Revision notes on Definite Integral

If then the equation f(x) = 0 has at least one root lying in (a, b) provided f is a continuous function in (a, b).
If the function f is same then
dx where c is any point lying inside or outside [a, b].

This holds true only when f is piecewise continuous in (a, b)

if f(x) = f(-x) i.e. f is an even function

If n is a positive rational number, then the improper integral

$\int_{0}^{\infty}e^{-x}x^{n-1}dx = \Gamma n$

is defined as the gamma function.

Γ(n+1) = n!

Γ1 = 1

Γ0 = ∞

Γ(1/2) = √π

where K = $\frac{\pi }{2}$ , if both m and n are even (m, n ϵ N)

= 1 otherwise

If h(x) and g(x) are differentiable functions of x then

In the above result, if n = 1, then
The definite integral $\int ^{b_{a}}$ f(x)dx is in fact a limiting case of the summation of an infinite series, provided f(x) is continuous on [a, b] i.e.,

where h = b – a/n.

The converse is also true i.e., if we have an infinite series of the above form, it can be expressed as definite integral.
Method to express the infinite series as definite integral:

1. Express the given series in the form Σ 1/n f (r/n)

2. Then the limit is its sum when n → ∞, i.e. lim_n→∞ h Σ 1/n f(r/n)

3. Replace r/n by x and 1/n by dx and lim_n→∞ Σ by the sign of ∫

4. The lower and the upper limit integration are the limiting values of r/n for the first and the last terms of r respectively.

$\int_{0}^{\pi/2}\frac{sin^{n}x}{sin^{n}x+cos^{n}x} dx = \frac{\pi}{4} = \int_{0}^{\pi/2}\frac{cos^{n}x}{sin^{n}x+cos^{n}x} dx$

$\int_{0}^{\pi/2}\frac{tan^{n}x}{1+tan^{n}x} dx = \frac{\pi}{4} = \int_{0}^{\pi/2}\frac{dx}{1+tan^{n}x}$

$\int_{0}^{\pi/2}\frac{dx}{1+cot^{n}x} = \frac{\pi}{4} = \int_{0}^{\pi/2}\frac{cot^{n}x}{1+cot^{n}x}dx$

$\int_{0}^{\pi/2}\frac{tan^{n}x}{tan^{n}x+cot^{n}x} dx = \frac{\pi}{4} = \int_{0}^{\pi/2}\frac{cot^{n}x}{cot^{n}x+tan^{n}x}dx$

$\int_{0}^{\pi/2}\frac{sec^{n}x}{sec^{n}x+cosec^{n}x} dx = \frac{\pi}{4} = \int_{0}^{\pi/2}\frac{cosec^{n}x}{sec^{n}x+cosec^{n}x}dx$

$\int_{0}^{\pi /2}log sin x dx = \int_{0}^{\pi /2}log cos x dx = \frac{-\pi }{2}log 2$

$\int_{0}^{\pi /2}log tan x dx = \int_{0}^{\pi /2}log cot x dx = 0$

$\int_{0}^{\pi /2}log sec x dx = \int_{0}^{\pi /2}log cosec x dx = \frac{\pi }{2}log 2$

$\int_{0}^{\infty }e^{-ax} sin bx dx = \frac{b}{a^{2}+b^{2}}$

$\int_{0}^{\infty }e^{-ax} cos bx dx = \frac{a}{a^{2}+b^{2}}$