Title: Default for 13.2 Definite Integrals
Course: HELM Workbooks 2023-2024

When you were first introduced to integration as the reverse of differentiation, the integrals you dealt with were indefinite integrals. The result of finding an indefinite integral is usually a function plus a constant of integration. In this Section we introduce definite integrals, so called because the result will be a definite answer, usually a number, with no constant of integration. Definite integrals have many applications, for example in finding areas bounded by curves, and finding volumes of solids.

Exercise

Evaluate

1. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x}=\)
2. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x}=\)
3. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x}=\)
4. \({\displaystyle\int_[value]^[value] ([value]) \, \mathrm{d}t}=\)
Feedback: If we represent the indefinite integral \({\displaystyle\int f(x) \, \mathrm{d}x}\) by \(F(x) \), then \({\displaystyle\int_{a}^{b}f(x) \, \mathrm{d}x}= F(b) - F(a)\), hence: 1. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x= \left[ [value] \right]_[value]^[value] = [value]}.\) 2. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x= \left[ [value] \right]_[value]^[value] =[value]}.\) 3. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x= \left[ [value] \right]_[value]^[value] = [value]}.\) 4. \({\displaystyle\int_[value]^[value] ([value]) \, \mathrm{d}t= \left[ [value] \right]_[value]^[value] = [value]}.\)

Definite integrals

We saw in the previous Section that \({\displaystyle\int{f(x)}\,\mathrm{d}x}=F(x)+c\) where \(F(x)\) is that function which, when differentiated, gives \(f(x)\). That is, \(\displaystyle{{{d F\over {d x}}=f(x)}}\).

For example, \({\displaystyle\int\sin(3x)\,\mathrm{d}x}=-\dfrac{\cos(3x)}{3}+c\).

Here, \(f(x)=\sin (3x)\) and \(F(x)=- \dfrac{1}{3}\cos (3x)\).
We now consider a definite integral which is simply an indefinite integral but with numbers written to the upper and lower right of the integral sign. The quantity \({\displaystyle\int_a^b f(x)\,\mathrm{d}x}\) is called the definite integral of \(f(x)\) from \(a\) to \(b\). The numbers \(a\) and \(b\) are known as the lower limit and upper limit respectively of the integral. We define \({\displaystyle\int_a^bf(x)\,\mathrm{d}x}=F(b)-F(a)\) so that a definite integral is usually a number. The meaning of a definite integral will be developed in later Sections. For the present we concentrate on the process of evaluating definite integrals.

Evaluating definite integrals

When you evaluate a definite integral the result will usually be a number. To see how to evaluate a definite integral consider the following Example.

Example

Find the definite integral of \(x^2\) from \(1\) to \(4\); that is, find \({\displaystyle \int_1^4 x^2 \,\mathrm{d}x}\)

Solution

\({\displaystyle \int{x^2} \,\mathrm{d}x} = \dfrac{x^3}{3}+c\). Thus according to our definition

\({\displaystyle \int_1^4x^2 \,\mathrm{d}x} = F(4)-F(1) = \dfrac{4^3}{3}-\dfrac{1^3}{3} = 21\)

Writing \(F(b)-F(a)\) each time we calculate a definite integral becomes laborious so we replace this difference by the shorthand notation  \(\left[\phantom{\dfrac{1}{1}}\!\!\!F(x)\right]_a^b\).

Thus \(\left[\phantom{\dfrac{1}{1}}\!\!\!F(x)\right]_a^b\equiv F(b)-F(a)\) Thus, from now on, we shall write \({\displaystyle\int_a^bf(x) \,\mathrm{d}x}=\left[\phantom{\dfrac{1}{1}}\!\!F(x)\right]_a^b\) so that, for example \({\displaystyle \int_1^4x^2 \,\mathrm{d}x} = \left[\dfrac{x^3}{3}\right]_1^4=\dfrac{4^3}{3}-\dfrac{1^3}{3}=21\)

Example

Find the definite integral of \(\cos x\) from \(0\) to \(\dfrac{\pi}{2}\); that is, find \({\displaystyle \int_0^{\frac{\pi}{2}}\cos(x) \,\mathrm{d}x}\)

Solution

\({\displaystyle \int_0^{\frac{\pi}{2}}\cos(x) \,\mathrm{d}x} = \left[\phantom{\dfrac{1}{1}}\!\!\!\sin(x)\right]_0^{\frac{\pi}{2}} = \sin\left(\dfrac{\pi}{2}\right)-\sin(0) = 1\).

Always remember, that if you use a calculator to evaluate any trigonometric functions, you must work in radian mode.

Note from the previous question that interchanging the limits of integration, changes the sign of the answer.

If you interchange the limits, you must change the sign:

Key Point

\({\displaystyle\int_a^bf(x) \,\mathrm{d}x}= -{\displaystyle\int_b^af(x) \,\mathrm{d}x}\)

The work done by a spring

Example

When a spring is fixed at one end and stretched at the free end it exerts a restoring force that is proportional to the displacement of the free end. The constant of proportionality \(k \ (\mathrm{Nm}^{-1})\) is known as the stiffness of the spring. Calculate the work done in stretching a spring with stiffness \(k\) from displacement \(x_1 \ \mathrm{m}\) to displacement \(x_2 \ \mathrm{m}\) (\(x_2 > x_1\)) given that the work done \((W)\) is the product of force and displacement.

Solution

The restoring force varies during the displacement. So the work done during the extension cannot be determined from a single simple product. Consider a small element \(\delta x\) of the extension beyond an arbitrary displacement \(x\) The element is sufficiently small that the force during the displacement can be regarded as constant and equal to the force at displacement \(x\) is \(kx\). So the work done \(\delta W\) in extending the spring from displacement \(x\) to displacement \(x+\delta x\) is approximately \(kx\delta x\).Using the idea of integration as a limit of a sum, in this case as \(\delta x\) tends to zero, \( W={\displaystyle\int_{x_1}^{x_2}kx \,\mathrm{d}x} = \left[\dfrac{1}{2}kx^2\right]_{x_1}^{x_2}=\dfrac{1}{2}k(x_2^2-x_1^2)\)

Engineering example 2

Example

Torsion of a mild-steel bar

For materials such as mild-steel, the relationship between applied shear stress and shear strain (deformation) can be described as follows.

- For small values of the shear strain, the shear stress (\(\tau\)) and shear strain (\(\omega\)) are proportional to one another, i.e. \(\omega = \dfrac{1}{G} \times \tau\ (1)\) (where \(G\) is the shear modulus).
This is known as elastic behaviour.

- There is a maximum shear stress that the material is capable of supporting. If the shear strain is increased further, the shear stress remains roughly constant.
This is known as plastic behaviour.

The figure below summarises the relationship between shear stress and shear strain; the point \((\omega_Y,\tau_Y)\) is known as the yield point.

[Image: A graph showing how shear stress increases linearly with shear strain up to the yield point, where it becomes constant.]

Now suppose that one end of a bar of circular cross section is twisted through an angle \(\theta\), then the shear strain on the surface is given by \(\omega_S = \dfrac{R\,\theta}{L}\ (2)\) (where \(R\) and \(L\) are the radius and length of the bar respectively), while the shear strain, at a distance \(r\) from the central core, is given by \(\omega = \dfrac{r\,\theta}{L}\ (3)\).

The torque transmitted by a bar is given by the integral \(T = \displaystyle \int^R_0 2 \pi r^2 \tau(r) \, \mathrm{d}r\ (4)\).

As the shear strain is a function of distance from the central axis of the bar, it may be that the shear strain on the surface is greater than the critical shear strain \(\omega_Y\). In this scenario the shear stress is given by:

\[\tau = \begin{cases}\displaystyle \frac{\tau_Y}{\omega_Y} \omega, & \omega \leq \omega_Y \\ \tau_Y, &\omega > \omega_Y \end{cases} \quad (5) \]

i.e. the regions near the central axis exhibit elasticity, but in those regions near the surface the elastic limit has been exceeded and the metal exhibits plasticity (see the figure below).

[Image: The outer section of the bar where the elastic limit is exceeded exhibit plasticity, which corresponds to the yield poin]

Problem in words Find an expression for the torque transmitted by a bar as a function of the angle \(\theta\) through which one end is turned.

Mathematical statement of problem

Using Equations \((3)\) to \((5)\), find a formula for \(T\) in terms of the variable \(\theta\).

Mathematical analysis

Substituting \((3)\) into \((5)\):

\[\begin{aligned} \tau &= \begin{cases}\displaystyle \frac{\tau_Y}{\omega_Y}\, \frac{r\theta}{L}, & \displaystyle\frac{r\theta}{L} \leq \omega_Y \\ \tau_Y, & \displaystyle\frac{r\theta}{L} > \omega_Y \end{cases} \\ &= \begin{cases}\displaystyle \frac{\tau_Y}{\omega_Y}\, \frac{r\theta}{L}, & \displaystyle r \leq \frac{L\omega_Y}{\theta} = r_e \\ \tau_Y, & r > \displaystyle\frac{L\omega_Y}{\theta} = r_e \end{cases} \end{aligned}\]

For small values of \(\theta\), \(r_e \geq R\) so that the whole of the bar will be in the elastic region, i.e. \(\tau = \dfrac{\tau_Y}{\omega_Y} \, \dfrac{r\theta}{L}\).

Now \((4)\) becomes:

\[\begin{aligned} T &= \int_0^R 2\pi r^2 \frac{\tau_Y}{\omega_Y}\,\frac{r\theta}{L}\,\mathrm{d}r \\ &= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \int_0^R r^3\,\mathrm{d}r \\ &= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \left[\frac{r^4}{4}\right]_0^R \\ &= \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} R^4, \quad (6) \end{aligned}\]

i.e. the torque is directly proportional to the twist, \(\theta\). For larger \(\theta\), \(r_e < R\), so that \((4)\) becomes:

\[\begin{aligned} T &= \int_0^{r_e} 2\pi r^2 \frac{\tau_Y}{\omega_Y}\,\frac{r\theta}{L} \,\mathrm{d}r + \int_{r_e}^R 2\pi r^2 \tau_Y \,\mathrm{d}r \\ &= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \int_0^{r_e} r^3 \,\mathrm{d}r + 2\pi\tau_Y \int_{r^e}^R r^2 \,\mathrm{d}r \\ &= 2\pi \frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} \left[\frac{r^4}{4}\right]_0^{r_e} + 2\pi \tau_Y \left[\frac{r^3}{3}\right]_{r_e}^R \\ &= \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} r_e^4 + \frac{2\pi}{3} \tau_Y (R^3 - r_e^3) \end{aligned}\]

But \(r_e = L\omega_Y / \theta\), so:

\[\begin{aligned} T &= \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L}\,\frac{L^4\omega_Y^4}{\theta^4} + \frac{2\pi}{3}\tau_Y R^3 - \frac{2\pi}{3} \tau_Y \frac{L^3\omega_Y^3}{\theta^3} \\ &= \frac{2\pi}{3}\tau_Y R^3 + \pi \left(\frac{1}{2}\tau_Y - \frac{2}{3}\tau_Y\right) \frac{L^3\omega_Y^3}{\theta^3} \\ &= \frac{2\pi}{3}\tau_Y R^3 - \frac{\pi}{6} \tau_Y \frac{L^3\omega_Y^3}{\theta^3} \quad (7) \end{aligned}\]

Equation \((6)\) will apply when \(r_e \geq R\), i.e. \(L\omega_Y/\theta \geq R\) or \(\theta \leq L\omega_Y/R\), so that combining \((6)\) and \((7)\) gives overall:

\[ T = \begin{cases}\displaystyle \frac{\pi}{2}\,\frac{\tau_Y}{\omega_Y}\,\frac{\theta}{L} R^4, & \displaystyle \theta \leq \frac{L \omega_Y}{R} \\ \displaystyle \frac{2\pi}{3} \tau_Y R^3 - \frac{\pi}{6}\tau_Y \frac{L^3\omega_Y^3}{\theta^3}, & \displaystyle \theta > \frac{L\omega_Y}{R} \end{cases} \quad (8) \]

At the critical value of \(\theta\), i.e. when the outer edge begins to exhibit plasticity, both formulae in \((8)\) give \(T_{\mbox{crit}} = \dfrac{\pi}{2} \tau_Y R^3\). Furthermore, the first derivatives are both \(\dfrac{dT}{d\theta} = \dfrac{\pi}{2} \dfrac{\tau_Y}{\omega_Y} \dfrac{R^4}{L}\), i.e. the curves join smoothly. The second derivatives, though, are not equal (zero in one case). In the theoretical limit as \(\theta \rightarrow \infty\), \(T = \dfrac{2\pi}{3} \tau_Y R^3\) so this is the total torsional torque which can be carried by the bar. (The critical torque above is three-quarters of this value.) However, clearly \(\theta \rightarrow \infty\) is merely a theoretical limit since the bar would, in fact, shear at a finite value of \(\theta\).

Some integrals with infinite limits

On occasions, and notably when dealing with Laplace and Fourier transforms, you will come across integrals in which one of the limits is infinite. We avoid a rigorous treatment of such cases here and instead give some commonly occurring examples.

Example

Find the definite integral of \(e^{-x}\) from \(0\) to \(\infty\); that is, find \({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\).

Solution

The integral is found in the normal way:

\({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\) = \(\bigg[- e^{-x}\bigg]_0^\infty\)

There is no difficulty in evaluating the square bracket at the lower limit. We obtain simply \(-e^{-0}=-1\). At the upper limit we must examine the behaviour of \(-e^{-x}\) as \(x\) gets infinitely large. This is where it is important that you are familiar with the properties of the exponential function. If you refer to the graph (see the figure below) you will see that as \(x\) tends to infinity \(e^{-x}\) tends to zero.

[Image: The graph of e to the negative x decreases towards zero for increasing x.]

Consequently the contribution to the integral from the upper limit is zero. So

\({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\) = \(\left[\phantom{\dfrac{1}{1}} - e^{-x}\right]_0^\infty\) = \((-e^{-\infty})-(e^{-0})\) = \((0) - (e^{-0})\) = \(1\)

Thus the value of \({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\) is 1.

Another way of achieving this result is as follows: We change the infinite limit to a finite limit, \(b\), say and then examine the behaviour of the integral as \(b\) tends to infinity, written as \({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x} = {\displaystyle\lim_{b\rightarrow\infty}}{\displaystyle\int_0^b e^{-x} \, \mathrm{d}x}\)

Now,

\({\displaystyle\int_0^b e^{-x} \, \mathrm{d}x} = \left[\phantom{\dfrac{1}{1}}e^{-x}\right]_0^b = \left( -e^{-b}\right) - \left(-e^{-0}\right) = -e^{-b} + 1\)

Then as \(b\) tends to infinity \(-e^{-b}\) tends to zero, and the resulting integral has the value 1, as before.

Many integrals having infinite limits cannot be evaluated in a simple way like this, and many cannot be evaluated at all. Fortunately, most of the integrals you will meet will exhibit the sort of behaviour seen in the last example.

Exercise

Evaluate

1. \({\displaystyle\int_[value]^{\infty} [value] \, \mathrm{d}x}=\)
2. \({\displaystyle\int_[value]^{\infty} [value] \, \mathrm{d}x}=\)
3. \({\displaystyle\int_{[value]}^{\infty} [value] \, \mathrm{d}x}=\)
4. \({\displaystyle\int_[value]^{\infty} [value] \, \mathrm{d}t}=\)
Feedback: Let \({\displaystyle\int f(x) \, \mathrm{d}x}= F(x)\) , then \({\displaystyle\int_{a}^{b} f(x) \, \mathrm{d}x}= F(b)-F(a)\). When \(b\) is infinity, we cannot directly evaluate the integral, rather we let \(b\) be finite and take the limit of the integral as \(b\) goes to infinity. Hence \(\displaystyle\int_{a}^{\infty} f(x) \, \mathrm{d}x = \lim_{b \rightarrow \infty} F(b) - F(a)\). 1. \(\displaystyle\int_[value]^{\infty} [value] \, \mathrm{d}x= \left[ [value] \right]_[value]^{\infty} = [value] \,\), as \(\displaystyle\, \lim_{x \to \infty} [value] = 0. \) 2. \(\displaystyle\int_[value]^{\infty} [value] \, \mathrm{d}x= \left[ [value] \right]_[value]^{\infty} = [value] \,\), as \(\displaystyle\, \lim_{x \to \infty} [value] = 0. \) 3. \(\displaystyle\int_{[value]}^{\infty} [value] \, \mathrm{d}x= \left[ [value] \right]_[value]^{\infty} = [value] \,\), as \(\displaystyle\, \lim_{x \to \infty} [value] = 0. \) 4. \(\displaystyle\int_[value]^{\infty} [value] \, \mathrm{d}t= \left[ [value] \right]_[value]^{\infty} = [value] \,\), as \(\displaystyle\, \lim_{t \to \infty} \dfrac{1}{t} = 0. \)

Exercise

Determine

1. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x}=\)
2. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x}=\)
3. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}t}=\)
Feedback: If we represent the indefinite integral \({\displaystyle\int f(x) \, \mathrm{d}x}\) by \(F(x) \), then \({\displaystyle\int_{a}^{b}f(x) \, \mathrm{d}x}= F(b) - F(a)\), hence 1. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x= \left[[value] \right]_[value]^[value] = [value]}.\) 2. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}x= \left[[value] \right]_[value]^[value] =[value]}.\) 3. \({\displaystyle\int_[value]^[value] [value] \, \mathrm{d}t= \left[[value] \right]_[value]^[value] = [value]}.\)

Exercise

Find the definite integral of [value] from [value] to [value], that is, find \(\displaystyle \int_[value]^[value] ([value]) \, \mathrm{d}x\).

First find the indefinite integral:

\(\displaystyle \int ([value]) \, \mathrm{d}x = \)

Now insert the limits of integration, the upper limit first, and hence evaluate the integral:

\(\displaystyle \int_[value]^[value] ([value]) \, \mathrm{d}x = \)

Now find \(\displaystyle \int_[value]^[value] ([value]) \, \mathrm{d}x\). Note that the limits have been interchanged.

\(\displaystyle \int_[value]^[value] ([value]) \, \mathrm{d}x = \)
Feedback: We are asked to find \(\displaystyle \int_[value]^[value] ([value]) \, \mathrm{d}x\). We begin by finding the indefinite integral: \[\int [value] = [value].\] This gives \(\displaystyle F(x) = [value]\). Next, we use the definition of the definite integral. \[\begin{aligned} \int_[value]^[value] ([value]) \, \mathrm{d}x &= F([value]) - F([value]) \\ &= [value] - [value] \\ &= [value]. \end{aligned}\] Lastly, we are asked to find \(\displaystyle\int_[value]^[value] ([value]) \, \mathrm{d}x\). Again, we use the definition. \[\begin{aligned} \int_[value]^[value] ([value]) \, \mathrm{d}x &= F([value]) - F([value]) \\ &= [value] - [value] \\ &= [value]. \end{aligned}\]