Further Integration Techniques
Introduction
What we are going to discuss in this article are some further techniques for integration mainly using the rules of indices.
The process discussed here is still for year 1 maths. It is important to ensure that you are fully aware of the rules of indices and here they are again as a summary:
Rules of indices:
\begin{aligned} & a^m \times a^n=a^{m+n} \\ & a^m \div a^n=a^{m-n} \\ & \left(a^m\right)^n=a^{m \times n} \\ & a^0=1 \\ & a^{-m}=\frac{1}{a^m} \\ & x^{\frac{a}{b}}=(\sqrt[b]{x})^a \end{aligned}And here is a reminder for the standard result of integration: \int x^n \mathrm{~d} x=\frac{1}{n+1} x^{n+1}+c
Further Integration Techniques - Examples
Let us consider the following questions:
a) \int \sqrt{x} \mathrm{~d} x
b) \int \frac{1}{x^3} \mathrm{~d} x
c) \int\left(\frac{2}{x^2}-\frac{3}{\sqrt{x}}\right) \mathrm{d} x
As you can see, before any integration can be done, they first need to be rewritten in a form that allows us to perform the actual integration.
Part a)
\begin{aligned} \int \sqrt{x} \mathrm{~d} x & =\int x^{\frac{1}{2}} \mathrm{~d} x \\ & =\frac{2}{3} x^{\frac{3}{2}}+c \end{aligned}
Take care when dividing by a fraction. In this case we have divided by \frac{3}{2} which gives \frac{2}{3}
Part b)
\begin{aligned} \int \frac{1}{x^3} \mathrm{~d} x & =\int x^{-3} \mathrm{~d} x \\ & =-\frac{1}{2} x^{-2}+c \end{aligned}Take care that you are using the rule of integration and not the rule of differentiation. The common mistake with integration is to subtract 1 away from the original power, which is associated with differentiation.
Part c)
\begin{aligned} \int\left(\frac{2}{x^2}-\frac{3}{\sqrt{x}}\right) \mathrm{d} x & =\int\left(2 x^{-2}-3 x^{-\frac{1}{2}}\right) \mathrm{d} x \\ & =-2 x^{-1}-3 \times 2 x^{\frac{1}{2}}+c \\ & =-\frac{2}{x}-6 \sqrt{x}+c \end{aligned}The integration first needs to be in the form xn and you can see the process that has been applied.
In all the above examples where you need to exercise further integration techniques, you will note that when the integral has been done, the integral symbol no longer appears. Plus, take note, all the examples are indefinite integrals and so the “+c” is required as part of the answer.
Further Integration Techniques - Definite Integrals
Let us consider the following questions
a)
\int_1^2\left(\frac{4 x-1}{x^4}\right) \mathrm{d} xb)
\int_1^4(3-x) \sqrt{x} \mathrm{~d} xPart a)
\begin{aligned} \int_1^2\left(\frac{4 x-1}{x^4}\right) \mathrm{d} x & =\int_1^2\left(4 x^{-3}-x^{-4}\right) \mathrm{d} x \\ & =\left[4 \times-\frac{1}{2} x^{-2}+\frac{1}{3} x^{-3}\right]_1^2 \\ & =\left[-\frac{2}{x^2}+\frac{1}{3 x^3}\right]_1^2 \\ & =\left(-\frac{1}{2}+\frac{1}{24}\right)-\left(-2+\frac{1}{3}\right) \\ & =\frac{29}{24} \end{aligned}The single algebraic fraction here has been split. You need to think of this as both terms in the numerator being divided by the numerator
\begin{aligned} &\frac{4 x}{x^4}=\frac{4}{x^3}=4 x^{-3}\\ &\frac{1}{x^4}=x^{-4} \end{aligned}Once this has been done, the integration can be done. Note, that we are dealing with definite integration and so no “+c” is required.
When substituting in the limits, you need to show the actual substitution being made. Take care when performing the actual calculations, as this is where most errors occur.
Part b)
\begin{aligned} \int_1^4(3-x) \sqrt{x} \mathrm{~d} x & =\int_1^4\left(3 x^{\frac{1}{2}}-x^{\frac{2}{2}}\right) \mathrm{d} x \\ & =\left[3 \times \frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}\right]_1^4 \\ & =\left[2 x \sqrt{x}-\frac{2}{5} x^2 \sqrt{x}\right]_1^4 \\ & =\left(2 \times 4 \times 2-\frac{2}{5} \times 16 \times 2\right)-\left(2 \times 1 \times 1-\frac{2}{5} \times 1 \times 1\right) \\ & =\frac{8}{5} \end{aligned}Once again the rules of indices have applied prior to any integration being performed and to allow us to perform any necessary further integration techniques. There is no actual need to go from fractional powers to terms involving square roots. But as mentioned before, you need to show the substitution being made and to take care when performing the actual calculations.
Some additional examples:
Now that you are able to apply the rules of indices correctly, we will look at some additional questions that require the use of indices.
Example:
This is a differential equation and the question is asking you to find the equation of the curve. This means we want to find an equation that starts “y = …”. The method is therefore to integrate. You need to be able to spot what the method is, as the question will not always involve an integral symbol.
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \sqrt{x}-\frac{1}{\sqrt{x}} \Rightarrow y & =\int\left(3 \sqrt{x}-\frac{1}{\sqrt{x}}\right) \mathrm{d} x \\ & =\int\left(3 x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right) \mathrm{d} x \\ & =3 \times \frac{2}{3} x^{\frac{3}{2}}-2 x^{\frac{1}{2}}+c \\ & =2 x^{\frac{3}{2}}-2 x^{\frac{1}{2}}+c \end{aligned}What we have obtained so far is the general equation of the curve. We want to find the particular solution, which means we need to find the value of “c”. We do this by using the points (4, 9), and this is shown as follows:
\begin{aligned} & 9=2 \times 8-2 \times 2+c \\ & c=9-16+4 \\ & c=-3 \end{aligned}And this means that the equation of the curve is y=2 x^{\frac{3}{2}}-2 x^{\frac{1}{2}}-3
Example
When you are required to find the area under a curve it is always a good idea to have a sketch. The sketch of the curve is as follows:
We want to determine the area under the curve between 0 and 4 which can be calculated as follows:
\begin{aligned} \text { Area under curve } & =\int_0^4(1+\sqrt{x}) \mathrm{d} x \\ & =\int_0^4\left(1+x^{\frac{1}{2}}\right) \mathrm{d} x \\ & =\left[x+\frac{2}{3} x^{\frac{3}{2}}\right]_0^4 \\ & =\left(4+\frac{2}{3} \times 8\right)-0 \\ & =\frac{28}{3} \end{aligned}Example
Part a) In order to find the coordinates of the straight line and the curve, you need to solve the equations simultaneously.
\frac{1}{x^2}-1=3 \Rightarrow \frac{1}{x^2}=4 \Rightarrow x^2=\frac{1}{4} \Rightarrow x= \pm \frac{1}{2}P\left(-\frac{1}{2}, 3\right)
Q\left(\frac{1}{2}, 3\right)
Part b) To perform the integral it is necessary to know what the limits are. From the sketch of the curve that is given, it does not indicate where the curve crosses the x-axis.
This can be done by setting y = 0 as this is the value of y on the x-axis. By doing this you will find that the curve crosses the x-axis at 1.However, we cannot just do an integral between 1 as the curve is not a continuous curve. You will see from the diagram that three separate areas need to be found.
The area of C is found by:
\begin{aligned} \int_{\frac{1}{2}}^1\left(\frac{1}{x^2}-1\right) \mathrm{d} x & =\left[-\frac{1}{x}-x\right]_{\frac{1}{2}}^1 \\ & =(-1-1)-\left(-2-\frac{1}{2}\right) \\ & =\frac{1}{2} \end{aligned}And from symmetry this is also going to be the same as A.
Area of B =3 \times 1=3
So the total area is \frac{1}{2}+\frac{1}{2}+3=4
Take lots of care when dealing with the area under a curve. Quite often you are not always doing a standard integral, so make sure you are using the diagram given or the diagram you have drawn to ensure you are finding the correct area and not something else.