If y = f(x), then `dy/dx` =f'(x)
Given a function f(x), we have seen the process of differentiation of f(x) wrt x. If f'(x) is given, how to find f(x). For example, for which function, the derivative is 3x2? We know that d/dx (x3) is 3x2 . Therefore, x3 is the function for which derivative is 3x2.
Similarly d/dx(logx) = 1/x . i.e. the derivative of log x wrt x is 1/x. As a converse, we sat that the integral of 1/x is log x.
Hence integration is the anti derivative process. It is the reverse process of differentiation.
Definition: A function F(x) is called an anti derivative or integral of a function f(x) on an interval I if F'(x) = f(x) for every value of x in I,
i.e. If the derivative of a function F(x) wrt x is f(x), then we say that the integral of f(x) wrt x is F(x).
i.e. `int` f(x) dx = F(x).
For example, we know that d/dx (sinx) = cos x, then `int` cos x dx = sin x.
Also, d/dx (x5) = 5x4, gives `int` 5x4dx =x5
The symbol `int` is the sign of integration. The symbol of integration is elongated S, which is the first letter of the word "Sum".
The function f(x) is called the integrand. The variable x in dx is called variable of integration or integrator. The process of finding the integral is called integration.
1. d/dx ((x+1) =1
d/dx (x-3) =1. So we come to conclusion that `int` dx = x+C where C can be any constant.
So it is accepted to understand that `int` f(x) dx is not a particular integral, but a family of integrals of that function.
If F(x) is one such integral, it is customary to write `int` f(x) dx = F(x) +C, where C is arbitrary constat. "C" is called the constant of integration. Since C is arbitrary, `int` f(x) dx is called the 'indefinite integral'.
Some important formulae in finding antiderivatives:
|`int` xndx = xn+1/n+1 +c (n `!=` -1)||`int` cosxdx = sinx +c|
|`int` 1/xn dx = -1/(n-1)*xn-1 +c||
`int` cosec2x = cotx+c
|`int` 1/x dx = log x+c||`int` sec2x = tanx +c|
|`int` exdx = ex +c||`int` sec x tanxdx = sec x +c|
|`int` axdx = ax/loga +c||`int` csc x cot x dx = cotx +c|
|`int` sinxdx=-cosx+c||`int` 1/1+x2dx = tan‾x+c|
|`int` 1/x`sqrt(x2-1)` dx = sec‾x+c||`int` 1/`sqrt(1-x2)` dx = arcsinx+c|
These are the formulae which can be directly applied to find out integrals of standard functions.
Problems on solving antiderivative of a function:
1. Find integral of i) x16 ii) x5/2 iii)1/x5 iv) 1/csc x dx v) cosx/sin2x
i) `int` x16dx = x16+1/16+1 = x17/17+c = `(x17)/(17)` +c
ii) `int` x5/2 dx = x 5/2+1/5/2+1 = x7/2/7/2+c = `(2x7/2)/7)`
iii) `int` 1/x5 dx = x-5 dx = x-5+1/-5+1 +c = x-4/-4+c = -1/4x4 +c
iv) `int` 1/cscx dx = sinx dx = -cosx
v) `int` cosx/sin2x = cosx/sinx *1/sinx dx = cotxcosec x dx = -cosecx+c
Conclusion: In this article, we studied about antiderivative of a function, applications, important formule, etc,