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# Indefinite Integrals

In Physics, we know that the rate of change of displacement is velocity.

Now, if we are given the velocity of an object at any time i.e. any given instant, can we determine the displacement of that object at that instant.

Similarly, the rate of change of velocity is acceleration. If acceleration at any given time is known to us, can we identify its velocity?

Differentiation helps us to find the rate of change of any quantity with respect to another. Similarly, we need to reverse the process to find the answers to the above questions.

The answers to these questions can be given by what is called Integration.

Here, we introduce the concept of Integration.

• Integration is the inverse process of differentiation.
• In Differentiation, we are given a function f(x) & we find its derivative i.e. f ‘ (x).
• In Integration, we are given the derivative f ‘  (xof a function f(x). We have to find the original function i.ef(x) or the anti derivative.

This process is called anti or Integration.

Indefinite Integral

Let F(x)  be function which is differentiable in an interval, I. f(x) is a function such

that  (F(x))= F ‘(x)=f(x), x  I Then we have,

f(x) dx = F(x) where,f(x)dx denotes the family or class of anti derivatives which is read as Indefinite integral of f with respect to x.

Note: C is a constant called the Constant of Integration.

For every distinct value of C, we get a different member of the family. Hence, it is called Indefinite (which is not fixed/certain) integral.

Properties of Indefinite Integrals

1.  f ‘ (x)dx= f(x)+C
2. [f(x)+g(x)]dx = f(x)dx +  g(x)dx
3. [f(x)-g(x)]dx = f(x)dx –  g(x)dx
4. kf(x)dx = k f(x)dx + C, where k is any non zero real number.
5. .[ f(x) + g(x)]dx =  f(x)dx +   g(x)dx .

Examples

Now let’s consider some examples on indefinite integrals.

Example 1Find the anti-derivative i.e. Integral of f(x) =  + 3x with respect to x.

( + 3x)dx = ()dx + 3()dx

+3+C =  + 3+C

where C is the constant of Integration.

Example 2: Evaluate:(7Sin x – 3Cos x)dx

(7Sin x-3Cos x)dx = 7Sin xdx – 3Cos xdx

= 7(-Cos x) – 3(Sin x) +C = -7Cos x – 3Sin x+C

Example 3: Evaluate: dx .

dx = dx

dx – dx =  – log|x|+ C =  – log|x|+C

Example 4:  Evaluate: dx

dx =  dx

(3 x + 4CosecxCotx)dx =3 xdx+4 CosecxCotxdx .

-3Cotx -4Cosecx + C

Example 5Evaluate: (3 -2Sin x+7)dx

(3 -2Sin x+7)dx

=3 dx -2Sin xdx+7 dx=3 -2(-Cosx) +7+C

=+2Cosx ++C= + 2Cosx+ x+C

## Check Point

Evaluate the following Integrals –

1. (15+20)dx
2. (7Cosx – 3 )dx
3. (7 + 34 +17)dx
4. (x-1)(x-2)dx
5. dx
1. (15+20)dx=  + 20x +C
2. (7Cosx – 3 )dx = 7Sin x – 3Tan x+C
3. (7 + 34 +17)dx = 7+2+17x+C
4. (x-1)(x-2)dx=  –  + 2x + C
5. dx  =    – 2  +C .

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