Pure Mathematics 1 (P1) - Integration

Objective: Techniques, Definite Integrals, Areas Under Curves

1. Integration Techniques

Integration is the process of finding the antiderivative of a function. The reverse of differentiation. We will explore several techniques for finding integrals.

1.1 Basic Integration Rules

• $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where n ≠ -1)
• $\int \frac{1}{x} dx = \ln|x| + C$
• $\int e^x dx = e^x + C$
• $\int \sin(x) dx = -\cos(x) + C$
• $\int \cos(x) dx = \sin(x) + C$
• $\int \sec^2(x) dx = \tan(x) + C$
• $\int \csc(x) dx = -\ln|\csc(x) + \cot(x)| + C$

1.2 Substitution

Substitution is used when the integrand is a composite function. The goal is to simplify the integral by replacing a part of the integrand with a new variable.

Example: $\int 2x \cos(x^2) dx$

Let $u = x^2$. Then $du = 2x dx$.

The integral becomes: $\int \cos(u) du = \sin(u) + C = \sin(x^2) + C$

1.3 Integration by Parts

Integration by parts is used when the integrand is a product of two functions.

Formula: $\int u dv = uv - \int v du$

Example: $\int x e^x dx$

Let $u = x$ and $dv = e^x dx$. Then $du = dx$ and $v = e^x$.

$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C$

1.4 Partial Fractions

Used to integrate rational functions where the denominator can be factored into distinct linear and irreducible quadratic factors.

Example: $\int \frac{1}{(x^2 - 1)} dx$

$\frac{1}{(x^2 - 1)} = \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$

$1 = A(x+1) + B(x-1)$

If $x = 1$, $1 = 2A \Rightarrow A = \frac{1}{2}$. If $x = -1$, $1 = -2B \Rightarrow B = -\frac{1}{2}$.

$\int \frac{1}{(x^2 - 1)} dx = \int (\frac{1/2}{x-1} - \frac{1/2}{x+1}) dx = \frac{1}{2} \ln|x-1| - \frac{1}{2} \ln|x+1| + C = \frac{1}{2} \ln|\frac{x-1}{x+1}| + C$

2. Definite Integrals

A definite integral represents the area under a curve between two limits of integration.

Notation: $\int_a^b f(x) dx$

Value: $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$.

Example: $\int_0^1 x^2 dx$

Antiderivative of $x^2$ is $\frac{x^3}{3}$.

Value: $\frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

3. Areas Under Curves

The area under a curve $y = f(x)$ between $x = a$ and $x = b$ is given by the definite integral $\int_a^b f(x) dx$.

If $f(x) \le 0$ on $[a, b]$, then the area is given by $-\int_a^b f(x) dx$.

Example: Find the area under the curve $y = x^2$ between $x = 0$ and $x = 2$.

Area = $\int_0^2 x^2 dx = [\frac{x^3}{3}]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}$

Example: Find the area under the curve $y = x^2$ between $x = 0$ and $x = 1$.

Area = $\int_0^1 x^2 dx = [\frac{x^3}{3}]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

Topic	Description
Integration Techniques	Substitution, Integration by Parts, Partial Fractions
Definite Integrals	Area under a curve between two limits
Areas Under Curves	Calculating the area between a curve and the x-axis