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

Power Rule
The power rule for differentiation states that:

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

The reverse of this is $$\int nx^{n-1} dx = x^n + c$$, 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:


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

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:


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

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 $$\int 4\cos^2 x dx$$.

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

Trapezium Rule


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 $$\Delta x$$: 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 $$A_i = \frac{f(x_i)+f(x_{i+1})}{2}\Delta x$$.

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

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

Thus, the trapezium rule states:

$$\int_a^b f(x) dx \approx \left(\frac{f(a) + f(b)}{2} + f(a+\Delta x) + f(a+2\Delta x) + \dots + f(b - \Delta x)\right)\Delta x$$

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 $h=1$

$$\int_{0}^{4}2\sqrt{4x-x^2}~dx\approx\frac{h}{2}\left[y_0+y_4+2\left(y_1+y_2+y_{3}\right)\right] $$

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

$$\approx4\sqrt{3}+4 $$

$$\approx10.93 $$ (to 2 decimal places)