Letf(t)be a con nuous func on defined on[a,b]. The definite integral∫b a f(x)dx is the “area underf” on[a,b]. We can turn this concept into a func on by le ng the upper (or lower) bound vary.
LetF(x) = ∫x
a f(t)dt. It computes the area underfon[a,x]as illustrated in Figure 5.22. We can study this func on using our knowledge of the definite integral. For instance,F(a) =0 since∫a
a f(t)dt=0.
We can also apply calculus ideas toF(x); in par cular, we can compute its deriva ve. While this may seem like an innocuous thing to do, it has far–reaching implica ons, as demonstrated by the fact that the result is given as an important theorem.
Theorem 39 The Fundamental Theorem of Calculus, Part 1 Letfbe con nuous on[a,b]and letF(x) =∫x
a f(t)dt. ThenFis a differ- en able func on on(a,b), and
F′(x) =f(x).
Ini ally this seems simple, as demonstrated in the following example.
Example 125 Using the Fundamental Theorem of Calculus, Part 1 LetF(x) =
∫ x
−5
(t2+sint)dt. What isF′(x)?
S Using the Fundamental Theorem of Calculus, we haveF′(x) = x2+sinx.
This simple example reveals something incredible: F(x)is an an deriva ve ofx2+sinx! Therefore,F(x) = 13x3−cosx+Cfor some value ofC. (We can findC, but generally we do not care. We know thatF(−5) =0, which allows us to computeC. In this case,C=cos(−5) +1253 .)
We have done more than found a complicated way of compu ng an an- deriva ve. Consider a func onfdefined on an open interval containinga,b andc. Suppose we want to compute∫b
a f(t)dt. First, letF(x) =∫x
c f(t)dt. Using
Notes:
the proper es of the definite integral found in Theorem 36, we know
∫ b a
f(t)dt=
∫ c a
f(t)dt+
∫ b c
f(t)dt
=−
∫ a c
f(t)dt+
∫ b c
f(t)dt
=−F(a) +F(b)
=F(b)−F(a).
We now see how indefinite integrals and definite integrals are related: we can evaluate a definite integral using an deriva ves! This is the second part of the Fundamental Theorem of Calculus.
Theorem 40 The Fundamental Theorem of Calculus, Part 2 Letfbe con nuous on[a,b]and letFbeanyan deriva ve off. Then
∫ b a
f(x)dx=F(b)−F(a).
Example 126 Using the Fundamental Theorem of Calculus, Part 2 We spent a great deal of me in the previous sec on studying∫4
0(4x−x2)dx.
Using the Fundamental Theorem of Calculus, evaluate this definite integral.
S We need an an deriva ve off(x) =4x−x2. All an deriva- ves offhave the formF(x) =2x2−13x3+C; for simplicity, chooseC=0.
The Fundamental Theorem of Calculus states
∫ 4 0
(4x−x2)dx=F(4)−F(0) =(
2(4)2−1 343)
−( 0−0)
=32−64
3 =32/3.
This is the same answer we obtained using limits in the previous sec on, just with much less work.
Nota on:A special nota on is o en used in the process of evalua ng definite integrals using the Fundamental Theorem of Calculus. Instead of explicitly writ- ingF(b)−F(a), the nota onF(x)b
a is used. Thus the solu on to Example 126 would be wri en as:
∫ 4 0
(4x−x2)dx= (
2x2−1 3x3)
4
0
=(
2(4)2−1 343)
−( 0−0)
=32/3.
Notes:
The ConstantC: Anyan deriva veF(x)can be chosen when using the Funda- mental Theorem of Calculus to evaluate a definite integral, meaning any value ofCcan be picked. The constantalwayscancels out of the expression when evalua ngF(b)−F(a), so it does not ma er what value is picked. This being the case, we might as well letC=0.
Example 127 Using the Fundamental Theorem of Calculus, Part 2 Evaluate the following definite integrals.
1.
∫ 2
−2
x3dx 2.
∫ π 0
sinx dx 3.
∫ 5 0
etdt 4.
∫ 9 4
√u du 5.
∫ 5 1
2dx S
1.
∫ 2
−2
x3dx= 1 4x4
2
−2
= (1
424 )
− (1
4(−2)4 )
=0.
2.
∫ π 0
sinx dx=−cosxπ
0 =−cosπ−(
−cos 0)
=1+1=2.
(This is interes ng; it says that the area under one “hump” of a sine curve is 2.)
3.
∫ 5 0
etdt=et5
0 =e5−e0=e5−1≈147.41.
4.
∫ 9 4
√u du=
∫ 9 4
u12 du= 2 3u32
9
4
= 2 3
(
932 −432)
= 2 3
(27−8)
= 38 3. 5.
∫ 5 1
2dx=2x5
1=2(5)−2=2(5−1) =8.
This integral is interes ng; the integrand is a constant func on, hence we are finding the area of a rectangle with width(5−1) =4 and height 2.
No ce how the evalua on of the definite integral led to 2(4) =8.
In general, ifcis a constant, then∫b
a c dx=c(b−a).
Understanding Mo on with the Fundamental Theorem of Calcu- lus
We established, star ng with Key Idea 1, that the deriva ve of a posi on func on is a velocity func on, and the deriva ve of a velocity func on is an ac- celera on func on. Now consider definite integrals of velocity and accelera on func ons. Specifically, ifv(t)is a velocity func on, what does
∫ b a
v(t)dtmean?
Notes:
The Fundamental Theorem of Calculus states that
∫ b a
v(t)dt=V(b)−V(a),
whereV(t)is any an deriva ve ofv(t). Sincev(t)is a velocity func on,V(t) must be a posi on func on, andV(b)−V(a)measures a change in posi on, or displacement.
Example 128 Finding displacement
A ball is thrown straight up with velocity given byv(t) =−32t+20 /s, where tis measured in seconds. Find, and interpret,
∫ 1
0
v(t)dt.
S Using the Fundamental Theorem of Calculus, we have
∫ 1 0
v(t)dt=
∫ 1 0
(−32t+20)dt
=−16t2+20t1
0
=4.
Thus if a ball is thrown straight up into the air with velocityv(t) =−32t+20, the height of the ball, 1 second later, will be 4 feet above the ini al height. (Note that the ball hastraveledmuch farther. It has gone up to its peak and is falling down, but the difference between its height att=0 andt=1 is 4 .)
Integra ng a rate of change func on gives total change. Velocity is the rate of posi on change; integra ng velocity gives the total change of posi on, i.e., displacement.
Integra ng a speed func on gives a similar, though different, result. Speed is also the rate of posi on change, but does not account for direc on. So inte- gra ng a speed func on gives total change of posi on, without the possibility of “nega ve posi on change.” Hence the integral of a speed func on givesdis- tance traveled.
As accelera on is the rate of velocity change, integra ng an accelera on func on gives total change in velocity. We do not have a simple term for this analogous to displacement. Ifa(t) = 5miles/h2 andtis measured in hours,
then ∫ 3
0
a(t)dt=15
means the velocity has increased by 15m/h fromt=0 tot=3.
Notes:
The Fundamental Theorem of Calculus and the Chain Rule
Part 1 of the Fundamental Theorem of Calculus (FTC) states that givenF(x) =
∫ x a
f(t)dt,F′(x) =f(x). Using other nota on, d dx
(F(x))
=f(x). While we have just prac ced evalua ng definite integrals, some mes finding an deriva ves is impossible and we need to rely on other techniques to approximate the value of a definite integral. Func ons wri en asF(x) = ∫x
af(t)dtare useful in such situa ons.
It may be of further use to compose such a func on with another. As an example, we may composeF(x)withg(x)to get
F( g(x))
=
∫ g(x) a
f(t)dt.
What is the deriva ve of such a func on? The Chain Rule can be employed to state
d dx
( F(
g(x)))
=F′( g(x))
g′(x) =f( g(x))
g′(x).
An example will help us understand this.
Example 129 The FTC, Part 1, and the Chain Rule Find the deriva ve ofF(x) =
∫ x2
2
lnt dt.
S We can viewF(x)as being the func onG(x) =
∫ x 2
lnt dt composed withg(x) =x2; that is,F(x) =G(
g(x))
. The Fundamental Theorem of Calculus states thatG′(x) =lnx. The Chain Rule gives us
F′(x) =G′( g(x))
g′(x)
=ln(g(x))g′(x)
=ln(x2)2x
=2xlnx2
Normally, the steps definingG(x)andg(x)are skipped.
Prac ce this once more.
Example 130 The FTC, Part 1, and the Chain Rule Find the deriva ve ofF(x) =
∫ 5 cosx
t3dt.
Notes:
...
f(x)
.
g(x)
.
a
.
b
.
x
.
y
...
f(x)
.
g(x)
.
a
.
b
.
x
.
y
Figure 5.23: Finding the area bounded by two func ons on an interval; it is found by subtrac ng the area undergfrom the area underf.
...
y=x2+x−5
.y=3x−2 .
−2
.
−1
.
1
.
2
.
3
.
4
.
5
.
10
.
15
.
x
.
y
Figure 5.24: Sketching the region en- closed byy=x2+x−5 andy=3x−2 in Example 131.
S Note thatF(x) =−
∫ cosx 5
t3dt. Viewed this way, the deriva- ve ofFis straigh orward:
F′(x) =sinxcos3x.
Area Between Curves
Consider con nuous func onsf(x)andg(x)defined on[a,b], wheref(x)≥ g(x)for allxin[a,b], as demonstrated in Figure 5.23. What is the area of the shaded region bounded by the two curves over[a,b]?
The area can be found by recognizing that this area is “the area underf− the area underg.” Using mathema cal nota on, the area is
∫ b
a
f(x)dx−
∫ b
a
g(x)dx.
Proper es of the definite integral allow us to simplify this expression to
∫ b a
(f(x)−g(x)) dx.
Theorem 41 Area Between Curves
Letf(x)andg(x)be con nuous func ons defined on[a,b]wheref(x)≥ g(x)for all xin [a,b]. The area of the region bounded by the curves y=f(x),y=g(x)and the linesx=aandx=bis
∫ b a
(f(x)−g(x)) dx.
Example 131 Finding area between curves
Find the area of the region enclosed byy=x2+x−5 andy=3x−2.
S It will help to sketch these two func ons, as done in Figure 5.24. The region whose area we seek is completely bounded by these two func ons; they seem to intersect atx=−1 andx=3. To check, setx2+x−5=
Notes:
...
1
.
2
.
3
.
4
.
x
.
y
Figure 5.25: A graph of a func onfto in- troduce the Mean Value Theorem.
...
1
.
2
.
3
.
4
.
x
.
y
(a)
...
1
.
2
.
3
.
4
.
x
.
y
(b)
...
1
.
2
.
3
.
4
.
x
.
y
(c)
Figure 5.26: Differently sized rectan- gles give upper and lower bounds on
∫4
f(x)dx; the last rectangle matches the
3x−2 and solve forx:
x2+x−5=3x−2 (x2+x−5)−(3x−2) =0
x2−2x−3=0 (x−3)(x+1) =0
x=−1, 3.
Following Theorem 41, the area is
∫ 3
−1
(3x−2−(x2+x−5)) dx=
∫ 3
−1
(−x2+2x+3)dx
= (
−1
3x3+x2+3x) 3
−1
=−1
3(27) +9+9− (1
3+1−3 )
=102 3=10.6
The Mean Value Theorem and Average Value
Consider the graph of a func onfin Figure 5.25 and the area defined by
∫4
1 f(x)dx. Three rectangles are drawn in Figure 5.26; in (a), the height of the rectangle is greater thanfon[1,4], hence the area of this rectangle is is greater than∫4
0 f(x)dx.
In (b), the height of the rectangle is smaller thanfon[1,4], hence the area of this rectangle is less than∫4
1 f(x)dx.
Finally, in (c) the height of the rectangle is such that the area of the rectangle isexactlythat of∫4
0 f(x)dx. Since rectangles that are “too big”, as in (a), and rectangles that are “too li le,” as in (b), give areas greater/lesser than∫4
1 f(x)dx, it makes sense that there is a rectangle, whose top intersectsf(x)somewhere on[1,4], whose area isexactlythat of the definite integral.
We state this idea formally in a theorem.
Notes:
...
..
1
.
2
.
1
.
c
.
π
.
sin 0.69
.
x
.
y
Figure 5.27: A graph ofy = sinxon [0,π] and the rectangle guaranteed by the Mean Value Theorem.
...
y=f(x)
.
a
.
b
.
c
.
f(c)
.
x
.
y
...
y=f(x)−f(c)
.
a
.
b
.
c
.
f(c)
.
x
.
y
Figure 5.28: On top, a graph of y = f(x)and the rectangle guaranteed by the Mean Value Theorem. Below,y=f(x)is shi ed down byf(c); the resul ng “area under the curve” is 0.
Theorem 42 The Mean Value Theorem of Integra on
Letfbe con nuous on[a,b]. There exists a valuecin[a,b]such that
∫ b a
f(x)dx=f(c)(b−a).
This is anexisten alstatement;cexists, but we do not provide a method of finding it. Theorem 42 is directly connected to the Mean Value Theorem of Differen a on, given as Theorem 27; we leave it to the reader to see how.
We demonstrate the principles involved in this version of the Mean Value Theorem in the following example.
Example 132 Using the Mean Value Theorem Consider∫π
0 sinx dx. Find a valuecguaranteed by the Mean Value Theorem.
S We first need to evaluate∫π
0 sinx dx. (This was previously done in Example 127.)
∫ π 0
sinx dx=−cosxπ
0 =2.
Thus we seek a valuecin[0,π]such thatπsinc=2.
πsinc=2 ⇒ sinc=2/π ⇒ c=arcsin(2/π)≈0.69.
In Figure 5.27 sinxis sketched along with a rectangle with height sin(0.69).
The area of the rectangle is the same as the area under sinxon[0,π].
Letfbe a func on on[a,b]withcsuch thatf(c)(b−a) =∫b
a f(x)dx. Consider
∫b a
(f(x)−f(c)) dx:
∫ b a
(f(x)−f(c)) dx=
∫ b a
f(x)−
∫ b a
f(c)dx
=f(c)(b−a)−f(c)(b−a)
=0.
Whenf(x)is shi ed by−f(c), the amount of area underfabove thex–axis on [a,b]is the same as the amount of area below thex–axis abovef; see Figure 5.28 for an illustra on of this. In this sense, we can say thatf(c)is theaverage valueoffon[a,b].
Notes:
The valuef(c)is the average value in another sense. First, recognize that the Mean Value Theorem can be rewri en as
f(c) = 1 b−a
∫ b a
f(x)dx,
for some value ofcin [a,b]. Next, par on the interval[a,b] intonequally spaced subintervals,a = x1 < x2 < . . . < xn+1 = band choose anyci in [xi,xi+1]. The average of the numbersf(c1),f(c2), …,f(cn)is:
1 n (
f(c1) +f(c2) +. . .+f(cn) )
=1 n
∑n i=1
f(ci).
Mul ply this last expression by 1 in the form of(b(b−−a)a): 1
n
∑n i=1
f(ci) =
∑n i=1
f(ci)1 n
=
∑n i=1
f(ci)1 n
(b−a) (b−a)
= 1 b−a
∑n i=1
f(ci)b−a n
= 1 b−a
∑n i=1
f(ci)∆x (where∆x= (b−a)/n)
Now take the limit asn→ ∞:
nlim→∞
1 b−a
∑n i=1
f(ci)∆x = 1 b−a
∫ b a
f(x)dx = f(c).
This tells us this: when we evaluatefatn(somewhat) equally spaced points in [a,b], the average value of these samples isf(c)asn→ ∞.
This leads us to a defini on.
Defini on 22 The Average Value offon[a,b]
Letfbe con nuous on[a,b]. Theaverage value of f on[a,b]isf(c), wherecis a value in[a,b]guaranteed by the Mean Value Theorem. I.e.,
Average Value offon[a,b] = 1 b−a
∫ b a
f(x)dx.
Notes:
An applica on of this defini on is given in the following example.
Example 133 Finding the average value of a func on
An object moves back and forth along a straight line with a velocity given by v(t) = (t−1)2on[0,3], wheretis measured in seconds andv(t)is measured in /s.
What is the average velocity of the object?
S By our defini on, the average velocity is:
1 3−0
∫ 3 0
(t−1)2dt= 1 3
∫ 3 0
(t2−2t+1) dt= 1
3 (1
3t3−t2+t) 3
0
=1 /s.
We can understand the above example through a simpler situa on. Suppose you drove 100 miles in 2 hours. What was your average speed? The answer is simple: displacement/ me = 100 miles/2 hours = 50 mph.
What was the displacement of the object in Example 133? We calculate this by integra ng its velocity func on:∫3
0(t−1)2dt=3 . Its final posi on was 3 feet from its ini al posi on a er 3 seconds: its average velocity was 1 /s.
This sec on has laid the groundwork for a lot of great mathema cs to fol- low. The most important lesson is this: definite integrals can be evaluated using an deriva ves. Since the previous sec on established that definite integrals are the limit of Riemann sums, we can later create Riemann sums to approximate values other than “area under the curve,” convert the sums to definite integrals, then evaluate these using the Fundamental Theorem of Calculus. This will allow us to compute the work done by a variable force, the volume of certain solids, the arc length of curves, and more.
The downside is this: generally speaking, compu ng an deriva ves is much more difficult than compu ng deriva ves. The next chapter is devoted to tech- niques of finding an deriva ves so that a wide variety of definite integrals can be evaluated. Before that, the next sec on explores techniques of approximat- ing the value of definite integrals beyond using the Le Hand, Right Hand and Midpoint Rules.
Notes:
Terms and Concepts
1. How are definite and indefinite integrals related?
2. What constant of integra on is most commonly used when evalua ng definite integrals?
3. T/F: Iffis a con nuous func on, thenF(x) =
∫ x
a
f(t)dtis also a con nuous func on.
4. The definite integral can be used to find “the area under a curve.” Give two other uses for definite integrals.
Problems
In Exercises 5 – 28, evaluate the definite integral.
5.
∫ 3
1
(3x2−2x+1)dx
6.
∫ 4
0
(x−1)2dx
7.
∫ 1
−1
(x3−x5)dx 8.
∫ π π/2
cosx dx
9.
∫ π/4
0
sec2x dx
10.
∫ e
1
1 xdx
11.
∫ 1
−1
5xdx
12.
∫ −1
−2
(4−2x3)dx
13.
∫ π 0
(2 cosx−2 sinx)dx
14.
∫ 3
1
exdx
15.
∫ 4
0
√t dt
16.
∫ 25
√1 dt
18.
1
1 xdx
19.
∫2
1
1 x2dx
20.
∫2
1
1 x3dx
21.
∫1 0
x dx
22.
∫1
0
x2dx
23.
∫1
0
x3dx
24.
∫1
0
x100dx
25.
∫4
−4
dx
26.
∫−5
−10
3dx
27.
∫2
−2
0dx
28.
∫ π/3
π/6
cscxcotx dx
29. Explain why:
(a)
∫1
−1
xndx=0, whennis a posi ve, odd integer, and (b)
∫1
−1
xndx = 2
∫ 1
0
xndxwhennis a posi ve, even integer.
In Exercises 30 – 33, find a valuecguaranteed by the Mean Value Theorem.
30.
∫2
0
x2dx
31.
∫2
−2
x2dx
32.
∫1 exdx
34. f(x) =sinxon[0,π/2]
35. y=sinxon[0,π]
36. y=xon[0,4]
37. y=x2on[0,4]
38. y=x3on[0,4]
39. g(t) =1/ton[1,e]
In Exercises 40 – 44, a velocity func on of an object moving along a straight line is given. Find the displacement of the object over the given me interval.
40. v(t) =−32t+20 /s on[0,5]
41. v(t) =−32t+200 /s on[0,10]
42. v(t) =2tmph on[−1,1]
43. v(t) =cost /s on[0,3π/2]
44. v(t) =√4
t /s on[0,16]
object’s velocity over the given me interval.
45. a(t) =−32 /s2on[0,2]
46. a(t) =10 /s2on[0,5]
47. a(t) =t /s2on[0,2]
48. a(t) =cost /s2on[0,π]
In Exercises 49 – 52, sketch the given func ons and find the area of the enclosed region.
49. y=2x,y=5x, andx=3.
50. y=−x+1,y=3x+6,x=2 andx=−1.
51. y=x2−2x+5,y=5x−5.
52. y=2x2+2x−5,y=x2+3x+7.
In Exercises 53 – 56, findF′(x).
53. F(x) =
∫ x3+x
2
1 t dt
54. F(x) =
∫ 0 x3
t3dt
55. F(x) =
∫ x2 x
(t+2)dt
56. F(x) =
∫ ex
lnx
sint dt
...
y=e−x2
. 0.5
. 1
.
0.5
.
1
.
x
.
y
...
y=sin(x3)
.
−1
.
1
. −0.5
.
0.5
.
1
.
x
.
y
...
y=sinx x
.
5
.
10
.
15
.
0.5
.
1
.
x
.
y
Figure 5.29: Graphically represen ng three definite integrals that cannot be evaluated using an deriva ves.