 3.4 3.0
(A) 29.6 (B) 31.2 (C) 34.3 (D) 37.4 (E) 39.0 Remember from Chapter 1:
• Tangent line approximations are an overestimate if the curve is concave down (since your “tangent lines” will be above the curve).
• Tangent line will be an underestimate if the curve is concave up (since your “tangent lines” will be below the curve).
Add to that:
• If a function is increasing, Right-hand Riemann Rectangle approximations are an overestimate, while Left-hand Rectangle are an under-approximation.
• If a function is decreasing, Right-hand Riemann Rectangle underestimate are an underestimate, while Left-hand Rectangle are an overestimate.
270
4.5 Homework Set A
1. The velocity of a car was read from its speedometer at 10-second intervals and recorded in the table. Use the Midpoint Rule to estimate the distance traveled by the car.
t(s) v (mi/h) t(s) v (mi/h)
0 0 60 56
10 38 70 53
20 52 80 50
30 58 90 47
40 55 100 45
50 51
2. The following table gives values of a continuous function. Estimate the average value of the function on x∈ 0, 8⎡⎣ ⎤⎦ using (a) Right-Hand Riemann rectangles, (b) Left-Hand Riemann rectangles, and (c) Midpoint Riemann rectangles.
x 0 1 2 3 4 5 6 7 8 F(x) 10 15 17 12 3 -5 8 -2 10
3. Below is a chart showing the rate of a rocket flying according to time in minutes. Use this information to answer each of the questions below.
t (in minutes) 0 10 20 30 40 50 60
v(t) (in km/min) 30 28 32 18 52 48 28
271
a) Find an approximation for ( )60
0
v t dt∫ using midpoint rectangles. Make sure you
express your answer in correct units.
b) Find an approximation for ( )30
0
v t dt∫ using trapezoids. Make sure you express
your answer in correct units.
c) Find an approximation for ( )60
30
v t dt∫ using left rectangles. Make sure you express
your answer in correct units.
d) Find an approximation for ( )40
0
v t dt∫ using right rectangles. Make sure you
express your answer in correct units.
4. Below is a chart showing the rate of water flowing through a pipeline according to time in minutes. Use this information to answer each of the questions below.
t (in minutes) 0 8 16 24 32 40 48
V(t) (in m3/min) 26 32 43 24 19 24 26
a) Find an approximation for ( )48
0
V t dt∫ using midpoint rectangles. Make sure you
express your answer in correct units.
b) Find an approximation for ( )16
0
V t dt∫ using right Riemann rectangles. Make sure
you express your answer in correct units.
5. Below is a chart of your speed driving to school in meters/second. Use the information below to find the values in a) and b) below.
272
t (in seconds) 0 30 90 120 220 300 360
v(t) (in m/sec) 0 21 43 38 30 24 0
a) Find an approximation for ( )360
0
v t dt∫ using left Riemann rectangles. Make sure
you express your answer in correct units.
b) Find an approximation for ( )220
0
v t dt∫ using trapezoids. Make sure you express
your answer in correct units. 6. Below is a chart showing the velocity of the Flash as he runs across the country. Use this information to answer each of the following.
t (in seconds) 0 4 8 12 16 20 24
W(t) (in km/second) 10 12 15 19 24 18 7
a) Find an approximation for ( )24
0
v t dt∫ using midpoint rectangles. Make sure
you express your answer in correct units. Describe what this integral means.
b) Find an approximation for ( )16
0
v t dt∫ using Trapezoids. Make sure you
express your answer in correct units.
273
7. The graph below shows energy use in kilowatts per hour for a community for June 25. Which of the following best approximates the number of kilowatts that were used on that day?
3 6 9 12 15 18 21 24 27
5
10
15
20
Hours
a) 30 b) 35 c) 360 d) 420 e) 720 8. Below is a chart showing the rate of sewage flowing through a pipeline according to time in minutes. Use this information to answer each of the questions below.
t (in minutes) 0 4 6 10 13 15 20
V(t) (in gallons/min) 83 68 82 40 38 30 68
a) Find an approximation for ( )20
0
V t dt∫ using trapezoids. Make sure you express
your answer in correct units.
b) Find an approximation for ( )20
0
V t dt∫ using left Riemann rectangles. Make sure
you express your answer in correct units.
Kilo
wat
ts p
er H
our
274
9. AP Handout: AB 1998 #3, AB 2001 #2, BC2007 #5
10. Use (a) the Trapezoidal Rule and (b) the Midpoint Rule to approximate the given integral with the specified value of n.
1+ x24
0
2
∫ dx,n = 8 11. Use (a) the Trapezoidal Rule and (b) the Midpoint Rule to approximate the
ln x1+ x
dx1
2
∫ with n =10 .
12. If w’(t) is the rate of growth of a child in pounds per year, what does
w '(t)
5
10
∫ dt represent? 13. If oil leaks from a tank at a rate of r(t) gallons per minute at time t, what does
r(t)dt
0
120
∫ represent? 14. If x is measured in meters and f(x) is measured in newtons, what are the units for
f (x)dx
0
100
∫ ? 4.6 Homework Set B
1. Use the trapezoidal rule with n = 4 to approximate the value of 24
0
xe dx∫ . Why
do you think that approximations would be necessary for evaluation of this integral? 2. Given the table of values below for velocities of a glider, find an approximation for the displacement of the glider using a left Riemann sum, a right Riemann sum, and a trapezoidal sum (make sure you include units in your answer). Which do you think is the most accurate? Explain.
275
t (minutes) 1 5 11 13 19 V(t) (m/min) 450 550 400 350 500
3. The following table is for a continuous function, ( )f x . Use the information in the table to find each of the following:
x 0 1 2 3 4 5 6 7 8 9 ( )f x 2.7 3.1 4.2 2.1 1.0 0.1 -0.3 -1.5 -0.4 0.2
a. Approximate ( )5
0
f x dx∫ using a right approximation
b. Approximate ( )5
0
f x dx∫ using a left approximation
c. Approximate
f x( )0
6
∫ dx using a midpoint approximation
d. Approximate
f x( )0
6
∫ dx using a trapezoidal approximation
e. Approximate
f x( )0
9
∫ dx using a right approximation
f. Approximate
f x( )0
9
∫ dx using a left approximation
g. Approximate
f x( )1
9
∫ dx using a midpoint approximation
h. Approximate
f x( )1
9
∫ dx using a trapezoidal approximation
4. Given the table of values below, find each of the following:
t in seconds R(t) in meters/second 0 12 5 15 10 17 15 13 20 16 25 18 30 21
276
a) Find a left Riemann approximation for ( )30
10
R t dt∫
b) Find a left Riemann approximation for ( )15
0
R t dt∫
c) Find a right Riemann approximation for ( )30
10
R t dt∫
d) Find a right Riemann approximation for ( )15
0
R t dt∫
e) Find a midpoint Riemann approximation for ( )30
0
R t dt∫
f) Find a midpoint Riemann approximation for ( )30
10
R t dt∫
g) Find a trapezoidal approximation for ( )30
0
R t dt∫
277
4.6 The Mean Value and Rolle’s Theorems The Mean Value Theorem is an interesting piece of the history of Calculus that was used to prove a lot of what we take for granted. The Mean Value Theorem was used to prove that a derivative being positive or negative told you that the function was increasing or decreasing, respectively. Of course, this led directly to the first derivative test and the intervals of concavity.
Mean Value Theorem
If f is a function that satisfies these two hypotheses 1. f is continuous on the closed interval
a,b⎡⎣ ⎤⎦
2. f is differentiable on the closed interval a,b( )
Then there is a number c in the interval a,b( ) such that
f ' c( ) = f b( )− f a( )
b− a .
Again, translating from math to English, this just says that, if you have a smooth, continuous curve, the slope of the line connecting the endpoints has to equal the slope of a tangent somewhere in that interval. Alternatively, it says that the secant line through the endpoints has the same slope as a tangent line.
a bc
MEAN VALUE THEOREM
278
Ex 1 Show that the function f x( ) = x3 − 4x2 +1, −1, 3⎡⎣ ⎤⎦ satisfies all the
conditions of the Mean Value Theorem and find c. Polynomials are continuous throughout their domain, so the first condition is
satisfied. Polynomials are also differentiable throughout their domain, so the second
condition is satisfied.
f ' c( ) = 3c2 −8c
f 3( )− f −1( )3− −1( ) =
−8− −4( )4 = −1
3c2 −8c = −13c2 −8c+1= 0
c =8± 82 − 4 3( ) 1( )
2 3( )
c = 2.535 or 0.131 Rolle’s Theorem is a specific a case of the Mean Value theorem, though Joseph-Louis Lagrange used it to prove the Mean Value Theorem. Therefore, Rolle’s Theorem was used to prove all of the rules we have used to interpret derivatives for the last couple of years. It was a very useful theorem, but it is now something of a historical curiosity.
Rolle’s Theorem
If f is a function that satisfies these three hypotheses 1. f is continuous on the closed interval
a,b⎡⎣ ⎤⎦
2. f is differentiable on the closed interval a,b( )
3. f(a) = f(b) Then there is a number c in the interval
a,b( ) such that
f ' c( ) = 0 .
279
Written in this typically mathematical way, it is a bit confusing, but it basically says that if you have a continuous, smooth curve with the initial point and the ending point at the same height, there is some point in the curve that has a derivative of zero. If you look at this from a graphical perspective, it should be pretty obvious.
6
4
2
–2
–4
–2 2
ba
Ex 2 Show that the function
f x( ) = x2 − 4x +1, 0,4⎡⎣ ⎤⎦ satisfies all the conditions
of the Mean Value Theorem and find c. Polynomials are continuous throughout their domain, so the first condition is
satisfied. Polynomials are also differentiable throughout their domain, so the second
condition is satisfied.
f ' c( ) = f b( )− f a( )
b− a → f ' c( ) = f 4( )− f 0( )4−1 = 0
f ' c( ) = 2c − 4 = 0
c = 2
Ex 3 BC 2004B #2
280
Ex 4 Suppose that f 0( ) = −3 , and
f ' x( ) ≤ 5 for all values of x. What is the largest
possible value for f 2( )?
We know that f is differentiable for all values of x and must, therefore, also
be continuous. So we just make up an interval to look at using the mean value theorem. The interval will be [0, 2] because we are looking at
f 0( )
and f 2( ) .
f ' c( ) = f 2( )− f 0( )
2−0 =f 2( )− −3( )
2
2 f ' c( ) = f 2( )+3
2 f ' c( )−3= f 2( ) Since we know the maximum value for
f ' c( ) is 5, plug in 5 for
f ' c( ) , and
we get
2 5( )− 3= f 2( )f 2( ) = 7
Therefore the maximum possible value for f 2( ) is 7.
Other Theorems that can be confused with
The Mean Value Theorem
The Average Value Theorem : The average value of a function f on a closed
interval [a, b] is defined as favg =
1b− a
f x( )dxa
b
∫ .
Average Rate of Change: The average value of a function f on a closed interval
[a, b] is defined as =
f b( )− f a( )b− a .
The Intermediate Value Theorem: If f is continuous on the closed interval [a, b] and f (b) and f (a) are opposite signs, there is a zero in the closed interval.
281
4.6 Homework Verify that the following functions fit all the conditions of Rolle’s Theorem, and then find all values of c that satisfy the conclusion of Mean Value Theorem. 1.
f x( ) = x3 − 3x2 + 2x +5, 0,2⎡⎣ ⎤⎦
2.
f x( ) = sin 2πx( ), −1,1⎡⎣ ⎤⎦
3.
g t( ) = t t + 6, −4,0⎡⎣ ⎤⎦
Given the graph of the function below, estimate all values of c that satisfy the conclusion of the Mean Value Theorem for the interval [1,8]. 4.
1 2 3 4 5 6 7 8 9 10
−4
−3
−2
−1
1
2
3
4
x
y
282
5. Given the graph of the function below, estimate all values of c that satisfy the conclusion of the Mean Value Theorem for the interval [1,9].
1 2 3 4 5 6 7 8 9 10
−4
−3
−2
−1
1
2
3
4
x
y
AP Handout 6. AP Calc AB/BC 2004B #3 7. AP Calc AB 2002 #6
283
4.7 Accumulation of Rates Beginning in 2002, AP shifted emphasis on understanding of the accumulation aspect of the Fundamental Theorem to a new kind of rate problem. Previously, accumulation of rates problems were mostly in context of velocity and distance, though the 1996 Cola Consumption problem hinted at the direction the test would take. The “Amusement Park Problem” of 2002 caught many students and teachers off guard. Almost every year since then, the test has included this kind of problem. Here is an example similar to the 2002 Amusement Park Problem: Ex 1 The rate at which people enter a park is given by the function
( ) 21560024 160
E tt t
=− +
, and the rate at which they are leaving is given by
( ) 29890 7638 370
L tt t
= −− +
. Both ( )E t and ( )L t are measured in people per hour
where t is the number of hours past midnight. The functions are valid for when the park is open, 8 24t≤ ≤ . At t = 8 there are no people in the park. a) How many people have entered the park at 4 pm (t = 16)? Round your answer to the nearest whole number. b) The price of admission is $36 until 4 pm (t = 16). After that, the price drops to $20. How much money is collected from admissions that day? Round your answer to the nearest whole number. c) Let
H t( ) = E x( )− L x( )dx
8
t∫ for 8 24t≤ ≤ . The value of ( )16H to the
nearest whole number is 5023. Find the value of ( )' 16H and explain the meaning of ( )16H and ( )' 16H in the context of the amusement park. d) At what time t, for 8 24t≤ ≤ , does the model predict the number of people in the park is at a maximum.
a) Since ( )E t is a rate in people per hour, the number of people who have entered the park will be an integral from 8 to 16.
284
Total entered = E t( )dt
8
16∫ = 6126 people
b) Since there is an entry fee per person, and we saw in the first part that the integral gave us the number of people the total revenue that the park gets from admissions just means multiplying the number of people by the admission charge.
Total entered at $36/person
= E t( )dt
8
16∫ = 6126 people
Total entered at $20/person
= E t( )dt
16
24∫ =1808 people
Total revenue = $36(6126) + $20(1808) = $256,696
c) Since ( ) ( ) ( )8
tH t E x L x dx= −∫ , we can use the Fundamental Theorem of
Calculus to determine the derivative.
H ′ t( ) = ddt
E x( )− L x( )dx8
t∫
= E t( )− L t( )= E 16( )− L 16( )=14
But we still need to interpret the meaning of the numbers.
( )16 5023H = people – since it was defined as an integral, we knew that
integrating the rates gave us people. This is how many people are in the park at that moment.
( )' 16 14H = people per hour – since the original equations we were provided
were rates, and ( ) ( ) ( )'H t E t L t= − , so this is how many people are entering the park at t = 16. Since the value is positive, this is the rate at which the number of people in the park is increasing.
d) To find the maximum number of people, we have to set the derivative,
( )'H t , equal to zero. We should also check the endpoints, which are also
285
critical values (we could just do a sign pattern instead to verify that the zero is the maximum).
( ) ( ) ( )' 0H t E t L t= − =
Using a graphing calculator, we find the zero is at t = 16.046 Therefore we have critical values at t = 0, 16.046, and 24.
When t = 0, H = 0 (this was given at the beginning of the problem) When t = 16.046, H = 5023 When t = 24, H = 1453
Objective Analyze the interplay between rates and accumulation in context. Ex 2 A certain industrial chemical reaction produces synthetic oil at a rate of
( ) 151 3tS tt
=+
. At the same time, the oil is removed from the reaction vessel by a
skimmer that has a rate of ( ) 42 5sin25
R t tπ⎛ ⎞⎜ ⎟⎝ ⎠
= + . Both functions have units of
gallons per hour, and the reaction runs from t = 0 to t = 6. At time of t = 0, the reaction vessel contains 2500 gallons of oil. a) How much oil will the skimmer remove from the reaction vessel in this six hour period? Indicate units of measure. b) Write an expression for ( )P t , the total number of gallons of oil in the reaction vessel at time t. c) Find the rate at which the total amount of oil is changing at t = 4. d) For the interval indicated above, at what time t is the amount of oil in the reaction vessel at a minimum? What is the minimum value? Justify your answers.
286
a) Amountremoved = 2+5sin 4π
25t⎛
⎝⎜⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥dt
0
6⌠
⌡⎮⎮ = 31.816 gallons
b) P t( ) = 2500+ S x( )− R x( )⎡
⎣⎤⎦0
t
∫ dx
c) S 4( )− R 4( ) = −1.909 gal/hr.
d) Critical values occur when
S t( ) = R t( ) and at the endpoints. Our
graphing calculator shows the time when S t( ) = R t( ) is t = 5.117 .
t
P t( )
0 2500 5.117 2492.367 6 2493.277
The minimum is 2492.367 gallons and occurs at t = 5.117 .
287
4.7 Homework Handout of AP Questions 1. AB/BC 2002B #2 2. AB 2005B #2 3. AB 2006 #2 4. BC 2011 #2 5. BC 2015 #1 6. The number of parts per million (ppm), C(t), of chlorine in a pool changes at the rate of ( ) 0.2' 1 3 tP t e−= − ounces per day, where t is measured in days. There are 50 ppm of chlorine in the pool at time t = 0. Chlorine should be added to the pool if the level drops below 40 ppm. a) Is the amount of chorine increasing or decreasing at t = 9? Why or why not? b) For what value of t is the amount of chlorine at a minimum? Justify your answer. c) When the value of chlorine is at a minimum, does chlorine need to be added? Justify your answer. 7. The basement of a house is flooded, and water keeps pouring in at a rate of ( ) ( )2
695 sin tw t t= gallons per hour. At the same time, water is being pumped
out at a rate of ( ) ( )23275sin tr t = . When the pump is started, at time t = 0, there is
1200 gallons of water in the basement. Water continues to pour in and be pumped out for the interval 0 18t≤ ≤ . a) Is the amount of water increasing at t = 15? Why or why not? b) To the nearest whole number, how many gallons are in the basement at the time t = 18?
288
c) At what time t, for 0 18t≤ ≤ , is the amount of water in the basement at an absolute minimum? Show the work that leads to this conclusion. d) For t > 18, the water stops pouring into the basement, but the pump continues to remove water until all of the water is pumped out of the basement. Let k be the time at which the tank becomes empty. Write, but do not solve, an equation involving an integral expression that can be used to find a value of k.
8. A tank at a sewage processing plant contains 125 gallons of raw sewage at time t = 0. During the time interval 0 12t≤ ≤ hours, sewage is pumped into the
tank at the rate ( ) ( )102
1 ln 1E t
t= +
+ +. During the same time interval, sewage is
pumped out at a rate of ( )2
12sin47tL t
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= .
a) How many gallons of sewage are pumped into the tank during the time interval 0 12t≤ ≤ hours? b) Is the level of sewage rising or falling at t = 6? Explain your reasoning. c) How many gallons of sewage are in the tank at t = 12 hours? d) At what time t, for 0 12t≤ ≤ , is the volume of the sewage at an absolute maximum? Show the analysis that leads to your answer. If the sewage level ever exceeds 150 gallons, the tank overflows. Is there a time at which the tank overflows? Explain. 9. At an intersection in San Francisco, cars turn left at the rate
( ) 250 sin3tL t t ⎛ ⎞
⎜ ⎟⎝ ⎠
= cars per hour for the time interval 0 18t≤ ≤ .
a) To the nearest whole number, find the total number of cars turning left on the time interval given above. b) Traffic engineers will consider turn restrictions if ( )L t equals or exceeds 125 cars per hour. Find the time interval where ( ) 125L t ≥ , and find the average value of ( )L t for this time interval. Indicate units of measure.
289
c) San Francisco will install a traffic light if there is a two hour time interval in which the product of the number of cars turning left and the number of cars travelling through the intersection exceeds 160,000. In every two hour interval, 480 cars travel straight through the intersection. Does this intersection need a traffic light? Explain your reasoning.
290
Definite Integral Chapter Test
1. Find 6x1
4⌠⌡⎮
dx
a. 12 b. 8 c. 47 d. 24 e. 48
2. dx3x +10
5⌠⌡⎮
a. 12 b. 2
3 c. 1 d. 2 e. 6
3. 1
1− 14 t2
1
2⌠
⌡
⎮⎮
dt
a. π
2 b. π3 c. π
6 d. π e. π4
4. The solution to the differential equation dydx
= 8xy with the initial condition
y(0) = 5 is a. ln 4x2 + 5( ) b. e4x2 + 5 c. e4x2 + 4
d. 5ln 4x2( ) e. 5e4x2
291
5. Which of the following is equal to cosx0
π∫ dx ?
a. sin x
0
π∫ dx b. cosx−π
2
π2∫ dx c. sin x−π
2
π2∫ dx
d. sin xπ
2π∫ dx e. cosxπ
2
3π2∫ dx
6. Consider the function f x( ) is shown below.
If the Trapezoidal Rule is used with n = 4 to approximate f x( )
1
9∫ dx the result is
a. 21 b. 22 c. 23 d. 24 e. 25 1. The figure below shows the graph of f ′ , the derivative of a twice-differentiable function f, on the closed interval 0 ≤ x ≤8 . The graph of has horizontal tangent lines at x =1, x = 3 , and x = 5 . The areas of the regions between the graph of and the -axis are labeled in the figure. The function is defined for all real numbers and satisfies f 8( ) = 4 .
292
a. Find all values of x on the open interval 0 < x <8 for which the function has a local minimum. Justify your answer. b. Determine the absolute minimum value of f on the closed interval 0 ≤ x ≤8 . Justify your answer. c. On what open intervals contained in 0 < x <8 is the graph of f both concave down and increasing? Explain your reasoning.
d. The function g is defined by g x( )= f x( )( )3 . If
f 3( )=− 5
2 find the slope of the
line tangent to the graph of at x= 3 .
293
t hours 0 2 5 7 8
E(t) (hundreds of entries)
0
4
13
21
23
2. A zoo sponsored a one-day contest to name a new baby elephant. Zoo visitors deposited entries in a special box between noon (t = 0) and 8 P.M. (t = 8). The number of entries in the box t hours after noon is modeled by a differentiable function E for 0 ≤ t ≤ 8 . Values of E(t), in hundreds of entries, at various times t are shown in the table above. a. Use the data in the table to approximate the rate, in hundreds of entries per hour, at which entries were being deposited at time t = 6. Show the computations that lead to your answer. b. Use a trapezoidal sum with the four subintervals given by the table to
approximate the value of 18 E t( )dt0
8∫ . Using correct units, explain the meaning of
18 E t( )dt0
8∫ in terms of the number of entries.
c. At 8 P.M., volunteers began to process the entries. They processed the entries at a rate modeled by the function P, where P t( ) = t 3 − 30t2 + 298t − 976 hundreds of entries per hour for 8 ≤ t ≤12 . According to the model, how many entries had not yet been processed by midnight (t = 12) ? d. According to the model from part (c), at what time were the entries being processed most quickly? Justify your answer.
294
3. The number of parts per million (ppm), C(t), of chlorine in a pool changes at the rate of
P ' t( ) =1−3e−0.3 t ounces per day, where t is measured in days. There are
50 ppm of chlorine in the pool at time t = 0. Chlorine should be added to the pool if the level drops below 40 ppm.
a. Is the amount of chorine increasing or decreasing at t = 9? Why or why not? b. For what value of t is the amount of chlorine at a minimum? Justify your answer. c. When the value of chlorine is at a minimum, does chlorine need to be added? Justify your answer.
295
4.1 Homework:
1.
3643
2. 107.5 3. 0 4. 280.605
5.
78
6. 0 7. 1 8. 3ln3
9. 257 10.
163
11.
2 33
12.
4π
13.
12
ln3 14. 225.753°F 15. 8.201 hrs 16. 63.045°F
17. y
2 sin y 18. 1+ 2x 19. −cos(x2)
20. −
1x2 arctan 1
x 21.
cos x2x
4.2 Homework
1.
1829
2. 4 3. e− e1
2 4. 1
5. 2 6. 12 ln3 7. ln3 8. ln ln4
ln2⎛⎝⎜
⎞⎠⎟
9. tan−1 2 − π4 10.
9128
11. -2 12. 10710
13. 0 14. −1
8e−25 − e−1( ) 15. −
325 16. 2
17
4.3 Homework:
1. 11.833 2. 4.917 3. 35.5 4. 3
296
5. ln2 6. 2.704 7. 1.627 8. 9.519
9. 2.707 10. 9.804 11. (a) − 32m and (b) 416 m
12. (a) −103 m and (b) 983 m
13.
e−x2 − x( )dx0
a⌠⌡⎮
− e−x2 − x( )dxa
2⌠⌡⎮
= 3.080 , where a = 0.65291864
14.
e−x2 − 2x( )dx0
b⌠⌡⎮
− e−x2 − 2x( )dxb
2⌠⌡⎮
= 5.305 , where b = 0.41936482
15. 2.235
16. 13.030
4.4 Homework: 1-5. See the AP Website
6a. g 3( ) = 3; g′ 3( ) = f 3( ) = 2; g″ 3( ) = f ′ 3( ) = − 12
6b. 73 6c. two 6d. x = 2 and 5 , because g′ x( ) = f x( ) and f x( )
changes from increasing to decreasing and vice versa.
7a. g 4( ) =π − 8; g′ 2( ) = f 2( ) = 0; g″ 4( ) = f ′ 4( ) = dne
7b. 0
7c. x∈ 2, 5( )because f x( ) = g′ x( ) and f x( ) is negative and increasing.
7d. x = −7
297
8a. g 0( ) = 2π − 8; g′ 0( ) = f 0( ) = −2; g″ 0( ) = f ′ 0( ) = dne
8b. y− 2π − 8( ) = −2 x− 0( )
8c. x∈ −4, − 2( ) because f x( ) = g′ x( ) and f x( ) is negative and increasing.
8d. x = 7
9a. g 4( ) =14; g′ 4( ) = f 4( ) = −2; g″ 4( ) = f ′ 4( ) = 0
9b. 1 9c. 1 9d. g 0( ) =10 4.5 Homework:
1. 4,940 miles 2a. 7.25 2b. 7.25 2c. 30
3a. 1,880 3b. 840 3c. 1,180 3d. 1,300
4a. 1,280 4b. 600 5a. 10,190 5b. 6,850
6a. 392 6b. 252 7. C
8a. 1,130 8b. 1,134
9. See AP Website
10a. 2.414 10b. 2.411 11a. .147 11b. .147 12.
w '(t)
5
10
∫ dt represents the change of the child’s weight in pounds from ages 5 to 10 years 13.
r(t)dt
0
120
∫ represents how much oil, in gallons, has leaked from the tank in the first 120 minutes.
298
14. Newtons motors, or joules 4.6 Homework 1. x = .423, 1.577 2. ⇒ x = 14, 34 3. t = −2.150
4. f 8( )− f 1( )8−1 ≈ ′f 3.2( ) 5.
f 9( )− f 1( )9−1 ≈ ′f 2.4( ) = ′f 4.6( ) = ′f 6.9( )
6. See AP Central 7. See AP Central 4.7 Homework
1-5. See AP Central
6a. Decreasing; P′ t( ) < 0 6b. 30.173
6c. 35.104, yes 7a. No, w− r < 0 7b. 1310
7c. 6.495 7d. 1310+ R t( )0
k
∫ dt 8a. 70.571
8b. Falling; E − L < 0 8c. 122.026 gallons
8d. t=4.790; max=149.408, so, no.
9a. 1382 cars 9b. t ∈ 12.42831,16.121657⎡⎣ ⎤⎦
9c. Yes. Between t =13 and t =15 , 360 cars turn left and 480 cars come straight, for a product of 172,800, which exceeds the 160,000 allowed.
