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 ![]()
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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
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
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If f(x) ≤ φ[x] for a ≤ x ≤ b then![]()
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Gamma Function:
If n is a positive rational number, then the improper integral

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]}](https://files.askiitians.com/cdn1/cms-content/common/latex.codecogs.comgif.latex_int_0_pi2sinmxcosnxdx_fracm-1m-3.2or1n-1n-3.2or1mnmn-22or1.jpg)
If f(x) ≥ 0 on the interval [a, b] then ![]()
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where K =
, 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
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For a monotonically increasing function in (a, b)
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Where f(x) is a continuous function on [a, b] and F’(x) = f(x).
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If f(x) = f(a – x), then![]()
If f(x) = – f (a – x), then![]()
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If f(x) is a periodic function with period T, then
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In the above result, if n = 1, then![]()
The definite integral
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:


















