Preliminary Mathematics for online MSc programmes in Data AnalyticsUnit 3: Integration in 1D
Integration
Introduction to integration and area under the curve
When a function , is known, we can differentiate it to obtain the derivative . The reverse process is to obtain from knowledge of its derivative. This process is called integration.
However, integration is much more than just differentiation in reverse. It can be applied to finding areas under curves. By area under the curve we mean the area above the -axis and below the graph of , assuming that is positive. To give an example from Physics, the area underneath a graph of the velocity of an object against time represents the distance travelled by the object.
It is also important to regard integration as a process of adding up, or summation. Often, a physical quantity can be obtained by summing lots of small contributions or elements. For example, the position of the centre of mass of a solid body can be found by adding the contributions from all the small parts of which the body is composed.
Find the area under between and . You can see the graph of the function below with the thick solid line and the area under the function with the grey triangle. The area forms a right triangle, with base and height both equal to . The area is therefore . More generally, the area underneath between and is a right triangle with base and height both equal to . Thus, the area is .
Some of you may have already figured out that we just solved our first integral. Specifically, we calculated (and found out that the value of the integral is ). Do not worry if you do not understand the previous sentence; I haven't explained the notation yet.
Terminology and notation
The topic of integration can be approached in several different ways. Probably, the simplest way to think of it is as differentiation in reverse. In some applications we will know the derivative of the function, but not the function from which it was derived. This is why we need knowledge of integration.
Suppose we differentiate the function (meaning that and are equivalent; some times we will use one instead of the other). We will obtain . Integration reverses this process, and we say that the integral of is . Unfortunately, the situation we are in is actually a bit more complicated than that because there are lots of functions that we can differentiate to give . Here are some of them: All these functions have the same derivative, , because when we differentiate the constant term we obtain zero. As a result, when we reverse the process, we have no idea what the original constant term might have been. Because of this we include in our answer an unknown constant, say , called the constant of integration. We state that the integral of is . There is nothing special about the letter . We might use (or something else), but we avoid using letters from the end of the alphabet like , , and which are used for variables.
The symbol for integration is , known as an integral sign. Formally we write where the term within the integral is called the integrand, and the term indicates the name of the variable involved, in this case .
We say that is integrated with respect to to give . Technically, integrals of this sort are called indefinite integrals, to distinguish them from definite integrals, which we will see (really) soon. When we find an indefinite integral our answer should always contain a constant of integration.
Can you now calculate the integral ?
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 the one seen in Table 1.
| Function | Indefinite integral |
|---|---|
| constant | |
| where | |
What is ?
The function is of the form with and , and thus
We can check whether we have calculated the integral correctly by differentiating the answer.
which is the function we were meant to integrate.
What is ?
We can rewrite the integrand as . It is thus of the form with and . Thus,
We can again check whether we have calculated the integral correctly by differentiating the answer.
What is ?
The function is of the form with , and thus
We can again check whether we have calculated the integral correctly by differentiating the answer.
Ok, that is a good start but what do we do with functions like or ?
The first function involves adding two functions (both of them being of the form ).
The second function, , involves multiplication of two functions ( and ).
We need to introduce some simple rules to enable us to extend the range of functions that we can integrate.
Rules of integration
- Integration is linear: For any functions and and any real numbers and the integral of the function is
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Substitution rule: This rule involves making a substitution in order to simplify an integral and make it easier to calculate. For example, we may let a new variable, say , equal a more complicated part of the function we are trying to integrate. The choice of which substitution to make often relies upon experience. Do not worry if at first you cannot see an appropriate substitution. However, it is not simply a matter of changing the variable, care must be taken with the term since the new integral will now include the variable and not .
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Integration by parts: The following formula shows us how we can calculate an integral of the form . We note that the previous formula replaces the original integral with a different integral, as seen on the right-hand side. This is because the second integral is easier to compute.
| Function | Indefinite integral of |
|---|---|
For calculate
We note that involves adding two functions (the first one is of the form while the second one is ). Thus, the integral of is
In the final line, has been replaced with a new constant . It's just a tidier way of including a constant of integration in the final line of the calculated integral.
For calculate
We note that this involves multiplication of two functions, and . This example requires the use of the integration by parts rule.
Using this rule, we can calculate an integral that involves a product of two functions taking the form , with the first term being the derivative of and the second one being . In this case, the two functions within the integral are and . If and are denoted as the functions and , we would be able to calculate the integral
Now that we have and , we can calculate the right-hand side of the integration by parts rule.
Note that in the final line has been replace with a new constant .
Using integration by parts we were able to 'convert' a relatively complicated integral, , into an 'easier' integral.
For calculate .
This is a similar integral to the previous example, except that we have to deal with instead of . We will use integration by parts in this example as well.
Let and , then and . Thus
For calculate .
Here we have an example of a composite function. It is a function that raises to the sixth power. We could expand the power to obtain a sum, and then integrate each term one-by-one, but this would be a lot of work.
We will substitute with a new variable, say . In that case, instead of having we will have . This gives us a much simpler function to integrate. However, there is a slight complication. The new function of must be integrated with respect to and not with respect to . This means that we need to take care of the term .
If we differentiate with respect to we get:
It follows that we can write .
We now have that
To finish off we must rewrite this answer in terms of the original variable and we do this by replacing with . So:
For calculate .
We have another composite function to integrate and to make our life easier, we will substitute with .
To rewrite the integral in terms of instead of we take the substitution and differentiate with respect to , to get
thus and .
By the way, we could have also rewritten as and calculated
thus , thus .
Either way, we can now write
Definite integrals
When integration was introduced as the reverse of differentiation the integrals we 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 can be recognised by numbers written to the upper and lower right of the integral sign. The quantity is called the definite integral of from to . The numbers and are known as the lower and upper limits of the integral respectively.
The definite integral is the (signed) area under the graph of between and , as illustrated in the figure below.
These integrals have many applications (mentioned briefly in earlier sections), for example in finding areas bounded by curves or estimating probabilities. In the latter case will not just be a function, it will be a density (you will see more of this in Probability and Stochastic Models or Probability and Sampling Fundamentals).
If the indefinite integral , i.e. , then the definite integral can be calculated as
When you evaluate a definite integral the result will usually be a number. Most importantly, when you integrate over a variable like in a definite integral, the answer will never involve . For an indefinite integral the answer would be likely to involve .
What is ?
This is the same integral as the one we solved earlier in Example 8 with the only difference being that now it is a definite integral. The procedure we have to follow is exactly the same as before with one small, but significant, addition. The limits of the new integral must be in terms of and not in terms of . This means that since , when then ; and when then . Thus, we have:
You can stop right there, there is no need to calculate and . Off the top of my head they should be close to and respectively.
What is ?
This is a slightly more complicated example as we integrate all the way up to infinity.
We have already seen a similar indefinite integral in Example 6 using integration by parts with and :
This time the difference is that we have instead of . Hence we will be using and , yielding and .
You might feel that it is slightly dodgy notation to use infinity in a calculation as if it was a number. A cleaner way of writing would be . You might also be wondering why this quantity is 0. The reason is that, in the limit, decays faster to zero than any polynomial (including just ) could increase.
Tasks
If , then what is ?
Using that itegration is a linear operator, we can rewrite
Thus which we can re-arrange to
Calculate the value of
Evaluate
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To calculate the integral we will use the substitution .
Differentiating gives , thus .
Thus
Alternatively, we could have also solved for and differentiated giving , thus , yielding
Calculate the value of
The integral becomes easier to calculate if we first expand the integrand.