Page 86 - DMTH202_BASIC_MATHEMATICS_II

Unit 6: Properties of Definite Integral

Notes
Did u know? The properties of definite integral make the integration simpler.

6.1.1 Integrability on all Subintervals

When a function involves a calculus integral on an interval it must also contain a calculus
integral on all subintervals.
Theorem (integrability on subintervals): If f is integrable on a closed, bounded interval [a,b] then
f is integrable on any subinterval [c,d]  [a,b].

6.1.2 Additivity of the Integral

When a function involves a calculus integral on two contiguous intervals it must also contain a
calculus integral on the amalgamation of the two intervals.

Did u know? The integral on the large interval is the sum of the other two integrals.
Theorem (additivity of the integral): Suppose f and g be integrable functions on the interval
[a; b]. Subsequently f +g is also integrable on [a; b] and we have
b b b
a   f  g   x dx  a  f   x dx  a  g   x dx .

Proof:
Let f and g are integrable on [a; b]. Select an arbitrary n N. Then we observe that there exists a
b 1 b 1 1
I
step function s  f so that  s n   x dx  I   f   a  f x , or else   f  would be an
 dx 
n a 4n 4n 4n
b
 
a 
upper bound to S = { s x dx s  f ; s is a step function} less than   f , in destruction of the
I
 
I
definition   f  sup S


Similarly we locate t s and with t ,  b t   x dx  a  b f   x dx  1 , a  b  n s   x dx  a  b g   x dx 1 .
n , n n a n 4n 4n
b  b 1
and  n t   x dx  a  g   x dx  .
a 4x
Merging these equations, and by means of the additivity for step functions provides us
That
b  b b 1
s 
a   n n s   x dx  a  f   x dx  a  g   x dx 2n ,
b  b b 1
 dx 
t 
a   n n t   x dx  a  f   x dx  a  g x 2n .
Furthermore we know that s   n s  f  g  t   n t (for all n  N), so s   n s is a step function
n n n
bounding f + g from below, and t   n t is a step function bounding f + g from above. Therefore
n
we locate that (still, for all n  N) we have
b b 1 b b 1
a  f   x dx  a  g   x dx  2n   I f  g  I f   g  a  f x a  g x 2n
 dx 
 dx 
for all n  N.

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