+91-705-589-3577

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 odd function

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

  • where f is a periodic function with period a

  • where f(a+x) = f(x), i.e. a is the period of the function f

  • If f(x) ≤ φ[x] for a ≤ x ≤ b then

  • Gamma 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

\int_{0}^{\pi/2}sin^{m}x cos^{n}x dx = \frac{[(m-1)(m-3)... 2 or 1][(n-1)(n-3)... 2 or 1]}{[(m+n)(m+n-2).... 2 or 1]}

  • If f(x) ≥ 0 on the interval [a, b] then

  • Walli’s Formula

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

                        = 1 otherwise

  • Leibnitz’s Rule:

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

  • For a monotonically increasing function in (a, b)

  • Where f(x) is a continuous function on [a, b] and F’(x) = f(x).

  • If f(x) = f(a – x), then

  • If f(x) = – f (a – x), then

  • If f(x) is a periodic function with period T, 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. limn→∞ h Σ 1/n f(r/n)

3. Replace r/n by x and 1/n by dx and limn→∞ 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.

  • Some particular case of the above are:

  • Important Results:

\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}}

Book a Live classLive coursesAsk question