# The Relation Between Integration and Differentiation - Part 2

#### 5.6 The Leibniz notation for primitives

We return now to a further study of the relationship between integration and differentiation. First we discuss some notation introduced by Leibniz.
We have defined a primitive P of a functionfto be any function for which P’(x) =f(x). Iff is continuous on an interval, one primitive is given by a formula of the form

and all other primitives cari differ from this one only by a constant. Leibniz used the symbol jf(x) dx to denote a general primitive off. In this notation, an equation like
(5.12)          = P(x) + c
is considered to be merely an alternative way of writing P’(x) =f(x). For example, since the derivative of the sine is the cosine, we may write
(5.13)
Similarly, since the derivative of xn + 1/(n + 1) is xn, we may write
(5.14)
for any rational power n ≠ - 1. The symbol C represents an arbitrary constant so each of Equations (5.13) and (5.14) is really a statement about a whole set of functions.
Despite similarity in appearance, the symbol jf(x) dx is conceptually distinct from the integration symbol . The symbols originate from two entirely different processes-differentiation and integration. Since, however, the two processes are related by the fundamental theorems of calculus, there are corresponding relationships between the two symbols.
The first fundamental theorem states that any indefinite integral off is also a primitive of f. Therefore we may replace P(x) in Equation (5.12) by Jz f(t) dt for some lower limit c and write (5.12) as follows:
(5.15)            .
This means that we cari think of the symbol as representing some indefinite integral of f, plus a constant.
The second fundamental theorem tells us that for any primitive P off and for any constant C, we have

If we replace P(x) + C by , this formula may be written in the form

The two formulas in (5.15) and (5.16) may be thought of as symbolic expressions of the first and second fundamental theorems of calculus.
Because of long historical usage, many calculus textbooks refer to the symbol as an “indefinite integral” rather than as a primitive or an antiderivative. This is justified, in part, by Equation (5.15), which tells us that the symbol jf(x) dx is, apart from an additive constant C, an indefinite integral off. For the same reason, many handbooks of mathematical tables contain extensive lists of formulas labeled “tables of indefinite integrals” which, in reality, are tables of primitives. TO distinguish the symbol from , the latter is called a dejnite integral. Since the second fundamental theorem reduces the problem of integration to that of finding a primitive, the term “technique of integration” is used to refer to any systematic method for finding primitives. This terminology is widely used in the mathematical literature, and it Will be adopted also in this book. Thus, for example, when one is asked to “integrate” , it is to be understood that what is wanted is the most general primitive off.
There are three principal techniques that are used to construct tables of indefinite integrals, and they should be learned by anyone who desires a good working knowledge of calculus. They are (1) integration by substitution (to be described in the next section), a method based on the chain rule; (2) integration byparts, a method based on the formula for differentia ting a product (to be described in Section 5.9); and (3) integration bypartial fractions, an algebraic technique which is discussed at the end of Chapter 6. These techniques not only explain how tables of indefinite integrals are constructed, but also they tell us how certain formulas are converted to the basic forms listed in the tables.

#### 5.7 Integration by substitution

Let Q be a composition of two functions P and g, say Q(x) = P[g(x)] for all x in some interval Z. If we know the derivative of P, say P'(x) =f(x), the chain rule tells us that the derivative of Q is given by the formula Q'(x) = P'[g(x)]g'(x). Since P' = f, this states that Q'(x) =f[g(x)]g'(x). In other words,
(5.17)          P'(x) = f(x) implies Q'(x) = f[g(x)]g'(x).
In Leibniz notation, this statement cari be written as follows: If we have the integration formula
(5.18)          ,
then we also have the more general formula
(5.19)          .
For example, if f(x) = cosx, then (5.18) holds with P(x) = sinx, so (5.19) becomes
(5.20)          .
In particular, if g(x) = x3, this gives us
cosx3.3x2dx = sinx3 + C,
a result that is easily verified directly since the derivative of sinx3 is 3x2cosx3.
Now we notice that the general formula in (5.19) is related to (5.18) by a simple mechanical process. Suppose we replace g(x) everywhere in (5.19) by a new symbol u and replace g'(x) by du/dx, the Leibniz notation for derivatives. Then (5.19) becomes

At this stage the temptation is strong to replace the combination (du/dv)dx by du. If we do this, the last formula becomes
(5.21)
Notice that this has exactly the same form as (5.18), except that the symbol u appears everywhere instead of x. In other words, every integration formula such as (5.18) can be made to yield a more general integration formula if we simply substitute symbols. We replace x in (5.18) by a new symbol u to obtain (5.21), and then we think of u as representing a new function of x, say u = g(x). Then we replace the symbol du by the combination g’(x) dx, and Equation (5.21) reduces to the general formula in (5.19).
For example, if we replace x by ZJ in the formula , we obtain

In this latter formula, u may be replaced by g(x) and du by g'(x)dx, and a correct integration formula, (5.20), results.
When this mechanical process is used in reverse, it becomes the method of integration by substitution. The abject of the method is to transform an integral with a complicated integrand, such as , into a more familiar integral, such as . The method is applicable whenever the original integral cari be written in the form

since the substitution
u = g(x), du = g'(x)dx,
transforms this to . If we succeed in carrying out the integration indicated by , we obtain a primitive, say P(u), and then the original integral may be evaluated by replacing u by g(x) in the formula for P(u).
The reader should realize that we have attached no meanings to the symbols dx and du by themselves. They are used as purely forma1 devices to help us perform the mathematical operations in a mechanical way. Each time we use the process, we are really applying the statement (5.17).
Success in this method depends on one’s ability to determine at the outset which part of the integrand should be replaced by the symbol u, and this ability cornes from a lot of experience in working out specific examples. The following examples illustrate how the method is carried out in actual practice.

EXAMPLE 1. Integrate .
Solution. Let us keep in mind that we are trying to Write x3cosx4 in the form f[g(x)]g'(x) with a suitable choice off and g. Since cosx4 is a composition, this suggests that we take f(x) = cosx and g(x) = x4 so that cosx4 becomes f [g(x)]. This choice of g gives g'(x) = 4x3, and hence f[g(x)]g'(x) = (cosx4) (4x3). The extra factor 4 is easily taken care of by multiplying and dividing the integrand by 4. Thus we have
x3cosx4 = (cosx4)(4x3)/4 = {f[g(x)]g'(x)}/4.
Now, we make the substitution u = g(x) = x4, du = g'(x)dx = 4x3dx, and obtain

Replacing u by x4 in the end result, we obtain the formula

which can be verified directly by differentiation.
After a little practice one cari perform some of the above steps mentally, and the entire calculation can be given more briefly as follows: Let u = x4. Then du = 4x3dx, and we obtain

Notice that the method works in this example because the factor x3 has an exponent one less than the power of x which appears in COS x4.

EXAMPLE 2. Integrate .
Solution. Let u = COS x. Then du = -sin x dx, and we get

Again, the final result is easily verified by differentiation.

ExAMPLE 3. Evaluate
Solution. Let u = x2 + 2x + 3. Then du = (2x + 2)dx, so that

Now we obtain new limits of integration by noting that u = 11 when x = 2, and that u = 18 when x = 3. Then we write
The same result is arrived at by expressing everything in terms of x.Now we prove a general theorem which justifies the process used in Example 5.

THEOREM 5.4. SUBSTITUTION THEOREM FOR INTEGRALS. Assume g has a continuous derivative g' on an open interval I. Let J be the set of values taken by g on I and assume that f is continuous on J. Then for each x and c in I, we have
(5.22)

Proof. Let a = g(c) and define two new functions P and Q as follows:
if x is from J
if x is from I.
Since P and Q are indefinite integrals of continuous functions, they have derivatives given by the formulas
P'(x) = f(x), Q'(x) = f[g(x)]g'(x).
Now let R denote the composite function, R(x) = P[g(x)]. Using the chain rule, we find
R'(x) = P'[g(x)]g'(x) = fg(x)g'(x) = Q'(x).
Applying the second fundamental theorem twice, we obtain

and

This shows that the two integrals in (5.22) are equal.

5.8 Exercises

In Exercises 1 through 20, evaluate the integrals by the method of substitution.

21. Deduce the formulas in Theorems 1.18 and 1.19 by the method of substitution.

22. Let

where a > 0, and p and q are positive integers. Show that F(x, a) = ap + 1 - 2qF(x/a, 1).

23. If m and n are positive integers, show that

24. Show that

#### 5.9 Integration by parts

We proved in Chapter 4 that the derivative of a product of two functions f and g is given by the formula
h'(x) = f(x)g'(x) + f'(x)g(x),
where h(x) =f(x)g(x). When this is translated into the Leibniz notation for primitives, it becomes , usually written as follows:
(5.23)           .
This equation, known as the formula for integration by parts, provides us with a new integration technique.
To evaluate an integral, say , using (5.23), we try to find two functions f and g such that k(x) can be written in the formf(x)g'(x). If we cari do this, then (5.23) tells us that we have
jk(x)dx = f(x)g(x) - jg(x)f'(x)dx + C, and the difficulty has been transferred to the evaluation of . If f and g are properly chosen, this last integral may be easier to evaluate than the original one. Sometimes two or more applications of (5.23) Will lead to an integral that is easily evaluated or that may be found in a table. The examples worked out below have been chosen to illustrate the advantages of this method. For definite integrals, (5.23) leads to the formula

If we introduce the substitutions u =f(x), u = g(x), du =f'(x) dx, and dv = g'(x) dx, the formula for integration by parts assumes an abbreviated form that many people find easier to remember, namely
(5.24)

EXAMPLE 1. Integrate
Solution. We choose f(x) = x and g’(x) = COS x. This means that we have f'(x) = 1 and g(x) = sin x,so (5.23) becomes
(5.25)            .
Note that in this case the second integral is one we have already calculated.
TO carry out the same calculation in the abbreviated notation of (5.24) we write
u = x,      dv = cosxdx,
du = dx,      v = jcosxdx = sinx,

Had we chosen u = COS x and du = x dx, we would have obtained du = -sin x dx, v = x2/2, and (5.24) would have given us

Since the last integral is one which we have not yet calculated, this choice of u and u is not as useful as the first choice. Notice, however, that we cari salve this last equation for and use (5.25) to obtain

As an application of the method of integration by parts, we obtain another version of the weighted mean-value theorem for integrals (Theorem 3.16).

THEOREM 5.5. SECOND MEAN-VALUE THEOREM FOR INTEGRAL~. Assumegiscontinuouson [a, b], and assume f has a derivative which is continuous and never changes sign in [a, b].Then, for some c in [a, b], we have
(5.28)

Proof. Let G(x) = . Since g is continuous, we have G'(x) = g(x). Therefore, integration by parts gives us
(5.29)
since G(a) = 0. By the weighted mean-value theorem, we have

for some c in [a, b]. Therefore, (5.29) becomes

This proves (5.28) since .

#### 5.10 Exercises

Use integration by parts to evaluate the integrals in Exercises 1 through 6.

7. Use integration by parts to deduce the formula

In the second integral, write cos2x = 1 - sin2x and thereby deduce the formula

7. Derive the following formulas.

8. Use integration by parts and the results of Exercises 7 to deduce the following formulas.

9. Use integration by parts to show that

Write x2 = x2 - 1 + 1 in the second integral and deduce the formula

10. If , use integration by parts to show that

11. Evaluate the integral , given that = 11.35. Leave the answer in terms of √3 and √31.

#### *5.11 Miscellaneous review exercises

1. Let f be a polynomial with f(0) = 1 and let g(x) = PJ'(x). Compute g(O), g'(O), . . . , g(n)(0).
2. Find a polynomial P of degree ≤ 5 with P(0) = 1,P(1) = 2,P'(0) = P''(0) = P'(1) = P''(1) = 0.
3. If f(x) = cosx and g(x) = sinx, prove that
,f(n)(x) = cos(x + nπ/2) and g(n)(x) = sin(x + nπ/2).
4. If h(x) = f(x)g(x), prove that the nth derivative of h is given by the formula

This is called Leibniz’s formula.
5. Given two functions f and g whose derivatives f' and g' satisfy the equations
(5.30)           f'(x) = g(x),       g'(x) = -f(x), f(0) = 0, g(0) = 1,
for every x in some open interval J containing 0. For example, these equations are satisfied when f (x) = sinx and g(x) = cosx.
(a) Prove that f2(x) + g2(x) = 1 for every x in J.
(b) Let F and G be another pair of functions satisfying (5.30). Prove that F(x) =f(x) and G(x) = g(x) for every x in J. [Hint: Consider h(x) = [F(x) -,f(x>]2 + [G(x) - g(x)]2.]
(c) What more cari you say about functionsfand g satisfying (5.30)?
6. A function f, defined for all positive real numbers, satisfies the equation f(x2) = x3 for every x > 0. Determine f’(4).
7. A function g, defined for all positive real numbers, satisfies the following two conditions: g(1) = 1 and g'(x2) = x3 for all x > 0. Compute g(4).
8. Let C1 and C2, be two curves passing through the origin as indicated in Figure 5.2. A curve C is said to “bisect in area” the region between C1 and C2 if, for each point P of C, the two shaded regions A and B shown in the figure have equal areas. Determine the Upper curve C2,given that the bisecting curve C has the equation y = x2 and that the lower curve C1 has the equation y = x2/2.

In Exercises 11 through 20, evaluate the given integrals. Try to simplify the calculations by using the method of substitution and/or integration by parts whenever possible.

21. Show that the value of the integral is 2n for some integer n.
22. Determine a pair of numbers a and b for which
23. Let F(m, n) = , m > 0, n > 0. Show that
(m + 1)F(m, n) + nF(m + 1, n - 1) = xm + 1(1 + x)n.
Use this to evaluate F(10, 2).
24. Let A denote the value of the integral

Compute the following integral in terms of A:

The formulas in Exercises 28 through 33 appear in integral tables. Verify each of these formulas by any method.

34. (a) Find a polynomial P(x) such that P'(x) - 3P(x) = 4 - 5x + 3x2. Prove that there is only one solution.
(b) If Q(x) is a given polynomial, prove that there is one and only one polynomial P(x) such that P'(x) - 3P(x) = Q(x).
35. A sequence of polynomials (called the Bernoullipolynomials) is defined inductively as follows:
P,(x) = 1; P'n(x) = nPn - 1(x) and if n ≥ 1.
(a) Determine explicit formulas for P1(x), P2(x), . . . , P5(x).
(b) Prove, by induction, that Pn(x) is a polynomial in x of degree n, the term of highest degree being xn.
(c) Prove that Pn(O) = Pn(1) if n ≥ 2.
(d) Prove that Pn(x + 1) - Pn(x) = nxn - 1 if n ≥ 1.
(e) Prove that for n ≥ 2 we have

(f) Prove that Pn(1 - x) = (-1)nPn(x) if n ≥ 1.
(g) Prove that P2n + 1(0) = 0 and P2n - 1(1/2) = 0 if n ≥ 1.
36. Assume that |f''(x)| ≤ m for each x in the interval [0, a], and assume that f takes on its largest value at an interior point of this interval. Show that |f'(0)| + |f'(a)| ≤ am. You may assume that f'' is continuous in [0, a].

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