Part 6: Properties of Integrals | Beginner’s Guide to Year 12 Ext 2 Maths

You have to know the properties of integrals to im-prove your marks. In this article, we'll show you how to simplify those definite integrals

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Unsure of the properties of integrals and how to make complex integrals solvable? In this article, we’ll illustrate the various techniques for transforming definite integrals into solvable ones with practice questions so you can test your skills.



In this article, we’ll discuss


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Year 12 Mathematics Extension 2: Properties of Integrals

Often time students will find that many definite integrals are difficult to evaluate directly. However, by using and applying certain properties of integrals we can transform a definite integral into an easier and solvable integral. Therefore, it becomes extremely important for students to be able to prove general properties of definite integrals and apply these properties in order to solve them.


NESA Syllabus Outcomes

NESA has provided the following outcomes for this topic:

  • Understand and uses different representations of numbers and functions to model, prove results and find solutions to problems in a variety of contexts
  • Applies techniques of integration to structured and unstructured problems
  • Applies various mathematical techniques and concepts to model and solve structured, unstructured and multi-step problems
  • Communicates and justifies abstract ideas and relationships using appropriate language, notation and logical argument



Assumed Knowledge

Student should also be familiar with basic properties of definite integrals from 2/3-unit such as reversing integrals, integrals of even and odd functions.

We will be covering further properties of integrals throughout this article and therefore, it is recommended for readers to revise on the basic properties.



Basic Properties of Integrals

Let us first review some basic properties of integrals from 2/3 unit.


1. Linearity of Integrals


\( \int_a^b (f(x) ± g(x)) \ dx = \int_a^b f(x) \ dx ± \int_a^b g(x) \ dx\)



\( \int_a^b kf(x) \ dx = k \int_a^b f(x) \ dx \)
for some real constant \(k\).


2. Additivity of Integrals

\( \int_a^b f(x) \ dx = \int_a^c f(x) \ dx +  \int_c^b f(x) \ dx\)

for \(a\) < \(c\) < \(b \).


3. Reversing an integral

\( \int_a^b f(x) \ dx = \ – \int_b^a f(x) \ dx\)


4. Integrals of Even Functions

If \(f(x)\) is an even function then,

\(\int_{-a}^a f(x) \ dx = 2\int_0^a f(x) \ dx\)


5. Integrals of Odd Functions

If \(f(x)\) is an odd function then,

\(\int_{-a}^a f(x) \ dx=0\)


6. Dummy Variable Property

\( \int_a^b f(x) \ dx =\int_a^b f(t) \ dt\)


7. Fundamental Theorem of Calculus

\( \int_a^b f(x) \ dx = F(b) \ – F(a)\)

Where \(F(x)\) is the antiderivative of \(f(x)\) (i.e. \(F'(x)=f(x)\)).

Now, let us use these basic properties to prove some further properties of integrals.


Further Properties of Integrals

Example 1:

Prove that

\(\int_0^a f(x) \ dx=\int_0^a f(a-x) \ dx\)

Hence, evaluate

\(\int_0^π \frac{xsinx}{1+cos^2x} \ dx\)




As with many proof questions, it is usually best to start with the ‘harder’ side and then slowly transform it into the ‘easier’ side.

In this case, we will start with the left-hand side and use a substitution to transform the integral into the right-hand side.

Let \(u=a-x\) then \(dx=-du\). When \(x=0\), \(u=a\) and when \(x=a\), \(u=0\).



\int_0^a f(a-x) \ dx &= \int_a^0 f(u) \ (-du)\\
&= \int_0^a f(u) \ du\\
&= \int_0^a f(x) \ dx\\


Now, let us use the previous formula to re-write the integral and simplify it

\int_0^π \frac{xsinx}{1+cos^2x} \ dx &= \int_0^π \frac{(π-x)sin(π-x)}{1+cos^2(π-x)} \ dx\\
&= \int_0^π \frac{(π-x)cosx}{1+cos^2x} \ dx\\


Now adding the new and the old integral, we have twice the old integral so

2 \int_0^π \frac{xsinx}{1+cos^2x} \ dx &= \int_0^π \frac{xsinx}{1+cos^2x} \ dx + \int_0^π \frac{(π-x)cosx}{1+cos^2x} \ dx\\
&= π \int_0^π \frac{sinx}{1+cos^2x} \ dx\\
&= -π[tan^{-1}(cosx)]_0^π\\
&= \frac{π^2}{2}\\
∴ \int_0^π \frac{xsinx}{1+cos^2x}dx &= \frac{π^2}{4}\\



Example 2:


\( \int_0^π xsinx \ dx \)



Using the property from Example 1, we have

\int_0^π xsinx \ dx &= \int_0^π (π-x)sin(π-x) \ dx \\
&= \int_0^π (π-x)sinx \ dx \\
&= \int_0^π πsinx \ dx – \int_0^π xsinx \ dx \\
2 \int_0^π xsinx \ dx &= \int_0^π πsinx \ dx \\
&= π[-cosx]_0^π\\
&= 2π\\
\int_0^π xsinx \ dx &= π\\



Example 3:


\(\int_0^1 x(1-x)^n \ dx\)



Again, we use one application of the property from example 1 and we have

\int_0^1 x(1-x)^n \ dx &= \int_0^1 (1-x)(1-(1-x))^n \ dx\\
&= \int_0^1 (1-x)x^n \ dx\\
&= \int_0^1 (x^n – x^{n+1}) \ dx\\
&= \left[ \frac{1}{n \ + \ 1}x^{n+1} – \frac{1}{n \ + \ 2} x^{n+2} \right]_0^1\\
&= \frac{1}{n \ + \ 1} \ – \frac{1}{n \ + \ 2}\\
&= \frac{1}{(n \ + \ 1)(n \ + \ 2)}\\



Harder Integration Proofs

Sometimes in school or HSC exams, students can be asked to produce theoretical proofs of integrals like the property from example 1.

Harder integration proofs typically involve asking students to prove a particular property of an integral and then applying the property to evaluate a more difficult integral.

Most proofs follow a similar method where we use a substitution on ‘harder’ side of the integral and gradually transform it into the ‘easier; side.

In the following example we begin on the left hand side and aim to obtain the integral on the right hand side.


Example 1:

Prove that

\(\int_a^b f(a+b-x) \ dx =\int_a^b f(x) \ dx\)

For real values of \(a\) and \(b\).



Usually, the hardest step is picking a suitable substitution.

The best method to determine the correct substitution is to focus on the function inside the integral and try to find a substitution that will transform it into our required function in the other integral.

Let \(u=a+b-x\) then \(du=-dx\). When \(x=a\), \(u=b\) and when \(x=b\), \(u=a\). So,

\int_a^b f(a+b-x) \ dx &= \int_b^a f(u) \ (-du)\\
&= \int_a^b f(u) \ du\\
&= \int_a^b f(x) \ dx\\



Example 2:

Show that

\(\int_{-a}^a f(x) \ dx =\int_0^a (f(x)+f(-x)) \ dx\)

For all \(a > 0\)



Hence, evaluate

\(\int_{-\frac{π}{2}}^{\frac{π}{2}} \frac{e^x \ cosx}{1 \ + \ e^x} \ dx\)


Using some basic properties of integrals, we have

\(\int_{-a}^a f(x) \ dx = \int_{-a}^0 f(x) \ dx + \int_0^a f(x) \ dx\)


For \(\int_{-a}^0 f(x) \ dx\), let \(u=-x\) then \(du=-dx\). When \(x=-a\), \(u=a\) and when \(x=0\), \(u=0\).


\int_{-a}^0 f(x) \ dx &= \int_{a}^0 f(-u) \ (-du)\\
&= \int_0^a f(-u) \ du\\
&= \int_0^a f(-x) \ dx\\
∴ \ \int_{-a}^a f(x) \ dx &= \int_0^a f(-x) \ dx + \int_0^a f(x) \ dx\\
&= \int_0^a \left( f(x) + f(-x) \right) \ dx\\


Now, using this property we can transform the integral into the following

\int_{-\frac{π}{2}}^{\frac{π}{2}} \frac{e^x \ cosx}{1 \ + \ e^x} \ dx &= \int_{0}^{\frac{π}{2}} \frac{e^x \ cosx}{1 \ + \ e^x} + \frac{e^{-x} \ cos(-x)}{1 \ + \ e^{-x}}\ dx\\
&= \int_{0}^{\frac{π}{2}} \frac{e^x \ cosx}{1 \ + \ e^x} + \frac{e^{-x} \ cosx}{1 \ + \ e^{-x}}\ dx\\
&= \int_{0}^{\frac{π}{2}} \frac{e^x \ cosx}{1 \ + \ e^x} + \frac{cosx}{1 \ + \ e^{x}}\ dx\\
&= \int_{0}^{\frac{π}{2}} \frac{(1 \ + \ e^x) \ cosx}{1 \ + \ e^x}\ dx\\
&= \int_{0}^{\frac{π}{2}} cosx \ dx\\
&= \ [sinx]_0^{\frac{π}{2}}\\
&= 1



Concept Check Questions

1. Evaluate the following integrals:

(i)     \( \int_0^1 x(1-x)^{99} \ dx\)

(ii)    \( \int_0^π xsin^3x \ dx\)

(iii)   \( \int_0^{\frac{π}{2}} \frac{sinx}{sinx \ + \ cosx} \ dx \)


2. Let \(f(x)\) be an odd function and \(b > a > 0\).

(i)     Prove that

\( \int_{-a}^a f(x) \ dx = 0 \)

(ii)    Hence, show that

\( \int_{-a}^b f(x) \ dx = \int_a^b f(x) \ dx \)


3. Let \(f(x)\) be an even function and \(b > a > 0\).

(i)     Prove that

\( \int_{-a}^a f(x) \ dx = 2 \int_0^a f(x) \ dx \)

(ii)    Hence, show that

\( \int_{-a}^b f(x) \ dx = \int_0^a f(x) \ dx + \int_0^b f(x) \ dx \)



Concept Check Solutions


(i)     \( \frac{1}{10 \ 100}\)

(ii)    \(\frac{2π}{3}\)

(iii)   \( \frac{π}{4} \)



(i)     Proof

(ii)    Proof



(i)     Proof

(ii)    Proof


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