what is antiderivative | simple techniques of integration | concepts of antiderivatives | Knowledge of physics

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 What do you mean by integration of a function?

Suppose,  F(x) is a function of x which is continuous at the open interval [a,b], that is x `\epsilon` [a,b] then f(x) is said to be the integration of function F(x) if at each point of [a,b] where b > a, 

`\frac{d}{dx} F(x) = f(x), x  \epsilon  [a,b]` ------(1)

This means f(x) is nothing but the derivative of F(x) and hence F(x) is nothing but the antiderivative or integral of f(x).  According to equation (1), F(x) will be antiderivative of f(x) if the derivative of F(x) becomes f(x). That is, the antiderivative of f(x) can be written as

`\int f(x) dx = F(x)` --------(2)

Thus, the integral value of F(x) at the interval [a,b] can be written as

`\int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)` ------------(3)


 Click HERE for the Proof of equation (3).


Here `x  \epsilon [a,b]` means that x belongs to interval [a,b], that is, x can have values from a to b.

Here integration of the form (2) is called indefinite integral because range of value of x is not defined and one can not determine definite value of integral in this case. The integration of the form (3) is called definite integral because the variable x is defined on interval [a,b] and integral value is given by F(b) - F(a).

Again as the derivative of a constant function or value (c) is zero, then F(x) + c is also an antiderivative of f(x).  That is

`\frac{d}{dx}(F(x) +c) = \frac{d}{dx}F(x) + \frac{d}{dx}c`

                                     `= \frac{d}{dx}F(x) + 0`

`\therefore` `\frac{d}{dx} F(x) + c) = \frac{d}{dx} F(x) = f(x)` 

Inversely, `\int f(x) dx = F(x) + c` --------(4)

Therefore, one needs to add a constant value (c) on integral value of a function when the integration is indefinite, as in equation (4). This c is called constant of integration or just an integration constant and it is adjusted with integral value or disappears when the integration is definite.


To learn rules of finding derivatives of functions: Click Here


Some formula for indefinite integration is listed as below:

Note that x is variable that may have different values. Other parameters such as n, p, q and c are constant values in the following integrations.

  • `\int x^n dx = \frac{x^{n+1}}{n+1} + c`
  • `\int cos(x) dx = sin(x) + c`
  • `\int sin(x) dx = - cos(x) + c`
  • `\int sec(x) tan(x) dx = sec(x) + c`
  • `\int sec^2(x) dx = tan(x) + c`
  • `\int cosec(x) cot(x) dx = - cosec(x) + c`
  • `\int cosec^2(x) dx = - cot(x) + c`
  • `\int cos(nx) dx = \frac{sin(nx)}{n} + c`
  • `\int sin(nx) dx = - \frac{cos(nx)}{n} + c`
  • `\int e^{nx} dx = \frac{e^{nx}}{n} + c`
  • `\int (px + q)^n dx = \frac{(px + q)^{n + 1}}{p(n+1)} + c`
  • `\int \frac{1}{x} dx = log(x) + c`
  • `\int \frac{1}{px + q} dx = \frac{1}{p} log(px + q) + c`
Where c is integration constant.




Now,
Some examples of definite integration:
Note that definite integration gives definite integral values. For this, two steps need to be followed.
  • first of all, perform indefinite integration
 `\int f(x) dx = F(x)` 
  • then use the given limit of variable x, to get the definite integral values.
For `x   \epsilon   [a,b]`,  `[F(x)]_a^b = F(b) - F(a)`
 
Examples:

Q.1.) Evaluate: `\int_0^1 x^2 dx`.

Solution:

Let the integral value be denoted by I then

I = `\int_0^1 x^2 dx`

Using the relation, `\int_a^b x^n dx = [\frac{x^{n+1}}{n+1}]_a^b`, we have

I = `[\frac{x^{2+1}}{2+1}]_0^1`

or, I = `[\frac{x^3}{3}]_0^1`

or, I = `\frac{1^3 - 0^3}{3}`

or, I = `\frac{1-0}{3}`

or, I = `\frac{1}{3}`

`\therefore` `\int_0^1 x^2 dx = \frac{1}{3}`.



Q.2.) Evaluate: `\int_{-1}^2 (x+2)^2 dx`.

Solution:

Here,
`\int_{-1}^2 (x+2)^2 dx`

= `\int_{-1}^2 (x^2 + 2x + 4) dx`

=`[\int (x^2 + 2x + 4) dx]_{-1}^2`

= `[\int x^2 dx + 2 \int  x dx + 4 \int dx ]_{-1}^2`

= `[\frac{x^3}{3} + 4 \frac{x^2}{2} + 4x]_{-1}^2`

Now,
= `[\frac{1}{3}(2^3 - (-1^3)) + 2 (2^2 - (-1)^2) + 4 (2 - (-1))]`

=`\frac{1}{3} (8+1) + 2 (4-1) + 4(2+1)`

=`\frac{1}{3} \times 9 + 2 \times 3 4 \times 3`

= `3 + 6 + 12`

= `21`

`\therefore` `\int_{-1}^2 (x+2)^2 dx = 21`.


Other examples:

1. Find definite integral: `\int_0^2 \frac{x}{\sqrt{x^2 + 4}}dx`.

Solution:

Here, 
I = `\int_0^2 \frac{x}{\sqrt{x^2 + 4}}dx` ------(1)

Suppose `y = x^2 +4` ------ (2), so that

`\frac{dy}{dx} = \frac{d}{dx}x^2 + \frac{d}{dx}4`

or, `\frac{dy}{dx} = 2x + 0`

`\therefore` `dy = 2x dx \Rightarrow x dx = \frac{dy}{2}`. 

Using these in (1), we have

I = `\int_0^2 \frac{1}{2} \frac{dy}{\sqrt{y}}`

  = `\int_0^2 \frac{1}{2} \frac{dy}{y^{\frac{1}{2}}}`

 = `\frac{1}{2}[\int y^{-\frac{1}{2}} dy]_0^2 `

= `\frac{1}{2}[\frac{y^{-\frac{1}{2} + 1}}{-\frac{1}{2}+1}]_0^2 `

= `\frac{1}{2}[\frac{y^{\frac{1}{2}}}{\frac{1}{2}}]_0^2`

= `[y^{\frac{1}{2}}]_0^2`

Using (2), 

= `[(x^2 + 4)^{\frac{1}{2}}]_0^2`

= `[\sqrt{x^2+4}]_0^2`

= `\sqrt{2^2 + 4} - \sqrt{0^2 + 4}`

= `\sqrt{4+4} - \sqrt{0+4}`

= `\sqrt{8} - \sqrt{4}`

= `2 \sqrt{2} - 2`

= `2(\sqrt{2} - 1)`

`\therefore` `\int_0^2 \frac{x}{\sqrt{x^2 + 4}}dx = 2 (\sqrt{2} - 1)`. 



2. `\int_0^{\pi/4} sec^2\theta d\theta`.

Solution:

Here,

`\int_0^{\pi/4} sec^2\theta d\theta`

= `[\int sec^2\theta d\theta]_0^{\pi/4} `

= `[tan\theta]_0^{\pi/4}`

= `tan\frac{\pi}{4} - tan0`

= `1 - 0`

=`1`

`\therefore` `\int_0^{\pi/4} sec^2\theta d\theta = 1`.



3. `\int_0^a \sqrt{a^2 - x^2} dx`

Solution:

Here,

I = `\int_0^a \sqrt{a^2 - x^2} dx` -----(1)

Suppose, `x = a sin\theta` -----(2) so that

`\frac{dx}{d\theta} = a \frac{ sin\theta}{d\theta}`

or, `\frac{dx}{d\theta} = a cos\theta`

`\therefore` `dx = a cos\theta d\theta`----(3)

Using (2) and (3) in (1), we have

I = `int_0^a \sqrt{a^2 - a^2 sin^2\theta}   a cos\theta d\theta`

I = `[\int \sqrt{a^2 - a^2 sin^2\theta}   a cos\theta d\theta ]_0^a`

I = `[\int a \sqrt{1 - sin^2\theta}   a cos\theta d\theta ]_0^a`

`therefore` Use of  `sin^2\theta + cos^2\theta = 1`//

I = `[\int a \sqrt{cos^2\theta}   a cos\theta d\theta ]_0^a`

I = `a^2[\int \sqrt{cos^2\theta}   cos\theta d\theta ]_0^a`

I = `a^2[\int cos\theta   cos\theta d\theta ]_0^a`

I = `a^2[\int cos^2\theta  d\theta ]_0^a` 

I = `a^2[\int  \frac{2}{2} cos^2\theta  d\theta ]_0^a` 

`therefore` Use of  `cos2\theta = 2cos^2\theta -1`//

I = `\frac{a^2}{2}[\int  (1+ cos2\theta) d\theta ]_0^a`

I = `\frac{a^2}{2}[\theta + \frac{sin2\theta}{2} ]_0^a`

`therefore` Use of  `sin2\theta = 2sin\theta cos\theta`//

I = `\frac{a^2}{2}[\theta + sin\theta cos\theta]_0^a`

Using (2),

I = `\frac{a^2}{2} [sin^{-1}(\frac{x}{a}) + \frac{x}{a} \sqrt{1 - (\frac{x}{a})^2}]_0^a`

I = `\frac{a^2}{2}[sin^{-1}\frac{a}{a} + \frac{a}{a}\sqrt{1-\frac{a}{a}} ]`

          `- \frac{a^2}{2}[sin^{-1}\frac{0}{a} + \frac{0}{a} \sqrt{1 - \frac{0}{a}} ]`

I = `\frac{a^2}{2}[sin^{-1}(1) + (1)\sqrt{1-1} ]`

          `- \frac{a^2}{2}[sin^{-1}0 + 0 \times \sqrt{1 - 0} ]`

I = `\frac{a^2}{2}[\frac{\pi}{2}+ 0] - \frac{a^2}{2}[0 + 0]`

I = `\frac{a^2}{2} \frac{\pi}{2} - 0`

I = `\frac{a^2}{2}\frac{\pi}{2} = \frac{\pi a^2}{4}`


`\therefore` `\int_0^a \sqrt{a^2 - x^2} dx =  \frac{\pi a^2}{4}`.

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