Example.1
Evaluate
.
Solution
Let ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image002.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image003.gif)
=
Integrating by parts,
ln L = ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image006.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image007.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image008.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image010.gif)
Hence L = eln2+1/2(pp -4) = 2 e1/2(p-4) .
Example.2
Evaluate
.
Solution
Let I = ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image012.gif)
Substituting x =
+t , we get
dx = dt and 2x - p
= 2t
I = ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image014.gif)
=![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image015.gif)
= 0 +
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image017.gif)
Now, let ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image018.gif)
&⇒
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
.
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image025.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image026.gif)
Alternative Solution :
Let I (a) = ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image027.gif)
&⇒
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image028.gif)
&⇒
I is independent of a .
Example.4
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image029.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image031.gif)
Solution
(i)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image033.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image035.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image036.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image037.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image038.gif)
(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 ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image039.gif)
&⇒![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image040.gif)
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 ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image045.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image046.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image047.gif)
(Since {t}= t - n for n £ t £ n + f)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image049.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image050.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image052.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image053.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image054.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image055.gif)
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image056.gif)
Hence the given integral I = I1
+ I2 = ![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image057.gif)
Alternatively a graphical method can be
used as follows:
![](http://www.quizsolver.com/radix/dth/notif/SUB%201_files/image058.gif)
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))
=
=
=
.