Title: Default for 13.1 Basic Concepts of Integration
Course: HELM Workbooks 2023-2024

Basic Concepts of Integration

When a function \(f(x)\) is known we can differentiate it to obtain its derivative \(\dfrac{df}{dx}\). The reverse process is to obtain the function \(f(x)\) from knowledge of its derivative. This process is called integration. Applications of integration are numerous and some of these will be explored in subsequent Sections. First, what is important is to practise basic techniques and learn a variety of methods for integrating functions.

Integration as differentiation in reverse

Suppose we differentiate the function \(y=x^2\). We obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x\). Integration reverses this process and we say that the integral of \(2x\) is \(x^2\). Pictorially we can regard this as shown in the figure below:

[Image: Diagram showing integration as the reverse process of differentiation.]

The situation is just a little more complicated because there are lots of functions we can differentiate to give \(2x\). Here are some of them: \(x^2+4, x^2 -15, x^2+0.5\)

All these functions have the same derivative, \(2x\), because when we differentiate the constant term we obtain zero. Consequently, when we reverse the process, we have no idea what the original constant term might have been. So we include in our answer an unknown constant, \(c\) say, called the constant of integration. We state that the integral of \(2x\) is \(x^2+c\).

When we want to differentiate a function, \(y(x)\), we use the notation \(\dfrac{\mathrm{d}}{\mathrm{d}x}\) as an instruction to differentiate, and write \(\dfrac{\mathrm{d}}{\mathrm{d}x} \bigl(y(x)\bigr)\). In a similar way, when we want to integrate a function we use a special notation: \(\int y(x)\,\mathrm{d}x\).

The symbol for integration, \(\int\), is known as an integral sign. To integrate \(2x\) we write

[Image: Description of integration]

Note that along with the integral sign there is a term of the form \(\mathrm{d}x\), which must always be written, and which indicates the variable involved, in this case \(x\). We say that \(2x\) is being integrated with respect to \(\pmb x\). The function being integrated is called the integrand. Technically, integrals of this sort are called indefinite integrals, to distinguish them from definite integrals which are dealt with subsequently. When you find an indefinite integral your answer should always contain a constant of integration.

Exercise

Write down the derivatives of each of the following functions with respect to \(x\):

1. if \(f(x)=[value]\) then \(f'(x)=\)
2. if \(g(x)=[value]\) then \(g'(x)=\)
3. if \(h(x)=[value]\) then \(h'(x)=\)
Feedback: The derivatives are: 1. \(\dfrac{d}{dx}[value]= [value]\) 2. \(\dfrac{d}{dx}([value])= [value]+0= [value]\) 3. \(\dfrac{d}{dx}([value])= [value]-0=[value]\) We deduce that \( {\displaystyle \int{[value]x^[value]\mathrm{d}x}} = x^[value] + c.\) Remember, when finding an indefinite integral, a constant of integration is always needed. Note how each of above suggest that that \({\displaystyle\int{[value]x^[value]\mathrm{d}x}} = x^[value] + c\).

A table of integrals

We could use a table of derivatives to find integrals, but the more common ones are usually found in a ‘Table of Integrals’ such as that shown below. You could check the entries in this table using your knowledge of differentiation. Try this for yourself.

Table 1: Integrals of Common Functions

\(f(x)\) | \[\int\,f(x)\,\mathrm{d}x\]
constant \( k\) | \(kx+c\)
\(x\) | \(\dfrac{1}{2}x^2+c\)
\(x^2\) | \(\dfrac{1}{3}x^3+c\)
\(x^n,\,n\neq-1\) | \(\dfrac{x^{n+1}}{n+1}+c\),  \(n\ne-1\)
\(x^{-1}\) \(\left(\text{or }\dfrac{1}{x}\right)\) | \(\ln|x|+c\)
\(\cos(x)\) | \(\sin(x)+c\)
\(\sin(x)\) | \(-\cos(x)+c\)
\(\cos(kx)\) | \(\dfrac{1}{k}\sin(kx)+c\)
\(\sin(kx)\) | \(-\dfrac{1}{k}\cos(kx)+c\)
\(\tan(kx)\) | \(\dfrac{1}{k}\ln|\sec(kx)|+c\)
\(e^x\) | \(e^x+c\)
\(e^{-x}\) | \(-e^{-x}+c\)
\(e^{kx},\,k\neq0\) | \(\dfrac{1}{k}e^{kx}+c\)

When dealing with the trigonometric functions the variable \(x\) must always be measured in radians and not degrees. Note that the fourth entry in the Table, for \(x^n\), is not valid for \(n=-1\) but is valid for any other value \(n\), positive or negative, whole number or fractional. When \(n=-1\) use the fifth entry in the Table.

Example

Use Table 1 to find the indefinite integral of \(x^7\): that is, find \[\int{x^7}\ \mathrm{d}x\]

Solution

From Table 1 note that \[\int{x^n}\ \mathrm{d}x = \dfrac{x^{n+1}}{n+1}+c\]

In words, this states that to integrate a power of \(x\), increase the power by 1, and then divide the result by the new power. With n = 7 we find \[\int{x^7}\ \mathrm{d}x = \dfrac{x^{8}}{8}+c\]

Example

Find the indefinite integral of \(\cos (5x)\): that is, find \[\int \cos (5x)\ \mathrm{d}x\]

Solution

From Table 1 note that \[\int{\cos(kx)}\ \mathrm{d}x = \dfrac{1}{k}\sin(kx)+c\]

With \(k=5\) we find \[\int{\cos(5x)}\ \mathrm{d}x = \dfrac{1}{5}\sin(5x)+c\]

Example

Find \[\int \cos (5t)\ \mathrm{d}t\]

Solution

We integrated \(\cos(5x)\) in the previous example. Now the independent variable is \(t\), so simply use Table 1 and replace every \(x\) with a \(t\). With \(k = 5\) we find \[\int{\cos(5t)}\ \mathrm{d}x = \dfrac{1}{5}\sin(5t)+c\]

It follows immediately that, for example,

\[\int{\cos(5\omega)}\ \mathrm{d}\omega = \dfrac{1}{5}\sin(5\omega)+c,\] \[\int{\cos(5u)}\ \mathrm{d}u = \dfrac{1}{5}\sin(5u)+c\]

Example

Find the indefinite integral of \(\dfrac{1}{x}\): that is, find \[\int \dfrac{1}{x}\ \mathrm{d}x\]

Solution

This integral deserves special mention. You may be tempted to try to write the integrand as \(x^{-1}\) and use the fourth row of Table 1. However the formula \[\int{x^{n}}\ \mathrm{d}x = \dfrac{x^{n+1}}{n+1}+c\] is not valid when \(n = -1\) as Table 1 makes clear. This is because we can never divide by zero. Look to the fifth entry of the table and you will see \[\int{x^{-1}}\ \mathrm{d}x = \ln|x|+c.\]

Example

Find \[\int 12\,\mathrm{d}x\] and \[\int 12\,\mathrm{d}t.\]

Solution

In this example we are integrating a constant, \(12\). Using Table 1 we find

\[\int 12\,\mathrm{d}x = 12x+c\] and \[\int 12\,\mathrm{d}t=12t+c.\]

Exercise

Integrate each of the following functions with respect to \(x\), i.e. for each function \(f(x)\) calculate \(\int f(x) \ \mathrm{d}x\) and enter this in the answer box:

1. if \(f(x)=[value]\) then \(\displaystyle{\int f(x)\ \mathrm{d}x=}\)
2. if \(f(x)=[value]\) then \(\displaystyle{\int f(x)\ \mathrm{d}x=}\)
3. if \(f(x)= \displaystyle [value]\) then \(\displaystyle{\int f(x)\ \mathrm{d}x=}\)
4. if \(f(x)= \displaystyle [value]\) then \(\displaystyle{\int f(x)\ \mathrm{d}x=}\)
5. if \(f(x)=[value]\) then \(\displaystyle{\int f(x)\ \mathrm{d}x=}\)
6. if \(f(x)=[value]\) then \(\displaystyle{\int f(x)\ \mathrm{d}x=}\)
Feedback: Remember that \(\dfrac{1}{x^n}\) can be written as \(x^{-n}\) and \(\sqrt{x}\) can be written as \(x^{\frac{1}{2}}\). So: 1. \( \displaystyle \int [value]\ \mathrm{d}x= [value].\) 2. \( \displaystyle \int [value]\ \mathrm{d}x = [value].\) 3. \( \displaystyle \int [value]\ \mathrm{d}x= [value].\) 4. \( \displaystyle \int [value]\ \mathrm{d}x = [value].\) 5. \( \displaystyle \int[value]\ \mathrm{d}x = [value].\) 6. \( \displaystyle \int[value]\ \mathrm{d}x = [value].\)

Exercise

Find

1. \({\displaystyle \int[value]\ \mathrm{d}t}=\)
2. \({\displaystyle \int[value]\ \mathrm{d}t}=\)
3. \({\displaystyle \int[value]\ \mathrm{d}t}=\)
4. \({\displaystyle \int[value]\ \mathrm{d}t}=\)
Feedback: 1. \( \displaystyle \int [value]\ \mathrm{d}t= [value].\) 2. \( \displaystyle \int [value]\ \mathrm{d}t = [value].\) 3. \( \displaystyle \int [value]\ \mathrm{d}t = [value]+c.\) 4. \( \displaystyle \int [value]\ \mathrm{d}t = [value].\)

Some rules of integration

To enable us to find integrals of a wider range of functions than those normally given in a table of integrals we can make use of the following rules.

The integral of \(kf(x)\) where \(k\) is a constant

A constant factor in an integral can be moved outside the integral sign as follows:

\[\int{kf(x)}\ \mathrm{d}x=k\int{f(x)}\ \mathrm{d}x\]

Example

Find the indefinite integral of \(11 x^2\): that is, find \[\int 11x^2\ \mathrm{d}x\]

Solution

\[\int{11x^2}\ \mathrm{d}x = 11\int {x^2}\ \mathrm{d}x = 11\left(\dfrac{x^3}{3} + c\right) = \dfrac{11x^3}{3} + K\] where \(K\) is a constant.

Example

Find the indefinite integral of \(-5\cos(x)\): that is, find \[\int -5\cos(x)\ \mathrm{d}x\]

Solution

\[\int{-5\cos(x)}\ \mathrm{d}x = -5\int {\cos(x)}\ \mathrm{d}x = -5(\sin(x) + c) = -5\sin(x) + K\] where \(K\) is a constant.

The integral of \(f(x)+g(x)\) and \(f(x)-g(x)\)

When we wish to integrate the sum or difference of two functions, we integrate each term separately as follows:

\[\int[f(x)+g(x)]\ \mathrm{d}x=\int{f(x)}\ \mathrm{d}x+\int{g(x)}\ \mathrm{d}x\]

\[\int[f(x)-g(x)]\ \mathrm{d}x=\int{f(x)}\ \mathrm{d}x-\int{g(x)}\ \mathrm{d}x\]

Example

Find\[\int{(x^3+\sin(x))}\ \mathrm{d}x\]

Solution

\[\int{x^3+\sin(x)}\ \mathrm{d}x = \int{x^3}\ \mathrm{d}x+\int{\sin(x)}\ \mathrm{d}x = \dfrac{1}{4}x^4-\cos(x)+c\]

Note that only a single constant of integration is needed.

Further rules for finding more complicated integrals are dealt with in subsequent Sections.

Electrostatic charge

Introduction

Electrostatic charge is important both where it is wanted, as in the electrostatic precipitator plate systems used for cleaning gases, and where it is unwanted, such as when charge builds up on moving belts. This Example is concerned with a charged object with a particular idealised shape - a sphere. However, similar analytical calculations can be carried out for certain other shapes and numerical methods can be used for more complicated shapes.

The electric field at all points inside and outside a charged sphere is given by

\(E(r) =\dfrac{Qr}{4\pi\epsilon_0 a^3}\), if \(r < a\)   (1a)

\(E(r) = \dfrac{Q}{4\pi\epsilon_0 r^2}\), if \(r < a\)   (1b)

where \(\epsilon_0\) is the permittivity of free space, \(Q\) is the total charge, \(a\) is the radius of the sphere, and \(r\) is the radial distance between the centre of the sphere and a point of observation (see Figure below).

[Image: Charged sphere]

Figure: Geometry and symbols associated with the charged sphere

The electric field associated with electrostatic charge has a scalar potential. The electric field defined by (1a) and (1b) shows only a radial dependence of position.

Therefore, the electric scalar potential \(V (r)\) is related to the field \(E (r)\) by \(E(r)=-\dfrac{ \mathrm{d} V}{ \mathrm{d} r}\)  (2)

Problem in words

A sphere is charged with a uniform density of charge and no other charge is present outside the sphere in space. Determine the variation of electric potential with distance from the centre of the sphere.

Mathematical statement of problem

Determine the electric scalar potential as a function of \(r,\ V(r)\), by integrating (2).

Mathematical analysis

Equation (2) yields \(V(r)\) as the negative of the indefinite integral of \(E(r)\).

\[-\int\ \mathrm{d}V=\int E(r)\, \mathrm{d}r. \text{(3)}\]

Using (1a) and (1b) with (3) leads to

\[V(r) = -\dfrac{Q}{4\pi\epsilon_0a^3}\int r\ \mathrm{d}r\] if \(r<a\), (4a)

\[V(r) = - \dfrac{Q}{4\pi\varepsilon_0}\int\dfrac{ \mathrm{d} r}{r^2}\] if \(r\ge a\), (4b)

Using the facts that \[\int r\ \mathrm{d}r=\dfrac{r^2}{2}+c_1\] and \[\int\dfrac{ \mathrm{d} r}{r^2}\ \mathrm{d}r=-\dfrac{1}{r}+c_2,\]

(4a) and (4b) become

\[V(r)=-\frac{Qr^2}{8\pi\epsilon_0a^3}+c_1\] if \(r<a\) (5a),

\[V(r)=\dfrac{Q}{4\pi\varepsilon_0 r}+c_2\] if \(r\ge a\) (5b).

The integration constant \(c_2\) can be determined by assuming that the electric potential is zero at an infinite distance from the sphere: \[\lim_{r\to\infty} [V(r)]=0\] \[\Rightarrow \lim_{r\to\infty}\left[-\frac{Q}{4\pi\epsilon_0r}\right]+ c_2 = 0\] \[ \Rightarrow c_2=0.\]

The constant \(c_1\) can be determined by assuming that the potential is continuous at \(r=a.\)

From equation (5a)

\(V(a)= -\dfrac{Qa^2}{8\pi\epsilon_0a^3} + c_1\)

From equation (5b)

\(V(a)=\dfrac{Q}{4\pi\varepsilon_0a}\)

Hence \(c_1=\dfrac{Q}{4\pi\varepsilon_0a}+\dfrac{2Q}{8\pi\varepsilon_0a}=\dfrac{3Q}{8\pi\varepsilon_0a}.\)

Substituting for \(c_1\) in (5), the electric potential obtained for all space is:

\(V(r)=\dfrac{Q}{4\pi\epsilon_0}\)\(\left(\dfrac{3a^2-r^2}{2a^3}\right)\), if \(r< a.\)

\(V(r)=\dfrac{Q}{4\pi\epsilon_0r}\), if \(r\ge a\).

Interpretation

The potential of the electrostatic field outside a charged sphere varies inversely with distance from the centre of the sphere. Inside the sphere, the electrostatic potential varies with the square of the distance from the centre.

An Engineering Exercise in a later section derives the corresponding expressions for the variation of the electrostatic field and another Engineering Exercise calculates the potential energy due to the charged sphere.

Exercise

Find

1. \({\displaystyle \int ([value])\ \mathrm{d}x}=\)
2. \({\displaystyle \int[value]\ \mathrm{d}x}=\)
3. \({\displaystyle \int[value]\ \mathrm{d}x}=\)
4. \({\displaystyle \int[value]\ \mathrm{d}x}=\)
5. \({\displaystyle \int[value]\ \mathrm{d}x}=\)
Feedback: [[facts:calc_int_linearity_rule]] 1. \(\begin{aligned} \displaystyle \int ([value])\ \mathrm{d}x &= [value]\displaystyle \int x \ \mathrm{d}x - \displaystyle \int [value] \ \mathrm{d}x, \\ &= [value]. \end{aligned}\) 2. \( \displaystyle \int [value]\ \mathrm{d}x= [value]\displaystyle \int e^{[value]x} \ \mathrm{d}x = [value].\) 3. \(\begin{aligned} \displaystyle \int [value]\ \mathrm{d}x &= \dfrac{1}{[value]} \displaystyle \int \cos(2x) \ \mathrm{d}x + \dfrac{1}{[value]} \displaystyle \int x \ \mathrm{d}x, \\ &= [value] + [value] + c, \\ &= [value]. \end{aligned}\) 4. \( \displaystyle \int [value]\ \mathrm{d}x = [value].\) 5. \(\begin{aligned} \displaystyle \int[value]\ \mathrm{d}x &= \int ([value])\ \mathrm{d}x , \\ &= \int x^2 \ \mathrm{d}x + \int [value]x \ \mathrm{d}x + \int [value] \ \mathrm{d}x, \\ &= [value]. \end{aligned}\)

Exercise

Find \(\displaystyle \int[value] \,\mathrm{d}t=\)
Feedback: Solution \(\displaystyle \int [value] \ \mathrm{d}t = [value]\).

Exercise

Find \(\displaystyle \int [value] \ \mathrm{d}t\) using the laws of indices to rewrite the integrand as \( x^{-[value]} \).

\(\displaystyle \int [value] \ \mathrm{d}t=\)
Feedback: Solution \(\displaystyle \int [value] \ \mathrm{d}t = \int x^[value] \ \mathrm{d}t= \dfrac{x^{[value]+1}}{[value]+1} +c = [value]+c.\)

Exercise

Find \(\displaystyle \int e^{-[value]x}\) using the entry in the table of integrals for integrating \( e^{kx}\).

\(\displaystyle \int e^{-[value]x}=\)
Feedback: Solution With \( k = [value],\) we have \(\displaystyle \int e^{-[value]x} = [value] .\)

Exercise

Find \(\displaystyle \int \left([value] + [value]\right) \,\mathrm{d}t=\)
Feedback: Solution Note that \(\displaystyle \int [value]\ \mathrm{d}t = \displaystyle [value] + c\) and \(\displaystyle \int [value] \ \mathrm{d}t= \int t^[value] \ \mathrm{d}t =[value]+ c \). Then \(\displaystyle \begin{aligned} \int \left([value] + [value]\right) \ \mathrm{d}t &= [value] + [value]+ c, \\ &= [value] \end{aligned}.\) Note that there is only one constant of integration.

Exercise

The hyperbolic sine and cosine functions, \(\sinh x\) and \(\cosh x\), are defined as follows: \(\sinh x = \dfrac{e^x - e^{-x}}{2},   \cosh x = \dfrac{e^x+e^{-x}}{2}\)
Note that they are combinations of the exponential functions \(e^x\) and \(e^{-x}\).

Find the indefinite integrals:

1. \(\displaystyle \int \sinh(x) \ \mathrm{d}t=\)
2. \(\displaystyle \int \cosh(x) \ \mathrm{d}t=\)
Feedback: Solution 1. \(\begin{aligned} \int{\sinh(x)} \ \mathrm{d}x = \int{\dfrac{e^x-e^{-x}}{2}}\ \mathrm{d}x = \int{\dfrac{e^x}{2}}\ \mathrm{d}x-\int{\dfrac{e^{-x}}{2}} \mathrm{d}x = \dfrac{e^x}{2} + \dfrac{e^{-x}}{2} + c = \cosh(x) + c\end{aligned}\). 2. \(\begin{aligned} \int{\cosh(x)} \ \mathrm{d}x = \int{\dfrac{e^x+e^{-x}}{2}}\ \mathrm{d}x = \int{\dfrac{e^x}{2}}\ \mathrm{d}x+\int{\dfrac{e^{-x}}{2}}\ \mathrm{d}x = \dfrac{e^x}{2} - \dfrac{e^{-x}}{2} + c = \sinh(x) + c\end{aligned}\).