A Level Maths: How To Master Integration
Master Integration – Introduction
You will have met differentiation and you will know the standard result. In this article you are going to look at Integration and some various techniques to help you solve integral questions. This article is going to cover the techniques met in the first year of A Level maths only. We will have another article which will look at year 2 integration techniques in the future.
Here is the standard result for integration:
\begin{aligned} & \frac{d y}{d x}=x^n \\ & y=\frac{x^{n+1}}{n+1}+c \end{aligned}You will also have met the standard symbol for integration having done definite and indefinite integration at school or college:
\begin{aligned} & \int f(x) d x=g(x)+c \\ & \int_b^a f(x) d x=[g(x)]_b^a=g(a)-g(b) \end{aligned}Master Integration - Example of a Definite Integral
Here you need to perform a definite integral between the limits shown and you will obtain a specific numerical value. It is important to remember that when you doing definite integration there is no +c.
Our solution is as follows:
\begin{aligned} & =\int_1^4 5+x^{-\frac{1}{2}} d x \\ & =\left[5 x+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_1^4 \\ & =\left[5 x+2 x^{\frac{1}{2}}\right]_1^4 \\ & =\left[5(4)+2(4)^{\frac{1}{2}}\right]-\left[5(1)+2(1)^{\frac{1}{2}}\right] \\ & =(20+4)-(7)=24-7=17 \end{aligned}You need to show all the working out for questions like these. Most modern day calculators actually perform integration, but if you are simply relying on your calculator then you are not going to achieve the marks.
Master Integration - Example of a Differential Equation
Here is the solution for this question:
\begin{aligned} y & =\frac{10 x^5}{5}-5 x+c \\ y & =2 x^5-5 x+c \\ 30 & =2(2)^5-5(2)+c \quad \therefore c=-24 \\ \therefore y & =2 x^5-5 x-24 \end{aligned}The methods required for the above question are:
Integrate – it is a indefinite integral so there should be a +c
Use the information that you are given to find the value of c
Once you have the value of c you need to complete the question as shown above.
Dealing with an indefinite integral
In order to find the integral, which is an indefinite integral, you need to first rewrite the expression that you wish to integrate.
The final answer is shown below. Three answers are presented to you because there are three ways in which you can express your answer:
\begin{aligned} & \frac{8}{3} x^{\frac{3}{2}}-\frac{1}{x}+10 x+c \\ & \frac{8}{3} x^{\frac{3}{2}}-x^{-1}+10 x+c \\ & \frac{8}{3}(\sqrt{x})^3-\frac{1}{x}+10 x+c \end{aligned}It is important that you can obtain an answer in any format and it is important that you are able to recognise that answers can be written differently. This is especially so when you are doing questions from a book or a past paper. What you have written may not be what is written in the answers or within the mark scheme. This does not mean you are wrong.
Integration – Finding Areas Under A Curve
The main part of integration is to determine the area under a curve. In order to do this successfully you need to have a sketch of the general shape of the curve in question. It is important to note if there are any areas above the horizontal or indeed below it, but you will not know until you actually sketch the curve.
Here is an example to highlight this:
From the sketch there is an area above and below the horizontal. These two areas need to be calculated separately using the intersection with the x-axis as the limits. The area above the horizontal is calculated as follows:
\begin{aligned} & =\int_{-2}^0 y d x \\ & =\int_{-2}^0\left(x^3+x^2-2 x\right) d x \\ & =\left[\frac{x^4}{4}+\frac{x^3}{3}-x^2\right]_{-2}^0 \\ & =0-\left[\frac{(-2)^4}{4}+\frac{(-2)^3}{3}-(-2)^2\right] \\ & =0-\left[4-\frac{8}{3}-4\right] \\ & =0–\frac{8}{3}=\frac{8}{3} \end{aligned}The area below the horizontal can be calculated as follows:
\begin{aligned} =\int_0^1 y d x & =\left[\frac{x^4}{4}+\frac{x^3}{3}-x\right]_0^1 \\ & =\left[\frac{1}{4}+\frac{1}{3}-1\right]-0=-\frac{5}{12} \end{aligned}For the area below, you can see that there is a negative answer. Nothing has been wrong, the negative is simply telling you that the area is below the horizontal.
To find the total area of the curve simply add the two areas found, ignoring any minus signs which will give:
\frac{8}{3}+\frac{5}{12}=\frac{37}{12}For part b) of the question you are asked to perform the integral between the limits 1 and -2 which are the points you will notice on the sketch.
\begin{aligned} \int_{-2}^1 y d x & =\left[\frac{x^4}{4}+\frac{x^3}{3}-x\right]_{-2}^1 \\ & =\left[\frac{1}{4}+\frac{1}{3}-1\right]-\left[\frac{(-2)^4}{4}+\frac{(-2)^3}{3}-(-2)\right] \\ & =-\frac{5}{12}–\frac{8}{3}=\frac{9}{4} . \end{aligned}Here we have obtained a completely different answer. This is because the calculation does not take into account the negative. In other words, the negative is not ignored and this is how a calculator will perform the same calculation, which is why you cannot just enter the values onto your calculator.
Exam Question
This is quite a big question as you can see by the total number of marks.
There are two parts of the question a part a) and a part b)
Part a) In order to find the coordinates of the two points it is a case of solving the equation of the curve and the straight line simultaneously.
The solution is as follows:
\begin{aligned} 10-x & =10 x-x^2-8 \\ 0 & =x^2-11 x+18 \\ 0 & =(x-9)(x-2) \\ \therefore x & =9 \text { or } x=2 \\ y & =1 \quad y=8 \end{aligned}So A(2.8) \quad B(9,1)
For part b) you need to determine the area of the shaded region. You need to remember that when doing integration in order to determine the area, you are finding the area under the curve.
If you look at the diagram below and in particular at the lines in red:
You will see a trapezium. So to find the area you need to first find the area under the curve and then the area under the trapezium and to subtract these two answers from each other.
The area under the curve is found as follows:
\begin{aligned} & \int_2^a y d x=\text { Area under curve } \\ & \begin{aligned} \int_2^a\left(10 x-x^2-8\right) d x & {\left[5 x^2-\frac{x^3}{3}-8 x\right]_2^9 } \\ & =90-\frac{4}{3}=\frac{266}{3} \end{aligned} \end{aligned}The area of the trapezium is found as follows:
\begin{aligned} & =\frac{1}{2} h(a+b) \\ & =\frac{1}{2}(7)(8+1) \\ & =\frac{63}{2} \end{aligned}The shaded area is then found by subtraction:
\frac{266}{3}-\frac{63}{2}=\frac{343}{6}Always give, where possible and where required, answers as a fraction even if the fraction is top heavy. This is sometimes better than giving a decimal or to round a decimal number, which can result in the loss of accuracy.
Here we have essentially summarised the types of year 1 A Level Integration maths questions that you could still expect to see in your final year exam papers. It is important not to forget what you have learnt in year 1 of your course as they can still be examined on an A Level paper as well as what you have learnt in your final year of A Level maths at school or college.