# PROPERTIES OF DEFINITE INTEGRALS

By the second fundamental theorem of integral calculus, the following properties of definite integrals hold. They are stated here without proof.

Property 1 :

Definite integral is independent of the change of variable.

Property 2 :

The value of the definite integral changes by minus sign if the limits are interchanged

Property 3 :

Property 4 :

Property 5 :

Property 6 :

If f(x) is even function, then

Property 7 :

If f(x) is odd function, then

Property 8 :

If f(2a-x) = f(x)

Property 9 :

If f(2a-x) = -f(x)

Property 10 :

If f(a-x) = f(x)

This property help us to remove the factor x present in the integrand of the LHS.

### Example Problems Using Properties of Definite Integral

Problem 1 :

Evaluate the following integrals using properties of integration :

Solution :

Let f(x) = x cos [(ex-1)/(ex+1)]

Apply x = -x

f(-x) = -x cos [(e-x-1)/(e-x+1)]

f(-x) = -f(x)

The given function f(x) is odd function. So, the value is 0.

Problem 2 :

Solution :

Let f(x) = x5 + x cos x + tan3x + 1

Apply x = -x

f(-x) = -x5 -x cos x - tan3x + 1

 f(x) = x5f(-x) = (-x)5f(-x) = -x5 f(x) = x cos xf(-x) = -x cos (-x)f(-x) = -x cos x

f(x) = tan3x

f(-x) = [tan(-x)]3

f(-x) = -tan3x

So, x5, x cos x and tan3x are odd functions. By integrating them we will get 0.

= π/2+π/2

= 2π/2

π

Problem 3 :

Solution :

Let f(x) = (sin x)2

Apply x = -x

f(-x) = (sin (-x))2

f(-x) = (-sin x)2

f(-x) = sin2x

Problem 4 :

Solution :

Let f(x) = x log [(3+cosx)/(3-cosx)]

Let g(π-x) = 2π [log (3+cos(π-x))- log (3-cos(π-x))]

= 2π [log (3-cosx) - log (3+cosx)]

= -2π [log (3+cosx) - log (3+cosx)]

g(-x) = -g(x)

So, the value is 0.

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