Title: Default for 14.1 Integration as the Limit of a Sum
Course: HELM Workbooks 2023-2024

Exercise

Refer to the diagram below to help you answer the questions below.

[Image: Graph displaying the curve y=x^2 between 0 and 1, split into n rectangles.]

1. If the interval between \(x = 0\) and \(x = 1\) is divided into \(n\) rectangles what is the width of each rectangle?
2. What is the \(x\) coordinate at the left-hand side of the first rectangle?
3. What is the \(x\) coordinate at the left-hand side of the second rectangle ?
4. What is the \(x\) coordinate at the left-hand side of the third rectangle?
5. What is the \(x\) coordinate at the left-hand side of the \(k\)-th rectangle?
6. Given that \(y = x^2\), what is the \(y\) coordinate at the left-hand side of the \(k\)-th rectangle?
7. The area of the \(k\)-th rectangle is its height \(\times\) its width. Write down the area of the \(k\)-th rectangle.
Feedback: [Image: Graph displaying the curve y=x^2 between 0 and 1, split into n rectangles.] 1. The interval has length \(1\) and is divided into \(n\) equal-width rectangles. The width of each rectangle must then be \(\displaystyle [value]\). 2. The first rectangle starts at the beginning of the \([0,1]\) interval. Hence, the \(x\) coordinate of its left-hand side is [value]. 3. The second rectangle starts as soon as the first one ends. Since they all have width \(\displaystyle [value]\), the \(x\) coordinate of its left-hand side must be \(\displaystyle [value]\). 4. The third rectangle starts as soon as the second one ends. Hence, the \(x\) coordinate of its left-hand side must be \(\displaystyle [value]\). 5. After \(k-1\), the \(k\)-th rectangle's left-hand side will have \(x\) coordinate \(\displaystyle [value]\). 6. The height of each rectangle is the value of the function \(y = x^2\) at its left-hand side. Hence, the \(y\) coordinate at the left-hand side of the \(k\)-th rectangle is \(\displaystyle [value]\). 7. We obtain the width of the \(k\)-th rectangle by multiplying its width by its height, giving \(\displaystyle [value]\).

Exercise

Find the area under \(y = [value]\) from \(x = 0\) to \(x = 10\) using the limit of a sum.

Area \(=\)
Feedback: Let \(A\) denote the area under the curve that we aim to find. Using integration as the limit of a sum, we have \[A = \int^{10}_0([value]) \, \mathrm{d}x = \left[[value]\right]^{10}_0 = [value].\]

Exercise

Find the area under \(y = [value] \) from \(x = 0\) to \(x = 2\) using the limit of a sum.

(Give a rational answer).

Area =
Feedback: Let \(A\) denote the area under the curve that we aim to find. Using integration as the limit of a sum, we have \[A = \int^{2}_0 [value] \, \mathrm{d}x = \left[[value] \right]^{2}_0 = [value].\]

Exercise

Write down, but do not evaluate, the integral defined by the limit as \(\delta x \rightarrow 0\), or \(\delta t \rightarrow 0\) of the following sums.

Recall that by \(\displaystyle{ \sum_{x=a}^b y(x) \delta x}\) we mean to break the interval \([a,b]\) into subintervals of width at most \(\delta x\) and then sum the area of suitable rectangles of width \(\delta x\). Review an earlier key point if you are unsure of the meaning of the notation.

1. \(\displaystyle{\lim_{\delta x \rightarrow 0}[value] \delta x}\).
Write out your answer in the form \(\displaystyle[value]\). Then
\(f(x) =\)   \(a = \)   \(b = \)
2. \(\displaystyle{\lim_{\delta x \rightarrow 0}[value] \delta x}\).
Write out your answer in the form \(\;\displaystyle[value]\). Then
\(f(x) =\)   \(a = \)   \(b = \)
3. \(\displaystyle{\lim_{\delta t \rightarrow 0}[value] \delta t}\).
Write out your answer in the form \(\displaystyle[value]\). Then
\(f(t) =\)   \(a = \)   \(b = \)
4. \(\displaystyle{\lim_{\delta x \rightarrow 0}[value] \delta x}\).
Write out your answer in the form \(\:\displaystyle[value]\). Then
\(f(x) =\)   \(a = \)   \(b = \)
Feedback: 1. For \(\displaystyle{\lim_{\delta x \rightarrow 0}[value] \delta x},\) you should get the integral \(\displaystyle\int_{[value]}^{[value]} [value] \) \(\mathrm{d}x.\) 2. For \(\displaystyle{\lim_{\delta x \rightarrow 0}[value] \delta x},\) you should get the integral \(\displaystyle\int_{[value]}^{[value]}[value] \) \(\mathrm{d}x.\) 3. For \(\displaystyle{\lim_{\delta t \rightarrow 0}[value] \delta t},\) you should get the integral \(\displaystyle\int_{[value]}^{[value]} [value] \) \(\mathrm{d}t.\) 4. For \(\displaystyle{\lim_{\delta x \rightarrow 0}[value] \delta x},\) you should get the integral \(\displaystyle \int_{[value]}^{[value]} [value] \) \(\mathrm{d}x.\)

The Limit of a Sum

[Image: Graph showing the area under the function y(x) betwwen the points x=a and x=b.]

Consider the graph of the positive function \(y(x)\) shown in the figure above. Suppose we are interested in finding the area under the graph between \(x=a\) and \(x=b\). One way in which this area can be approximated is to divide it into a number of rectangles of equal width, find the area of each rectangle, and then add up all these individual rectangular areas. This is illustrated in the figure below on the left, which shows the area divided into \(n\) rectangles (with some small discrepancies at the tops), and the figure below on the right which shows the dimensions of a typical rectangle which is located at \(x=x_k\).

[Image: As described in the text above.]

We wish to find an expression for the area under a curve based on the sum of many rectangles. Firstly, we note that the distance from \(x=a\) to \(x=b\) is \(b-a\). In Figure 2a the area has been divided into \(n\) rectangles. If \(n\) rectangles span the distance from \(a\) to \(b\) the width of each rectangle is \(\dfrac{b-a}{n}\)

It is conventional to label the width of each rectangle as \(\delta x\), i.e. \(\delta x = \dfrac{b-a}{n}\). We label the \(x\) coordinates at the left-hand side of the rectangles as \(x_1\), \(x_2\) up to \(x_n\) (here \(x_1=a\) and \(x_{n+1}=b\)). A typical rectangle, the \(k\)th rectangle, is shown in the figure above on the right. Note that its height is \(y(x_k)\), so its area is \(y(x_k)\times \delta x\).

The sum of the areas of all \(n\) rectangles is then \[y(x_1)\delta x + y(x_2)\delta x + y(x_3)\delta x + \dots + y(x_n)\delta x\] which we write concisely using sigma notation as \[\sum_{k=1}^n y(x_k)\delta x\]

This quantity gives us an estimate of the area under the curve but it is not exact. To improve the estimate we must take a large number of very thin rectangles. So, what we want to find is the value of this sum when \(n\) tends to infinity and \(\delta x\) tends to zero. We write this value as \[ \lim_{n\rightarrow\infty} \sum_{k=1}^n y(x_k)\delta x\]

The lower and upper limits on the sum correspond to the first rectangle and last rectangle where \(x=a\) and \(x=b\) respectively and so we can write this limit in the equivalent form \[ \displaystyle{\lim_{\delta x\rightarrow 0}\sum_{x=a}^{x=b}y(x)\delta x}.\]

Here, as the number of rectangles increases without bound we drop the subscript \(k\) from \(x_k\) and write \(y(x)\) which is the value of \(y\) at a ‘typical’ value of \(x\). If this sum can actually be found, it is called the definite integral of \(y(x)\), from \(x=a\) to \(x=b\) and it is written \(\displaystyle{\int_a^by(x) \, \mathrm{d}x}\).

Therefore we have the following definition:

Key Point

The definite integral \(\displaystyle{ \int_a^b y(x) \, \mathrm{d}x }\) is defined as \(\displaystyle{ \lim_{\delta x\rightarrow 0}\sum_{x=a}^{x=b}y(x)\delta x }\).

Note that the quantity \(\delta x\) represents the thickness of a small but finite rectangle. When we have taken the limit as \(\delta x\) tends to zero to obtain the integral, we write \(\mathrm{d}x\), which reminds us of the variable of integration.

This process of dividing an area into very small regions, performing a calculation on each region, and then adding the results by means of an integral is very important. This will become apparent when finding volumes, centres of gravity, moments of inertia etc.

Example

The area under the graph of \(y=x^2\) between \(x=0\) and \(x=1\) is to be found by approximating it by a large number of thin rectangles and finding the limit of the sum of their areas. This is equivalent to \(\displaystyle{\lim_{\delta x\rightarrow 0}\sum_{x=0}^{x=1}y(x)\,\delta x}\). Write down the integral which this sum defines and evaluate it to obtain the area under the curve.

Solution

The limit of the sum defines the integral \(\displaystyle{\int_0^1y(x) \, \mathrm{d}x}\). Here \(y=x^2\) and so \( \displaystyle{\int_0^1 x^2 \, \mathrm{d}x = \left[\frac{x^3}{3}\right]_0^1=\frac{1}{3}}\).

To show that the process of taking the limit of a sum actually works we investigate the problem in detail. We use the idea of the limit of a sum to find the area under the graph of \(y=x^2\) between \(x=0\) and \(x=1\), as illustrated in the figure below.

[Image: Graph displaying the curve y=x^2 between 0 and 1, split into n rectangles.]

Example

We continue from the end of the pervious exercise and use the limit of a sum to calculate \(\displaystyle {\int_0^1 x^2 \, \mathrm{d}x}\).

To find the total area \(A_n\) of the \(n\) rectangles we must add up all these individual rectangular areas: \[A_n=\sum_{k=1}^n {(k-1)^2\over n^3}\]

This sum can be simplified and then calculated as follows. You will need to make use of the formulas for the sum of the first \(n\) integers, and the sum of the squares of the first \(n\) integers:

\[\sum_{k=1}^n1=n,\qquad \sum_{k=1}^nk=\frac{1}{2}n(n+1),\qquad \sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1)\]

Then, the total area of the rectangles is given by \[ \begin{aligned} A_n&=\sum_{k=1}^n {(k-1)^2\over n^3}\\&= {1\over n^3}\sum_{k=1}^n (k-1)^2\\ &= {1\over n^3}\sum_{k=1}^n (k^2 -2k +1) \\ &= {1\over n^3}\left(\sum_{k=1}^n k^2 -2\sum_{k=1}^nk +\sum_{k=1}^n1 \right)\\ &= {1\over n^3}\left({n\over 6}(n+1)(2n+1) - 2{n\over 2}(n+1) + n\right)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &= {1\over n^2}\left({(n+1)(2n+1)\over 6} - (n+1)+1\right)\\ &= {1\over n^2}\left({(n+1)(2n+1)\over 6}-n\right)\\ &= {1\over 6n^2}\left(2n^2-3n+1\right)= {1\over 3} - {1\over 2n} +{1\over 6n^2} \end{aligned} \]

Note that this is a formula for the exact total area of the \(n\) rectangles. It is an estimate of the area under the graph of \(y=x^2\). However, as \(n\) gets larger, the terms \(\dfrac{1}{2n}\) and \(\dfrac{1}{6n^2}\) become small and will eventually tend to zero. If we let \(n\) tend to infinity we obtain the exact answer of \({\dfrac13}\).

The required area is \(\dfrac13\). It has been found as the limit of a sum and of course agrees with that calculated by integration.

In the calculations which follow in future work the need to evaluate complicated limits like this is avoided by performing integration techniques. Nevertheless it will sometimes be necessary to go through the process of dividing a region into small sections, performing a calculation on each section and then adding the results, in order to formulate the integral required. When numerical methods of integration are studied this summation method will prove fundamental.

Engineering Example

Pulley Belt Tension

Consider that a belt is partially wound around a pulley so that there is a difference in the tension either side of the pulley (see the figure below). The pulley will be stationary as long as the friction between belt and pulley is sufficient. The frictional force on the pulley will depend on the extent of the contact between belt and pulley i.e. on the angle \(\theta\) shown in the figure below. Given that the tensions on either side of the belt are \(T_2\) and \(T_1\) and that the coefficient of friction between belt and pulley is \(\mu\), find an expression for \(T_2\) in terms of \(T_1\), \(\mu\) and \(\theta\).

Solution

Consider a small element of the belt, at angle \(\theta\) where the tension is \(T\). Changing the angle by a small amount \(\Delta \theta\) changes the tension from \(T\) to \(T+\Delta T\).

[Image: Pulley system as described in the text above. Changes in theta are shown.]

Take moments about the centre of the pulley, denoting the radius of the pulley by \(R\) and assuming that the frictional force is \(\mu T\) per unit length. For the pulley to remain stationary, \[R\Delta \theta\mu T=R(T+\Delta T)-RT \quad \text{or} \quad \Delta \theta=\dfrac{\Delta T}{\mu T}.\]

Using integration as the limit of a sum, \[\theta=\int_{T_1}^{T_2}\frac{\mathrm{d}T}{\mu T}=\frac{1}{\mu}\bigg[\ln T\bigg]_{T_1}^{T_2} =\frac{1}{\mu}\ln\left(\frac{T_2}{T_1}\right). \quad \; \text{So} \; T_2=T_1 e^{\mu\theta}.\]

Exercise

Refer to the diagram below to help you answer the questions.

1. If the interval between \(x = 0\) and \(x = 1\) is divided into \(n\) rectangles what is the width of each rectangle?
2. What is the \(x\) coordinate at the left-hand side of the first rectangle?
3. What is the \(x\) coordinate at the left-hand side of the second rectangle ?
4. What is the \(x\) coordinate at the left-hand side of the third rectangle?
5. What is the \(x\) coordinate at the left-hand side of the \(k\)-th rectangle?
6. Given that \(y = x^2\), what is the \(y\) coordinate at the left-hand side of the \(k\)-th rectangle?
7. The area of the \(k\)-th rectangle is its height \(\times\) its width. Write down the area of the \(k\)-th rectangle.
Feedback: 1. The interval has length \(1\) and is divided into \(n\) equal-width rectangles. The width of each rectangle must then be \(\displaystyle [value]\). 2. The first rectangle starts at the beginning of the \([0,1]\) interval. Hence, the \(x\) coordinate of its left-hand side is [value]. 3. The second rectangle starts as soon as the first one ends. Since they all have width \(\displaystyle [value]\), the \(x\) coordinate of its left-hand side must be \(\displaystyle [value]\). 4. The third rectangle starts as soon as the second one ends. Hence, the \(x\) coordinate of its left-hand side must be \(\displaystyle [value]\). 5. After \(k-1\), the \(k\)-th rectangle's left-hand side will have \(x\) coordinate \(\displaystyle [value]\). 6. The height of each rectangle is the value of the function \(y = x^2\) at its left-hand side. Hence, the \(y\) coordinate at the left-hand side of the \(k\)-th rectangle is \(\displaystyle [value]\). 7. We obtain the width of the \(k\)-th rectangle by multiplying its width by its height, giving \(\displaystyle [value]\).