4
Integral calculus
We now turn from differentiation to the other crucial ingredient of the infinitesimal calculus, namely integration. This may be approached in two ways: either as the inverse process of differentiation; or as the means of calculating the area under a given curve, using an argument that serves as a template for many other important applications. However, before discussing these, we must develop the above two approaches, and the relations between them.
4.1 Indefinite integrals
Given a function f(x), the indefinite integral F(x) is defined as the most general solution of the equation
It is not unique. Suppose we have a particular solution F0(x), with
Then the most general solution can be written
where G(x) is any function such that (4.1) is satisfied. On substituting (4.3) into (4.1) and using (4.2), one obtains 
, and so G(x) = c, where c is a constant. Hence the indefinite integral is given by
where F0(x) is any particular solution of (4.1) and c is an arbitrary constant. An alternative way of writing the indefinite integral is
for reasons that will become clearer in the next section. Thus (4.4) is often written in the form
The process of determining the indefinite integral of a given function f(x) is called integration, f(x) is called the integrand 1, and c the integration constant. In simple cases, it can be done using the standard table of derivatives, Table 3.1, together with the basic properties
or, more generally,
which follow directly from the definition (4.1), where a, b are arbitrary constants and f(x) and g(x) are arbitrary functions. For example, from Table 3.1 we have
and so deduce that
and hence, using (4.7a),
where 
. For the case n = −1, we can use the result
from Table 3.1, so that
In this way, one builds up the table of standard indefinite integrals shown in Table 4.1, from which other integrals may be deduced using (4.7). More complicated integrals will be considered later. In all cases, there is an undetermined integration constant c, whose value can only be determined given additional information, as illustrated in Example 4.2 below.
Table 4.1 Standard indefinite integrals
| Integrand | Integral |
| xn(n ≠ −1) | xn − 1/(n + 1) + c |
| ex | ex+c |
| 1/x | ln|x| + c |
| sin x | −cos x + c |
| cos x | sin x + c |
| sec2 x | tan x + c |
| cosh x | sinh x + c |
| sinh x | cosh x + c |
| sech2 x | tanh x + c |
4.2 Definite integrals
In this section we introduce the ‘definite integral’, defined in terms of areas, and relate it to the indefinite integral of the previous section.
4.2.1 Integrals and areas
Consider a function f(x) that is continuous in the interval a ≤ x ≤ b. Then the definite integral, written as,
is defined to be the area between the curve y = f(x) and the x-axis between the points x = a and x = b, where the areas above and below the axis are defined to be positive and negative, respectively. The quantities a and b are called the lower and upper limits of integration, respectively. Thus in Figure 4.1,
where A, B, C, are the magnitudes of the areas shown. Two simple results that follow directly from this definition are
and for a < b < c,
The definition of the definite integral can be extended to the case where a > b in a way consistent with (4.10a) and (4.10b) by defining
Figure 4.1 Integral of the function f(x) between the limits a and b.
However, the key result, which will be derived in the next section, is
where F(a) is the indefinite integral (4.6) and F0(x) is any solution of (4.2), since the constant c in (4.6) cancels on taking the difference F0(b) − F0(a). For example, the area under the curve y = sin x between x = 0 and x = π, shown in Figure 4.2, is given by
and since, from Table 4.1,
Figure 4.2 Integral of the function sin x between the limits 0 and π.
Equation (4.11) gives
Before proceeding to derive (4.11), it is worth mentioning here three further points. First, integrals like (4.9) are not functions of x, and the variable in the integrand is therefore not significant, i.e.
for any given function f. The symbols x, t, etc. are referred to as dummy variables. Second, one sometimes meets integrals in which one of the limits of integration is itself a variable. In other words, if we denote the dummy variable by t and the variable limit by x, then
are both functions of x, and on differentiating with respect x using (4.11) and (4.1) one obtains
(4.12) 
Finally, when evaluating definite integrals it is useful to introduce the notation
For example,
4.2.2 Riemann integration
It remains to derive (4.11), relating the indefinite integral (4.6), defined as the inverse of differentiation, to the definite integral (4.9), defined as the area under the curve f(x) in the given range a ≤ x ≤ b. To do this, we need to express the area under the curve explicitly in terms of the function f(x) itself. This is achieved by the construction of Figure 4.4, in which the range a ≤ x ≤ b has been divided into n strips of width
where x0 = a and xn = b. We then consider the quantity
Figure 4.4 Construction to derive Eq. (4.11).
where ζk is any point within the kth strip, i.e. xk − 1 ≤ ζk ≤ xk. For continuous functions, (4.14) is an approximation to the area under the curve, and tends to the exact value as n → ∞ with the widths of all the strips δxk → 0, irrespective of the precise ways in which the values ζk are chosen and the limit is taken. This leads to the Riemann definition of the definite integral:
where f(x) is, for the moment, assumed to be continuous in the interval a ≤ x ≤ b. It leads directly to the important general results (4.10a) and (4.10b), which we obtained in the previous section from the intuitive understanding of the area under the curve. In addition, it enables the proof of the crucial result (4.11). However, to do this, we must first prove another important result, called the first mean value theorem for integration.
Suppose fm and fM are the minimum and maximum values of f(x) in the interval a ≤ x ≤ b. Then clearly
and since f(x) varies continuously between fm and fM, there must be at least one value x = ζ in this range for which
This result is called the first mean value theorem for integration, and f(ζ) is the mean value of f(x) in the interval a ≤ x ≤ b.
We can now derive the key result (4.11) by defining
and considering
where a is kept fixed. By (4.10b) and the mean value theorem (4.16),
with b ≤ ζ ≤ b + δb. Hence f(ζ) → f(b) as δb → 0, and (4.17) gives
Integrating with respect to b using (4.4) then gives
where F0 is any indefinite integral of f and where the integration constant ca is independent of b. A similar argument, treating a as a variable with b fixed, leads to
and hence
where cb is independent of a. Equations (4.18) and (4.19) are only compatible if
where c is independent of a and b, and using (4.10a) one sees that c = 0. Hence, using (4.4), we obtain
This is the desired relation between the definite integral (4.9) and the indefinite integral (4.6).
4.3 Change of variables and substitutions
In this and the following two sections we will discuss techniques for integrating given functions, beginning with methods based on a change of variables. We start by summarising the approach in general and then consider specific applications. In all cases, the aim is to relate the given integral to the standard integrals of Table 4.1. The latter are well worth remembering and will be assumed in what follows. More complicated integrals may often be evaluated using the methods in Sections 4.3 and 4.4, together with the results of Table 4.1, but if this fails, there exist several useful reference books that include tables of integrals and there are also online resources2.
4.3.1 Change of variables
Suppose that
is the indefinite integral of a given function f(x), where x ≡ x[z] is itself a function of another variable z. Then, using the chain rule (3.29), we have
and so from (4.1),
is the original integral expressed in terms of the new variable z. In addition, in the case of a definite integral, we must also modify the limits of integration, i.e.
(4.21b) 
where z(a) and z(b) are the values of z at x = a and x = b, respectively.
Changes of variable are often used in evaluating integrals. As an example, consider
Substituting z = 3x + 2 and using (4.21a) gives
where c is an integration constant and we have used the standard integral for cosh z. Substituting for z gives the final result
In general, integration is a more difficult process than differentiation. So after evaluating an integral, it is always good practice to differentiate the result to check that the original function is recovered. This is easily done in this case, i.e.
but other cases can be more complicated.
4.3.2 Products of sines and cosines
Next we consider integrals of the form
where m and n are integers with m, n ≥ 0. If m is odd, then independent of whether n is odd or even, the integral can be evaluated by substituting z = cos x. Similarly, if n is odd, independent of the value of m, it can be evaluated by substituting z = sin x. As an example, consider
If we set z = cos x so that sin 2x = 1 − z2 and 
, then (4.21a) gives
where we have used the relation (4.7b). Hence
where, as usual, c is an integration constant.
The only remaining possibility is if m and n are both even in (4.22). In this case, neither of the substitutions z = sin x or z = cos x helps. However, in these cases the integrals can be evaluated by exploiting the relations
which follow from (2.37a). For example,
where we have used the substitution z = 2x to evaluate the second integral.
Finally, we note that integrals of the type
can be evaluated by similar methods, where the sines and cosines are replaced by their hyperbolic analogues, as we demonstrate in Example 4.6(a) below.
4.3.3 Logarithmic integration
Consider integrals of the form
where φ(x) is any differentiable function of x. The substitution z = φ(x) gives
provided z = φ(x) > 0; while substituting z = −φ(x) gives
providing z = −φ(x) > 0. Combining these results gives the logarithmic integral
(4.24) 
For the particular case φ(x) = x, this reduces to
which generalises the standard integral (4.8) to all non-zero x.
4.3.4 Partial fractions
Rational functions (i.e. ratios of polynomials) can often be decomposed into a sum of simpler functions called partial fractions. This was discussed in detail in Section 2.1.2. We shall not repeat that discussion here, but merely note its usefulness in evaluating integrals of rational functions. For example, consider the integral of the function
The denominator is x2 + 5x + 6 = (x + 2)(x + 3), so that
by (2.17). Hence
and setting x = −2 and then x = −3, gives A = 2 and B = 3, respectively. Hence the integral is
Other examples are given in later sections.
4.3.5 More standard integrals
Consider an integral of the form
(4.26a) 
where f(x) is a given function. If we can find a substitution x = g(z), such that 
, where α is a constant, then equation (4.21a) gives
where g− 1 is the inverse function of g (not 
).
Several ‘standard integrals’ may be evaluated in this way. An example is
(4.27a) 
Substituting x = asinh z, we have
so that (4.26b) gives
where we have used the notation sinh − 1 to denote the inverse function of sinh , as an alternative to arcsinh.
Other standard integrals of a similar type, together with the substitutions required to derive then, are given in Table 4.2. Their use is illustrated in Example 4.8 below.
Table 4.2 More standard integrals and the substitutions used to derive them
| Integrand | Substitution | Integral |
 |
x = asin z |  |
 |
x = asinh z |  |
 |
x = acosh z |  |
 |
x = atan z |  |
 |
 |
 |
 |
x = atanh z |  |
4.3.6 Tangent substitutions
It is sometime useful to convert integrals involving sin x and cos x into integrals over rational functions by the substitution
whence
(4.29a) 
(4.29b) 
and
(4.29c) 
Similarly, expressions involving sin 2x and cos 2x can sometimes be converted to rational functions by the substitutions
whence
(4.31a) 
and
(4.31b) 
The resulting integrals can then be evaluated by the standard methods for rational functions.
4.3.7 Symmetric and antisymmetric integrals
If an integrand has a definite symmetry, either even or odd, then this can be exploited to reduce the calculations involved in evaluating its integral. Thus, if f−(x) = −f−( − x) is any odd function of x, for example f−(x) = sin x, a simple result that is often useful is
which follows directly from the definition of the integral as the area under the curve. Formally, it is obtained by making the substitution z = −x in
where we have used f−( − z) = −f−(z), together with (4.10c). Equation (4.32a) then follows from (4.10b). The corresponding result for even functions f+(x) = f+( − x), obtained in the same way, is
(4.32b) 
4.4 Integration by parts
On integrating the product rule equation (3.20),
one obtains the formula
or its equivalent form for definite integrals,
(4.33b) 
where we have used the notation (4.13).
Equations (4.33) are the basic formulas for integration by parts. They are often useful for integrals where the integrand can be written as the product of two terms, at least one of which can be easily integrated. For example, consider the integral
On setting u = x and 
, (4.33a) gives,
Integration by parts is also sometimes useful to integrate functions that can be differentiated to give simpler functions. For example,
so that (4.33a) gives
Finally, it can also be used to derive relations, called reduction formulas (also called recurrence relations), between families of integrals In whose members are characterised by an integer n. For example, consider the integrals
Then
i.e.
This is a typical reduction formula, enabling In to be evaluated in terms of In − 1, and hence by repeated application, In for any n to be found, starting from
Although we have shown that reduction formulas can be obtained using integration by parts, they can sometimes be obtained in other ways, as illustrated in Example 4.12 below.
4.5 Numerical integration
In practice, one often needs to evaluate the definite integral (4.9) for functions f(x) where an explicit form for the corresponding indefinite integral (4.6) cannot be found. In these cases, one must resort to a numerical evaluation, usually with the aid of a computer.
There are many methods available for doing this. Here we shall consider only the two simplest, which are based on dividing the integral into strips, as shown in Figure 4.4, Section 4.4.2. We will assume that the widths of the strips are all equal, i.e. xn − xn + 1 = hfor all n. Then, if we approximate f(x) between x0 and x1 by a straight line, the area of the first strip is approximately
Repeating the procedure for the second strip gives its area as
and so on. When the contributions of all the strips are added, we have in this approximation
where
This approximation is called the trapezium rule. Its accuracy increases with n, becoming exact in the limit n → ∞.
An alternative to the trapezium rule is obtained by choosing n to be even, and approximating f(x) across each pair of strips by a quadratic form. Consider the first pair of strips as shown in Figure 4.4, spanning the interval a ≤ x ≤ b, where x0 = a and x2 = b. Then approximating
and changing the variable to y = x − x1 gives
In the same approximation, we have
and
so that
Substituting (4.36) into (4.35) gives
as the contribution from the first two strips. Similarly, the second pair contributes
and so on, giving finally
where, once again, fk ≡ f(xk) and n is even.
Equation (4.37) is called Simpson's rule and is usually more precise than the trapezium rule for a given fixed n. However, like other methods we will not discuss, they are both easy to implement on a computer and tend to the exact result as n → ∞. Hence using either method, one can simply keep increasing n until the resulting value of the integral is stable to the precision required.
4.6 Improper integrals
So far we have restricted the discussion of definite integrals to cases where the limits of the integration a, b are finite and the integrand f(x) is continuous in the range a ≤ x ≤ b. Here we consider whether it is possible to define integrals when these conditions are not satisfied. Such integrals are sometimes called improper integrals and often occur in physical applications.
4.6.1 Infinite integrals
We first consider the case
where f(x) is continuous in the range a ≤ x < ∞. Then (4.38) can be defined by
provided the limit is well-defined and finite. If it is, then the integral (4.38) is said to be convergent. If the limit is not well-defined or is infinite, then the integral is said to be divergent. Similar considerations apply in an obvious way when the lower limit a → −∞.
It is easiest to determine whether an integral converges when the corresponding finite integral I(b) can be evaluated explicitly. For example, consider the integral
The limit
is ill-defined, so that (4.40) is divergent. On the other hand,
so that
is a convergent integral, but diverges if α ≤ 0.
More generally, it should be clear from (4.39) that the convergence of (4.38) only depends on the behaviour of f(x) as x → ∞, that is, on its asymptotic behaviour. Some useful results then follow from the interpretation of the integral as the area under the curve y = f(x) as b → ∞, as shown in Figure 4.5. In particular, it is clear that (4.39) can only converge to a finite limit if f(x) → 0 as x → ∞. However, this is only a necessary, but not a sufficient condition. For example,
so that, even though f(x) → 0, the integral does not converge.
Figure 4.5 Infinite integrals over f(x) and g(x) (see text).
Suppose we have a function g(x) that is continuous in the range a ≤ x < ∞, and suppose that f and g are such that the conditions
are satisfied for large x. Then it is clear from Figure 4.5 that if (4.38) converges, then the integral
representing the area under the curve y = g(x), also converges. Furthermore, from (4.7a) and (4.39), it follows that the convergence or otherwise of an integral is not affected by multiplying the integrand by a constant. Hence the condition (4.42) for the convergence of (4.43) can be relaxed to
where k > 0 is a positive constant. Thus, for example, from the convergence of (4.41) we can immediately infer that integrals like
also converge if α > 0.
A similar argument shows that if (4.38) diverges and 0 ≤ f(x) ≤ kg(x) at large x, with k > 0, then the corresponding integral (4.43) also diverges. However, the converse results do not apply; that is, if (4.38) converges and (4.44) is not satisfied, then it does not follow that (4.43) is divergent. On the other hand, if g(x) → c f(x) as x → ∞, i.e. if
where c is any finite non-zero constant, then the asymptotic behaviours of f(x) and g(x) are identical, apart from an irrelevant constant. Hence in this case, (4.38) and (4.43) are either both convergent or both divergent.
4.6.2 Singular integrals
In Section 3.1.2, we introduced the idea of an infinite discontinuity, that is, a point where the value of a given function tended to infinity. In this section we consider integrals in which the integrand is not continuous, but has an infinite discontinuity, also called a singularity, at some point xs in the range of integration a ≤ x ≤ b. In this case we can write the integral in the form
provided the two integrals on the right-hand side are well-defined. Hence it is sufficient to investigate whether we can define integrals with a singular point at the end of the range of integration. This is done by a limiting process, as in the case of infinite integrals. For example, the second integral above is defined by
assuming a well-defined finite limit exists. If it does, the integral (4.46a) is said to be convergent and is well-defined. On the other hand, if a well-defined finite limit does not exist, the integral is said to be divergent and is ill-defined. Similar considerations apply to the first integral in (4.45), where the limit analogous to (4.46a) is
The integral (4.45) is only defined if both the integrals on the right-hand side are convergent.
To illustrate these points, consider the family of integrals
where α is a real parameter, and the integrand diverges as x → xs = 0. For α = 1 we have
so that the integral (4.47) is divergent for α = 1. For α ≠ 1, we have
From this and (4.48a) we see that the integral (4.47) is divergent for α ≥ 1, whereas for α < 1, it is convergent and given by
(4.48b) 
More generally, the divergence or otherwise of integrals like (4.46a) and (4.46b) depends solely on the behaviour of the integrand as the singularity is approached, that is, as x → xs. Hence it is not necessary to evaluate the entire integral explicitly to determine whether it converges. To illustrate this, consider the integral
whose integrand is singular at x = 0, but is well-behaved as x → 1. To determine its convergence, or otherwise, we therefore examine the behaviour of the integrand as x → 0. To do this we write
using (3.15). Hence (4.50) has the same convergence properties at x = 0 as (4.49) with 
, and, like (4.49), is convergent.
Finally, if both integrals on the right-hand side of (4.45) are divergent, it may be that the integral
is well-defined. If this is the case, (4.50) is called the principal value of the integral and is written
For example, the integral
is ill-defined, since both integrals on the right-hand side are divergent. However, the corresponding principal value integral
is well-defined, and in this case vanishes.
4.7 Applications of integration
At the beginning of this chapter, we said that the calculation of the area under a curve, which we subsequently based on the Riemann definition (4.15), served as a template for many applications of integration. In this section we illustrate this with some important examples in physics and geometry.
4.7.1 Work done by a varying force
Consider a force F(x), which acts in the x-direction and which varies continuously in the range a ≤ x ≤ b. To calculate the work done in moving from x = a to x = b, we divide the interval into a large number n of small steps δxk , k = 1, 2, …, as shown in Figure 4.4 for F(x) = f(x). If the strip widths δxk are small, the variation of F(x) across the strip can be neglected and the work done in moving from xk − 1 to xk is given by δWk = F(ζk)δxk, where x = ζk is any point within the strip. Hence,
which on comparing to the Riemann definition (4.14), is just the integral
(4.51) 
In other words, the work done by the force F(x) is the area under the curve y = F(x) between x = a and x = b.
4.7.2 The length of a curve
The length of any curve y = f(x) between any two points x = a and x = b > a may be found by reference to Figure 4.7. Let δlk be the contribution to the length L arising from the small interval δxk = xk − xk − 1. Then from Figure 4.7, we see that in the limit δxk → 0, this is given by
Figure 4.7 Construction used to obtain (4.52)
Hence, on summing over all segments δxk and letting δxk → 0, we immediately obtain
as the desired formula for the length of the curve. In many cases this integral will have to be evaluated numerically.
*4.7.3 Surfaces and volumes of revolution
Suppose we form a three-dimensional shape by taking the curve y = f(x), z = 0 in the range a ≤ x ≤ b and rotating it about the x-axis, as shown in Figure 4.8. Then the area swept out by the curve is called the area of revolution and the volume enclosed is called the volume of revolution. Any three-dimensional shape with an axis of rotation symmetry, chosen to be the x-axis, can be constructed in this way. For example, y = R produces a cylinder of radius R and length (b − a), while
Figure 4.8 The surface of revolution generated by y = f(x) for a ≤ x ≤ b and the interval x to (x + dx).
produces a cone of length h with radius R at the base.
To calculate the surface and volume of revolution, we divide the interval a ≤ x ≤ b into slices of infinitesimal width dx. The slice between an arbitrary x and x + dx is shown in Figure 4.8. The area of its edge is
where we have used the result of the previous section for dl, and its volume is dV = πy2dx. Summing over all strips in the manner of the previous sections then gives
for the volume of revolution and
for the area of revolution. To find the surface area of the whole shape, we need to add the areas of the circular discs at x = a and x = b.
*4.7.4 Moments of inertia
Moments of inertia play an important role in mechanical problems involving rotations. For a system of point particles, they are defined by
where ri is the perpendicular distance from the mass mi to the axis of rotation, as illustrated in Figure 4.10. The corresponding moments of inertia for extended objects, which are what is required for most practical applications, are then derived from (4.57) by integration, as we shall illustrate by some simple examples.
Figure 4.10 Definition of the moment of inertia (4.57) for a point particle.
Problems 4
-
Sketch the area between the curves
between the points 1.5 and 3.0 and find its value.
-
Prove that for any function f(x) that is differentiable in the interval a ≤ x ≤ b, there exists a value x = ζ in the range a ≤ ζ ≤ b, such that
is satisfied. This result is called the first mean value theorem, as distinct from the first mean value theorem of integration discussed in the text.
- Evaluate the mean value of the function
over the range 0 ≤ x ≤ 1 and verify the first mean value theorem for integration (4.16).
-
Use an appropriate substitution to integrate the following functions:
-
Integrate the following functions:
-
Evaluate the following indefinite integrals:
-
Evaluate the following integrals:
-
By using partial fractions, evaluate the following integrals:
-
Evaluate the following integrals:
-
Use appropriate ‘tangent substitutions’ to integrate the following functions:
-
Integrate the following functions:
-
Find the following indefinite integrals:
-
Find the following indefinite integrals:
-
Evaluate the following definite integrals:
-
If
where a is a constant, show by repeated integration by parts, that
-
If
show that
and hence find I3.
-
Find the following indefinite integrals:
-
The Euler gamma function Γ(x) is defined by
- Show that Γ(x + 1) = xΓ(x) and hence that Γ(n + 1) = n! for integer n ≥ 1.
- The relation Γ(x + 1) = xΓ(x) can be used to extend the definition of the gamma function to negative x. Locate the singularities of Γ(x) and evaluate
, given that 
.
-
A rocket fired vertically into the air at t = 0 has an acceleration profile
for five seconds, where α = 0.1 s− 1 and g = 9.8 ms− 2 is the acceleration due to gravity. Then the fuel runs out and it subsequently moves freely under gravity. What height does the rocket reach?
-
Examine the convergence of the following integrals. Evaluate those that are convergent.
-
Which of the following integrals are convergent?
-
For what values of α and β (if any) do the following integrals converge?
-
A particle moves along the x-axis subject to a force given by
where k and a are positive constants. Find the work done W in moving the particle from
to 3a. -
Find the length of the curve y = (1 − x2)1/2 between the points 0 and 1.
-
Find an integral expression for the length L of that portion of the hyperbola x2 − y2 = 1 for which 2.5 ≥ x ≥ 1.5 and use both the trapezium rule and Simpson's rule with 4 intervals to estimate the value of L.
-
Use Simpson's rule with n intervals to calculate the values of pi by integrating the function (1 − x2)− 1/2 between the limits 0 and 0.5. Find the value of n needed to ensure that the estimated value of π does not change by more than 0.1%.
-
The curve y = (1 − x2)1/2 between the points 0 and 1 is rotated about the x-axis. Find The area A and the volume V of rotation and the total surface area S of the resulting shape.
-
A prolate spheroid is obtained by rotating the ellipse
about the x-axis. Derive a formula for the volume of the spheroid and show that the surface area is given by
where the ellipticity e is defined by
. -
Calculate the moment of inertia of a thin square plate of mass m and sides of length a about (a) a line joining the mid-points of opposite sides, and (b) a diagonal.
-
Find the moment of inertia of a thin circular disc of mass M and radius r about an axis perpendicular to the disc and passing through its centre.
-
Use the result of Problem 4.29 to find the moment of inertia of a cone of mass M, height h and radius R at the base, for rotations about the axis of symmetry.