A-level Mathematics/CIE/Pure Mathematics 2/Integration

Integration as Reverse Differentiation

Power Rule

The power rule for differentiation states that:

$\frac{d}{dx}{x}^n=n{x}^{n-1}$

The reverse of this is $\int n{x}^{n-1}dx=x^n+c$, which can be written as $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$

This is the power rule for integration.

Exponentials & Logarithms

Regarding the derivatives of exponentials and logarithms, we know that:

• $\frac{d}{dx}{e}^x=e^x$
• $\frac{d}{dx}\ln x=\frac{1}{x}$

Reversing these, we get:

• $\int e^x=e^x+c$
• $\int \frac{1}{x}=\ln |x|$

The natural logarithm takes a modulus input so that it can handle negative numbers.

Note that similar rules apply to any linear expressions that may be composed with these functions:

• $\int e^{ax+b}=\frac{1}{a}{e}^{ax+b}+c$
• $\int \frac{1}{ax+b}=\frac{1}{a}\ln |ax+b|$

Trigonometric Functions

The derivatives of the trigonometric functions are:

• $\frac{d}{dx}\sin x=\cos x$
• $\frac{d}{dx}\cos x=-\sin x$
• $\frac{d}{dx}\tan x={\sec }^{2}x$
• $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$

Thus, the corresponding integrals are:

• $\int \cos x=\sin x+c$
• $\int \sin x=-\cos x+c$
• $\int {\sec }^{2}x=\tan x+c$
• $\int \frac{1}{1+x^2}={\tan }^{-1}x+c$

As with exponentials and logarithms, this applies to linear expressions that are composed with trigonometric functions:

• $\int \cos (ax+b)=\frac{1}{a}\sin (ax+b)+c$
• $\int \sin (ax+b)=-\frac{1}{a}\cos (ax+b)+c$
• $\int {\sec }^2(ax+b)=\frac{1}{a}\tan (ax+b)+c$
• $\int \frac{1}{1+(ax+b{)}^2}=\frac{1}{a}{\tan }^{-1}(ax+b)+c$

Using Trigonometry when Integrating

Sometimes, it is useful to use trigonometric identities when finding the integral of an expression.

e.g. Find the integral $\int 4{\cos }^{2}xdx$.

$\begin{array}{cc}\cos 2x& =2{\cos }^{2}x-1\text{Double Angle Identity}\\ {\cos }^{2}x& =\frac{\cos 2x-1}{2}\\ \int 4{\cos }^{2}xdx& =\int 4\frac{\cos 2x-1}{2}dx\\ & =\int \cos 2x-1dx\\ & =\frac{1}{2}\sin 2x-x+c\end{array}$

Trapezium Rule

The area under a curve can be approximated using the area of several trapeziums

The trapezium rule states that the area under a curve can be approximated by finding the sum of the areas of trapeziums. The area of a trapezium is given by $A=\frac{B+b}{2}h$ Where $B$ is the length of the longer side, $b$ is the length of the shorter side, and $h$ is the length of the perpendicular distance between the sides.

In the context of the trapezium rule, each trapezium's perpendicular distance $h$ is a constant $\mathrm{\Delta }x$: the width of each trapezium. The lengths of the sides are given by points on a curve. Thus, for a curve $f(x)$ approximated using trapeziums, each trapezium has an area $A_i=\frac{f(x_i)+f(x_{i+1})}{2}\mathrm{\Delta }x$.

The area under the curve between the bounds $a$ and $b$ is the sum of the trapeziums' areas: $A=\sum _{x=a}^b\frac{f(x)+f(x+\mathrm{\Delta }x)}{2}\mathrm{\Delta }x$.

Because $\mathrm{\Delta }x$ is constant, the expression for the area can be written $A=\frac{f(a)+2f(a+\mathrm{\Delta }x)+2f(a+2\mathrm{\Delta }x)+\dots +2f(b-\mathrm{\Delta }x)+f(b)}{2}\mathrm{\Delta }x$, which can be simplified to $A=(\frac{f(a)+f(b)}{2}+f(a+\mathrm{\Delta }x)+f(a+2\mathrm{\Delta }x)+\dots +f(b-\mathrm{\Delta }x))\mathrm{\Delta }x$

Thus, the trapezium rule states:

$\underset{a}{\overset{b}{\int }}f(x)dx\approx (\frac{f(a)+f(b)}{2}+f(a+\mathrm{\Delta }x)+f(a+2\mathrm{\Delta }x)+\dots +f(b-\mathrm{\Delta }x))\mathrm{\Delta }x$

Trapezium Rule Worked Example

Q. Use the trapezium rule with 4 intervals to estimate the value of ${\int }_0^{4}2\sqrt{4x-x^2}dx$, giving your answer correct to 2 decimal places.

A.

$a=0$, $b=4$, and $h=1$

Using $y=2\sqrt{4x-x^2}$

$x$	0	1	2	3	4
$y$	$y_0=0$	$y_1=2\sqrt{3}$	$y_2=4$	$y_3=2\sqrt{3}$	$y_4=0$

$\underset{0}{\overset{4}{\int }}2\sqrt{4x-x^2}dx\approx \frac{h}{2}[y_0+y_4+2(y_1+y_2+y_3)]$

$\approx \frac{1}{2}[0+0+2(2\sqrt{3}+4+2\sqrt{3})]$

$\approx 4\sqrt{3}+4$

$\approx 10.93$ (to 2 decimal places)Template:Chapnav