162 Chapter 6. Integration
Example Z 2
2
(1+2x−3x2) dx=0 since the function 1+2x−3x2is continuous on [0,3] (for example) and 2∈[0,3].
Example Z 0
1
x2dx = − Z 1
0
x2dx by convention
= −1
3 by (6.1.3)
Terminology In a definite integralRb
a f(x) dx,
• the function f is called theintegrand;
• the numbersaandbare called thelimits of integration;ais thelower limitandbtheupper limit.
Exercise 6.1
1. For each of the following definite integrals, use the results in this section to find its value:
(a) R1
0(1−3x2) dx (b) R1
2 4xdx 2. Use definition to find the definite integralR1
0 x3dx.
Given: 13+23+· · ·+n3= n2(n+1)2
4
6.2. Fundamental Theorem of Calculus 163
By the construction ofF, the required definite integral isF(b). If we can find a formula forF(x), then we can solve the problem. The following result gives a relation betweenFand f.
Fundamental Theorem of Calculus, Version 1Let f be a function that is continuous on a closed and bounded interval [a,b]. LetFbe the function from [a,b]intoRdefined by
F(x)= Z x
a
f(t) dt fora≤t≤b.
ThenFis continuous on [a,b]and differentiable on (a,b)withF0(x)= f(x)for allx∈(a,b).
ExplanationThe proof of this result will be given in the appendix. Below we explain how to “obtain”F0 = f on (a,b) intuitively for the case where f is nonnegative on [a,b]. Recall that
F0(x)= lim
h→0
F(x+h)−F(x)
h .
Forx∈(a,b) and for sufficiently smallh>0 (such thata+h≤b), we have F(x+h)−F(x) =
Z x+h
a
f(t) dt− Z x
a
f(t) dt by construction ofF
= Z x
a
f(t) dt+ Z x+h
x
f(t) dt
!
− Z x
a
f(t) dt Rule (Int3)
=
Z x+h
x
f(t) dt.
a x x+h b
Figure 6.8
Note thatR x+h
x f(t) dtis the area of the region below the graph of f (and above the horizontal axis) from x tox+h. If his small, then [x,x+h] is a short interval and the area of the small region under consideration can be approximated by the area of the rectangular region with base [x,x+h] on the horizontal axis and height equal to f(x). Thus we have
Z x+h
x
f(t) dt is close to f(x)·h ifhis small, from which we obtain
F(x+h)−F(x)
h =
R x+h
x f(t) dt
h is close to f(x) ifhis small.
Taking limit, we getF0(x)= f(x).
RemarkTo be more precise, the above argument gives lim
h→0+
F(x+h)−F(x)
h = f(x) only.
In view of the Fundamental Theorem of Calculus (Version 1), to findRb
a f(x) dx, we should look for func- tionsGsuch thatG0 = f.
Definition Let f be a function that is continuous on a closed and bounded interval [a,b]. Suppose thatGis a function that is defined on [a,b] such that the following two conditions are satisfied:
164 Chapter 6. Integration
(1) Gis continuous on [a,b];
(2) Gis differentiable on (a,b) andG0(x)= f(x) for allx∈(a,b).
Then we say thatGis aprimitivefor f on [a,b].
Example Let f(x) =3x2and letG(x)= x3. Note that f andGare continuous onRand thatG0(x) = f(x) for all x∈R. ThusGis a primitive for f on every closed and bounded interval [a,b].
RemarkPrimitive is not unique. For example, the functionG1(x) = x3+1 is also a primitive for f (on every closed and bounded interval). In fact, for every constantC, the function
x3+C (6.2.1)
is a primitive for f (on every closed and bounded interval). It is natural to ask whether there are any more primitives:
• IfF0(x)= f(x) for allx∈R, mustFbe in the form (6.2.1)?
Corollary 6.2.2, which is based on the following theorem, tells that the answer is affirmative.
Theorem 6.2.1 LetFandGbe functions that are defined on a closed and bounded interval [a,b]. Suppose that FandGare continuous on [a,b]and are differentiable on (a,b)withF0(x) =G0(x)for allx ∈(a,b). Then on [a,b], the functionsFandGdiffer by a constant, that is, there exists a constantCsuch that
F(x)−G(x)=C for allx∈[a,b].
ExplanationThe following is the geometry meaning of the result:
• The condition “F0(x)=G0(x)for all x∈(a,b)” means that at corresponding points (samex-coordinates), tangents to the graphs ofFandGare parallel.
• The conclusion is that the graph of F can be obtained from that ofGby moving it upward (C > 0) or downward (C<0).
Proof Let f be the function from [a,b] intoRdefined by f(x)= F(x)−G(x).
Note that f0(x) = F0(x)−G0(x) =0 for all x∈(a,b). Hence by Theorem 5.1.1, there exists a constantCsuch that
f(x)=C for allx∈(a,b).
Since f is continuous on [a,b], it follows that
f(x)=C for allx∈[a,b]
from which we get the required result.
Corollary 6.2.2 Let f be a function that is continuous on a closed and bounded interval [a,b]. Suppose thatF andGare primitives for f on [a,b]. Then on [a,b], the functionsFandGdiffer by a constant.
6.2. Fundamental Theorem of Calculus 165
Proof This is an immediate consequence of Theorem 6.2.1 since by the definition of primitive, the functions FandGare continuous on [a,b] and differentiable on (a,b) andF0(x)= f(x)=G0(x) for allx∈(a,b).
Example Find the value of the definite integral Z 2
1
3x2dx.
Solution Let f(x)=3x2and letG(x)= x3. ThenGis a primitive for f on [1,2].
By the Fundamental Theorem of Calculus (Version 1), the functionFgiven by F(x)=
Z x
1
3t2,dt, 1≤ x≤2.
is a primitive for f on [1,2].
By Corollary 6.2.2, on the interval [1,2], the functions F andG differ by a constant, that is, there exists a constantCsuch that
F(x)−x3=C for allx∈[1,2].
Puttingx=1 and using the construction ofF, we get 0−1=C.
which implies thatF(x)= x3−1 for allx∈[1,2]. Hence, we have Z 2
1
3x2dx = F(2) by construction ofF
= 23−1
= 7.
RemarkThe above procedure can be used to find the definite integral of f on any closed and bounded interval [a,b]. This is becauseGis a primitive for f on every [a,b].
From the above example, we see that given a function f that is continuous on a closed and bounded interval [a,b], if we can find a primitive forf over [a,b], then we can find the definite integralRb
a f(x) dx. The following result describe an alternative procedure for findingRb
a f(x) dx(there is no need to find the constantC).
Fundamental Theorem of Calculus, Version 2Let f be a function that is continuous on a closed and bounded interval [a,b]. Suppose thatGis a primitive for f on [a,b]. Then we have
Z b
a
f(x) dx=G(b)−G(a).
Proof LetF be the function from [a,b] intoRdefined by F(x)=
Z x
a
f(t) dt, a≤ x≤b.
By the Fundamental Theorem of Calculus, Version 1, the functionF is a primitive for f on [a,b]. Hence by Corollary 6.2.2, there exists a constantCsuch that
F(x)−G(x)=C for allx∈[a,b]. (6.2.2)
166 Chapter 6. Integration
Therefore, we have Z b
a
f(x) dx = F(b) by construction ofF
= F(b)−F(a) sinceF(a)=0
=
G(b)+C
−
G(a)+C
by (6.2.2)
= G(b)−G(a)
Below we redo the last example using the second version of the Fundamental Theorem of Calculus.
Example Evaluate Z 2
1
3x2dx
Solution Since the functionG(x) = x3is a primitive for the function 3x2on the interval [1,2], it follows from the Fundamental Theorem of Calculus (Version 2) that
Z 2
1
3x2dx = G(2)−G(1)
= 23−13
= 7.
FAQ Can we use other primitives for f?
Answer The Fundamental Theorem tells that any primitive will work.Try it yourselves.
Notation We will use the notationG(b)−G(a) quite often. For simplicity, it will be denoted by hG(x)ib
a or G(x) ba. Example Find
Z 5
3
2xdx.
ExplanationTo findRb
a f(x) dx, in applying the Fundamental Theorem of Calculus (Version 2), we have to find a functionGthat is continuous on [a,b] such thatG0 = f on (a,b). In this course, functions that we considered are “nice”—there is no need to check continuity; we just need to check thatG0 = f (usually valid on a much larger interval).
Solution By inspection, we see that the functionx2 is a primitive for the integrand 2x(on every closed and bounded interval). Thus by the Fundamental Theorem of Calculus (Version 2), we have
Z 5
3
2xdx = h x2i5
3
= 52−32
= 16.
Remark The definite integral is the area of the trapezoidal region that lies below the liney = 2x, above the x-axis and is bounded on the left and right by the vertical linesx = 3 and x = 5. Use formula to check the answer yourselves.