Differentiation
Under the Integral Sign
A.
Leibnitz’s Rule.
If
g is continuous on [a, b] and f1 (x) and f2 (x) are
differentiable functions whose values lie in [a , b], then
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image001.gif)
Example.1
Find
the points of maxima / minima of ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image002.gif)
Solution
Let![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image003.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image004.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image005.gif)
From the wavy curve, it is clear that f ¢(x)
changes its sign at x = ±2, ±1, 0 and hence the points of maxima are -1, 1 and of the
minima are –2, 0, 2 . ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image006.gif)
Example.2
If ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image007.gif)
Solution
Differentiating
w.r.t. x we have.
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image008.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image009.gif)
.
B. If F(t) =![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image011.gif)
, where
represents the
differentiation of g with respect to t keeping x constant.
Example.1
Evaluate
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image015.gif)
Solution
Let I (b) = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image016.gif)
&⇒
(b) =
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image019.gif)
or ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image020.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image021.gif)
If b = 0, then I (b) = 0
&⇒ c = 0. .
Hence, I (b) = ln (b +1)
Example.2
Prove
that
= p
sin–1 (b), where |b| < 1.
Solution
I(b)
=
.… (1)
I¢(b) =![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image023.gif)
Put
x =
– t
&⇒ dx = –dt
I¢ (b) = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image025.gif)
=
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image026.gif)
=
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%204_files/image027.gif)
I
(b) = psin–1 (b)
+ c
From
(1) when b = 0, I (b) = 0
&⇒
0 = p sin–1 0
+ c = 0 + c
&⇒ c = 0
Hence
I (b) = p sin–1
(b)