168 Chapter 6. Integration
Theorem 6.3.1 means that if we can find one antiderivative for a continuous function f on an open interval (a,b), then we can find all. More precisely, ifF is an antiderivative for f on (a,b), then all the antiderivatives for f on (a,b) are in the form
F(x)+C, a< x<b (6.3.1)
whereCis a constant.
Note that (6.3.1) represents a family of functions defined on (a,b)—there are infinitely many of them, with eachCcorresponds to an antiderivative for f and vice versa. We call the family to be theindefinite integralof
f (with respect tox) and we denote it by Z
f(x) dx.
That is, Z
f(x) dx=F(x)+C, a<x<b,
whereFis a function such thatF0(x)= f(x) for all x∈(a,b) andCis an arbitrary constant, calledconstant of integration.
Example Using the two results in the last example, we have the following:
(1) Z
x2dx= 1
3x3+C, −∞< x<∞, whereCis an arbitrary constant.
(2)
Z 1
√xdx=2√
x+C, x>0, whereCis an arbitrary constant.
Remark
• Sometimes, for simplicity, we writeR
x2dx= 1
3x3+Cetc.
♦ The intervalRis omitted because it can be determined easily.
♦ The symbolCis understood to be an arbitrary constant.
• Since we can use any symbol to denote the independent variable, we may also writeR
t2dt= 1
3t3+Cetc.
• Instead of a family of functions, sometimes we write R
f(x) dx to represent a function only. See the discussion in theAlternative Solutionon page 177.
Terminology
• Tointegratea function f means to find the indefinite integral of f (that is, to findR
f(x) dxifxis chosen to be the independent variable).
• Same as that for definite integrals, in the notationR
f(x) dx, the function f is called theintegrand.
Integration of Constant (Function)Letkbe a constant. Then we have Z
kdx=kx+C, −∞< x<∞.
ExplanationAs usual,Cis understood to be an arbitrary constant.
6.3. Indefinite Integrals 169
Proof The result follows from the Constant Multiple Rule for Differentiation and the Rule for Derivative of the Identity Function:
d
dxkx = k· d dxx
= k
Example Z
3 dx=3x+C
Power Rule for Integration (positive integer version)Letnbe a positive integer. Then we have Z
xndx= xn+1
n+1+C, −∞< x<∞.
Proof The result follows from the Constant Multiple Rule and Power Rule (positive integer version) for Dif- ferentiation:
d dx
xn+1
n+1 = 1
n+1 · d dxxn+1
= 1
n+1 ·(n+1)xn+1−1
= xn
Example Z
x3dx= x3+1
3+1+C= 1
4·x4+C
Remark In the formula for Integration of Constant, puttingk=1, we get Z
1 dx= x+C, −∞< x<∞.
By considering the constant function 1 as the functionx0, the above result can be written as Z
x0dx= x0+1
0+1+C, −∞< x<∞.
Thus the Power RuleR
xndx= xn+1
n+1 +Cis also valid for the case wheren=0.
Power Rule for Integration (negative integer version)Letn be a negative integer different from−1. Then
we have Z
xndx= xn+1
n+1 +C, x,0.
ExplanationThe result means that on the intervals (−∞,0) and (0,∞), the function xn+1
n+1 is an antiderivative for the functionxn.
Proof The result follows from the Constant Multiple Rule and Power Rule (negative integer version) for
Differentiation.
Example Z 1
x5 dx= Z
x−5dx= x−5+1
−5+1 +C = −1 4x4 +C FAQ What is
Z 1 xdx?
170 Chapter 6. Integration
Answer You can’t apply the Power Rule ifn= −1. Note that x−1+1
−1+1 is meaningless. You will learn a formula
in Chapter 8.
Power Rule for Integration (n+ 12 version)Letnbe an integer. Then we have Z
xn+12 dx= xn+32
n+ 32 +C, x>0.
Proof The result follows from the Constant Multiple Rule and Power Rule (n+12 version) for Differentiation.
Remark The above result can be written as
Z
xrdx= xr+1
r+1 +C, x>0, wherer=n+1
2 andnis an integer. In fact, the formula is valid for all real numbersr ,−1 (see Chapter 10).
Example Z 1
√xdx= Z
x−12dx= x−12+1
−12 +1+C=2x12 +C
Constant Multiple Rule for IntegrationLetkbe a constant and let f be a function that is continuous on an open interval (a,b). Then we have
Z
k f(x) dx=k Z
f(x) dx, a< x<b.
Proof The result follows from the Constant Multiple Rule for Differentiation.
Example Find Z
2x7dx.
ExplanationThe question is to find the family of functions that are antiderivatives for the integrand (on some open intervals). The answer should be given in the form “a function of x+C”. Usually, for integration problems, there is no need to mention the underlying open intervals. For the given problem, the function 2x7is continuous onRand so it has antiderivatives onR.
Solution Z
2x7dx = 2 Z
x7dx Constant Multiple Rule
= 2 x7+1 7+1 +C
!
Power Rule
= 1
4x8+2C
RemarkFrom the answer, we see that the function 1
4x8is an antiderivative for the function 2x7(onR). There-
fore, we can also write Z
2x7dx= 1 4x8+C.
Although the answers 1
4x8+2Cand 1
4x8+C look different, they represent the same family of functions. In general, to do integration, we can use rules and formulas to get an antiderivative for the integrand and then add a constant of integration.
6.3. Indefinite Integrals 171
Sum Rule for Integration (Term by Term Integration)Let f andgbe functions that are continuous on an open interval (a,b). Then we have
Z hf(x)+g(x)i dx=
Z
f(x) dx+ Z
g(x) dx, a< x<b.
Proof The result follows from the Sum Rule for Differentiation.
Example Find Z
(1+x3) dx.
ExplanationWe use rules and formulas for integration to obtain an antiderivative for the integrand and then add a constant of integration.
Solution Z
(1+x3) dx = Z
1 dx+ Z
x3dx Term by Term Integration
= x+ x4
4 +C Power Rule
Remark Using the Sum Rule together with the Constant Multiple Rule, we obtain the following:
Z hf(x)−g(x)i dx=
Z
f(x) dx− Z
g(x) dx.
More generally, Term by Term Integration can be applied to sum and difference of finitely many terms.
Example Perform the following integration:
(1) Z
x−11+ 3
√x
! dx (2)
Z
(2x−3)(x2+1) dx
ExplanationThe question is to find the given indefinite integrals. The answers should be given in the form “a function of x+C”.
Solution (1)
Z
x−11+ 3
√x
! dx =
Z
xdx− Z
11 dx+ Z
3x−12 dx Term by Term Integration
= x2
2 −11x+3 Z
x−12dx Power Rule, Integration of Constant
& Constant Multiple Rule
= x2
2 −11x+3· x12
12
+C Power Rule
= 1
2x2−11x+6√ x+C
RemarkIn the second step, there is no need to add a constant of integration (because there is an indef- inite integral in the third term). In the third step, we must add a constant of integration (otherwise, the expression represents a function but not a family of functions).
172 Chapter 6. Integration
(2) Z
(2x−3)(x2+1) dx = Z
(2x3−3x2+2x−3) dx Rewrite the integrand
= Z
2x3dx− Z
3x2dx+ Z
2xdx− Z
3 dx Term by Term Integration
= 2 Z
x3dx−3 Z
x2dx+2 Z
xdx−3x Constant Multiple Rule
& Integration of Constant
= 2· x4
4 −3· x3
3 +2· x2
2 −3x+C Power Rule
= x4
2 −x3+x2−3x+C
Caution Z h
f(x)·g(x)i dx,
Z
f(x) dx· Z
g(x) dx
FAQ Do we have a rule for integration that corresponds to the product rule in differentiation?
Answer In integration, corresponding to the product rule, there is a technique calledintegration by parts. A
brief introduction to this technique will be given in Chapter 10.
To close this section, we give an example to illustrate the steps for finding definite integrals using rules for integration.
Example Evaluate the following definite integrals:
(1) Z 2
−1
(x2−2x+3)dx
(2) Z 1
0
x(x2+1)dx Solution
(1) Z 2
−1
(x2−2x+3)dx =
"
x3
3 −2· x2 2 +3x
#2
−1
Term by Term Integration,
Power Rule, Constant Multiple Rule
& Fundamental Theorem of Calculus
= 8
3 −4+6
!
− −1
3 −1−3
!
= 9
(2) Z 1
0
x(x2+1)dx = Z 1
0
(x3+x2)dx Rewrite the integrand
=
"
x4 4 + x3
3
#1
0
Term by Term Integration, Power Rule
& Fundamental Theorem of Calculus
= 1 4 + 1
3
= 7
12