Page 305 - DMTH401_REAL ANALYSIS

Unit 25: Mean Value Theorem

Notes
1 1 1 2 1 1
x
x
x
ò
-
e
ò x e dx = x e x 1 ò e dx = and e dx = - .
- 1 - - 1 e - 1 e
2 æ 1ö 2 2 2
 =  ç e - ÷ i.e.,  = 2 = 2 =
e è e ø e - 1 (2.7) - 1 6.29
1
1
-
g.l.b. {f(x) 1 £ x £ 1} = - and l.u.b. {f(x) 1 £ x £ 1} = and, so,   [ 1,1]. First Mean Value
-
-
Theorem is verified.
Verification of Second Mean Value Theorem
As shown above, f and g are integrable in [–1, 1]. Also g is monotonically increasing in [-1, 1].
By second mean value theorem there is a points c [–1, 1] such that
1 c 1
ò
ò
ò f(x) g(x) dx = g( 1) f(x) dx g(1) f(x) dx
-
+
- 1 - 1
1 1
x
 ò x e dx = 'I' x dx e x dx
+
ò
- 1 c
2
æ
2 1 c 2 1ö æ 1 c ö
 - ç - ÷ + e ç - ÷ .
è
e e 2 2ø è 2 2 ø
2
e - 5 2.29 2 29
2
Therefore c = = i.e. c = ±  [ 1,1]
-
2
e - 1 6.29 6 29
Thus second mean value theorem is verified.
Now we show the use of mean value theorems to prove some inequalities.
Example: By applying the first mean value theorem of Integral calculus, prove that
1/2 1 p 1
p /6 £ ò × dx £ 2
2
2
2
-
-
0 é ë (1 x) (1 k x )ù û 6 (1 - 1 4 k )
1 1 é 1 ù
Solution: In the first mean value theorem, take f(x) = , g(x) = , x  ê 0, ú . Being
2
2
-
(1 k x ) 1 x 2 ë 2û
-
é 1 ù
continuous functions, f and g are integrable in 0, ú .
ê
ë 2 û
By the first mean value theorem, there is a number   [m,M] such that
1 1
2 1 2 dx
ò 2 2 2 dx = ò 2 =  p / ,
d
-
-
-
0 é ë (1 x )(1 k x )ù û 0 1 x
{ 1 } ì 1ü 1
x
where in = g.l.b. f(x) 0 £ £ 2 and M = l.u.b. f(x) 0 £ x £ 2 þ . ý Now m = 1 and M = 2 .
í
î
1 - k
4
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