Example.1
Evaluate 
.
Solution
Let 
=
Integrating by parts,
ln L = 
Hence L = eln2+1/2(pp -4) = 2 e1/2(p-4) .
Example.2
Evaluate 
.
Solution
Let I = 
Substituting x = 
+t , we get
dx = dt and 2x - p
= 2t
I = 
=
= 0 +
Now, let 
&⇒
I = 
= 
.
Example.3
Let
f(x) be a function such that f (t + x) = f (x) "
x, t ∈ R and f(x) is integrable everywhere on the real line.
Show that for any real a .
.
Solution
We Know that
.
Now consider 
.
Let x = u + t
&⇒
d x = du
.
Alternative Solution :
Let I (a) = 
&⇒
&⇒
I is independent of a .
Example.4
Solution
(i)
(ii) For 0 < x < p/4, we have 0 < tann x < tann-2
x so that 0 < In < In-2 ..
Therefore,
In + In+2 < 2In <
In + In-2 
&⇒
Example.5
Find
the value of 
, here {.} denotes fractional
part of ‘x’.
Solution
Here {t} = t-[t] and x = n + f, where n is
non negative integer and f = {x} i.e. 0 £
f <1.
Hence
= 
= n
.
Because {t} = t for 0 £ t < 1 and {t}
has period 1.
Now 
(Since {t}= t - n for n £ t £ n + f)
Hence the given integral I = I1
+ I2 = 
Alternatively a graphical method can be
used as follows:
If x = n + f then the given integral,
I = n (area of A with base and height ‘1’) + (area of A
with base and height (x -–n))
=
= 
= 
.
