Class 12 Maths Integrals Some Properties of Definite Integrals

Some Properties of Definite Integrals

There are some properties of definite integral which are very useful for calculating definite integral very

easily.

(i) aʃb f(x) dx = aʃb f(t) dt

(ii) aʃb f(x) dx = -bʃa f(t) dt

In particular, aʃa f(x) dx = 0

(iii) aʃb f(x) dx = aʃc f(x) dx + cʃb f(x) dx

(iv) aʃb f(x) dx = aʃb f(a + b - x) dx

(v) 0ʃa f(x) dx = 0ʃa f(a - x) dx

(vi) 0ʃ2a f(x) dx = 0ʃa f(x) dx + 0ʃa f(2a - x) dx

(vii) 0ʃ2a f(x) dx = 2 * 0ʃa f(x) dx, if f(2a - x) = f(x)

= 0, if f(2a - x) = -f(x)

(vii) -aʃa f(x) dx = 2 * 0ʃa f(x) dx, if f is an even function i.e. f(-x) = f(x)

= 0, if f is an odd function i.e. f(-x) = -f(x)

Problem: Evaluate the following integrals

(a) ʃ0π/2 cos2 x dx                                                                           (b) ʃ-π/2π/2 sin2 x dx

Solution:

(a) Let I = ʃ0π/2 cos2 x dx    …………..1

=> I = ʃ0π/2 cos2 (π/2 - x) dx                              [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]

=> I = ʃ0π/2 sin2 x dx      …………..2

Adding equation 1and 2, we get

2I = I = ʃ0π/2 (sin2 x + cos2 x) dx

=> 2I = I = ʃ0π/2 1 dx

=> 2I = I = [x]0π/2

=> 2I = π/2

=> I = π/4

So, ʃ0π/2 cos2 x dx = π/4

(b) Let I = ʃ-π/2π/2 sin2 x dx

Since sin2 (−x) = {sin(−x)}2 = (−sin x)2 = sin2 x, therefore, sin2 x is an even function.

It is known that if f(x) is an even function, then ʃ-aa f(x) dx = 2 * ʃ0a f(x) dx

So, I = 2 * ʃ0π/2 sin2 x dx

=> I = 2 * ʃ0π/2 {(1 – cos 2x)/2} dx

=> I = ʃ0π/2 (1 – cos 2x) dx

=> I = [x – sin 2x /2]0π/2

=> I = [π/2 – sin (2 * π/2) /2]

=> I = [π/2 – (sin π)/2]

=> I = π/2

So, ʃ-π/2π/2 sin2 x dx = π/2

.