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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.
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 has provided the following outcomes for this topic:
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.
Let us first review some basic properties of integrals from 2/3 unit.
a)
\( \int_a^b (f(x) ± g(x)) \ dx = \int_a^b f(x) \ dx ± \int_a^b g(x) \ dx\)
b)
\( \int_a^b kf(x) \ dx = k \int_a^b f(x) \ dx \)
for some real constant \(k\).
\( \int_a^b f(x) \ dx = \int_a^c f(x) \ dx + \int_c^b f(x) \ dx\)
for \(a\) < \(c\) < \(b \).
\( \int_a^b f(x) \ dx = \ – \int_b^a f(x) \ dx\)
If \(f(x)\) is an even function then,
\(\int_{-a}^a f(x) \ dx = 2\int_0^a f(x) \ dx\)
If \(f(x)\) is an odd function then,
\(\int_{-a}^a f(x) \ dx=0\)
\( \int_a^b f(x) \ dx =\int_a^b f(t) \ dt\)
\( \int_a^b f(x) \ dx = F(b) \ – F(a)\)
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.
Prove that
\(\int_0^a f(x) \ dx=\int_0^a f(a-x) \ dx\)
Hence, evaluate
\(\int_0^π \frac{xsinx}{1+cos^2x} \ dx\)
Proof:
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 \(x=a\), \(u=0\).
So,
\begin{align*}
\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\\
\end{align*}
Now, let us use the previous formula to re-write the integral and simplify it
\begin{align*} \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\\ \end{align*} |
Now adding the new and the old integral, we have twice the old integral so
\begin{align*} 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}\\ \end{align*} |
Evaluate
\( \int_0^π xsinx \ dx \)
Using the property from Example 1, we have
\begin{align*} \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\\ \end{align*} |
Evaluate
\(\int_0^1 x(1-x)^n \ dx\)
Again, we use one application of the property from example 1 and we have
\begin{align*} \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)}\\ \end{align*} |
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.
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,
\begin{align*} \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\\ \end{align*} |
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 \(x=0\), \(u=0\).
So,
\begin{align*} \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\\ \end{align*} |
Now, using this property we can transform the integral into the following
\begin{align*} \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 \end{align*} |
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
\( \int_{-a}^b f(x) \ dx = \int_0^a f(x) \ dx + \int_0^b f(x) \ dx \)
1.
(i) \( \frac{1}{10 \ 100}\)
(ii) \(\frac{2π}{3}\)
(iii) \( \frac{π}{4} \)
2.
(i) Proof
(ii) Proof
3.
(i) Proof
(ii) Proof
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