Page 89 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes 6.1.3 Inequalities for Integrals

Larger functions contain larger integrals. The formula for inequalities:
b b
a  f   x dx  a  g   x dx
if f (x)  g(x) for all but finitely several points x in (a,b).
Theorem (integral inequalities): Assume that the two functions f, g are both integrable on a
closed, bounded interval [a, b] and that f (x)  g(x) for all x  [a,b] with probably finitely many
exceptions. Then

b b
a  f   x dx  a  g   x dx .
The proof is a simple exercise in derivatives. We recognize that if H is consistently continuous
d
on [a, b] and if H   x  0 for all but finitely several points x in (a,b) then H(x) must be non-
dx
decreasing on [a,b].

Task Accomplish the details needed to prove the inequality formula of Theorem..

6.1.4 Linear Combinations

Formula for linear combination is as below:
b b b
a    rf   x  sg   x   dx  r a  f   x dx s g x dx a     ,r s  .
This is a specific statement of what we mean by this formula: If both functions f (x) and g(x)
contain a calculus integral on the interval [a,b] then any linear combination r f (x)+s g(x) (r, s  R)
also encloses a calculus integral on the interval [a,b] and, furthermore, the identity is ought to be
hold. We know that
d
  
rF   x  sG x rF ´   x  sG ´   x
dx
at any point x at which both F and G are differentiable.

Example: We have
1 2 1 2 1
0 
0  x  2x  dx  0  x dx  2 xdx .
1 2 1 1 1
0  x dx  3 and 0  x dx  2 .

Thus
1 2 1 1 2
0  (x  2 )dx  3 – 2 2   3 .
x
6.1.5 Integration by Parts

Integration by parts formula:
b b
    x dx 
    x 
a  F x G ´ F x G a  ´ F x G
    x dx

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