Example.1
Evaluate
.
Solution
I =
= I1 + I2
where I1
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image003.png)
Put x = – t
&⇒ dx = - dt
&⇒ I1 = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image004.png)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image005.png)
Hence I = I1
+ I2 =![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image006.png)
I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image007.png)
&⇒ 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 =
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image011.png)
= - ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image012.png)
7I = 3I + 4I
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image013.png)
= 5
(since f(a – x) g(a – x) = – f(x)
g(x) ).
&⇒ I = 0. .
Example.3
Evaluate (i)
(ii) ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image016.png)
where [.] denotes
the greatest integer function.
Solution
(i) I = 2
, 0 < x < p
&⇒ 0 < x2 < p2
&⇒ –3.14 < x2 –p < 6.72.
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image018.png)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image019.png)
= 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 = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image022.png)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image023.png)
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 = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image024.png)
x2 –3x +
3 = 3, when x2 –3x = 0, i.e. when x = 0, 3.
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image025.png)
Hence [x2 –3x +
3] = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image026.png)
I =
=
0
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image028.png)
Example.4
Evaluate
a
> b > 0.
Solution
I =
dq …. (1)
= 2
dq.
Also, I = 2
dq ….
(2)
Adding (1) and (2),
2I = 2
= 8a![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image034.png)
= 8a
= 8a
put a tanq = t
&⇒ I = 4
dt
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image038.png)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image039.png)
=
=
.
Example.5
Prove that
.
Solution
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image044.png)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image045.png)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image046.png)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image047.png)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%203_files/image048.png)
&⇒
.