Example.1
Evaluate
.
Solution
I = 
= I1 + I2
where I1
= 
Put x = – t
&⇒ dx = - dt
&⇒ I1 = 
Hence I = I1
+ I2 =
I = 
&⇒ 2I = 2p
= 2p
&⇒ I = p
Example.2
If f, g, h be
continuous functions on [0, a] such that f(a – x) = - f(x), g(a –x) = g(x) and
3h(x) – 4h(a –x) = 5, then prove that 
.
Solution
I = 
= 
= - 
7I = 3I + 4I
= 
= 5
(since f(a – x) g(a – x) = – f(x)
g(x) ).
&⇒ I = 0. .
Example.3
Evaluate (i) 
(ii) 
where [.] denotes
the greatest integer function.
Solution
(i) I = 2
, 0 < x < p
&⇒ 0 < x2 < p2
&⇒ –3.14 < x2 –p < 6.72.
= sin 4(p –3) + sin 3(1) + sin 2(1) + sin 1
(1) + ....+ sin (1) + .... + sin (1) + ....+ sin 6(p2 –p –6).
= 2 sin 1 + 2 sin 2
+ 2 sin 3 + (p –2) sin 4 + sin 5 +
(p2 –p –6) sin 6
(ii). x2
– 3x + 3 > 0, " x ∈ R (
D < 0)
It attains the
minimum value = 
at x = 
And the graph is
parabola
x2 –3x +
3 = 1, when x2 –3x + 2 = 0, i.e. when x = 1, 2.
x2 –3x +
3 = 2, when x2 –3x + 1 = 0, i.e. when x = 
x2 –3x +
3 = 3, when x2 –3x = 0, i.e. when x = 0, 3.
Hence [x2 –3x +
3] = 
I = 
=
0
Example.4
Evaluate 
a
> b > 0.
Solution
I = 
dq …. (1)
= 2 
dq.
Also, I = 2
dq ….
(2)
Adding (1) and (2),
2I = 2
= 8a
= 8a
= 8a
put a tanq = t
&⇒ I = 4
dt
= 
= 
= 
= 
.
Example.5
Prove that 
.
Solution
= 
&⇒ 
.
