f is bounded and integrable on [a,b], let c be a constatn
Prove that 
a) integral(c*f) fron a to b = c * integral(f) from a to b

b) show that f(x) +c is integrable with 
int(f+c) from a to b = int(f+c(b-a) fron a to b

Hi fmunshi!
If f is defined on [a,b], we can use the definition of integral to
solve this problem. In the first page, you will find that the integral
of f from a to b is defined as the limit of a Riemann sum as n goes to
infinity.

Definition Integrals
http://archives.math.utk.edu/visual.calculus/4/definite.1/

Riemann Sum
http://mathworld.wolfram.com/RiemannSum.html

Therefore, we can use propoerties of the limits and sums in order to
solve these problems:


a. integral(c*f) from a to b

= lim           Rsum(c*f(x))
  n->infinity

where Rsum is the Riemann sum as defined in the link above. Since it
is as summatory of terms, we can take c as the common factor.
Therefore,

= lim          c*Rsum(f(x))
  n->infinity

Then, since lim   Rsum(f(x)) exists (because f is integrable), we can
take the
             n->inf.
"c" out of the limit (property of limit). Then,

= c* lim          Rsum(f(x))
     n->infinity

= c*int(f)

So, we have arrived to the desired result. A proof of the property of
the limit used above is given in Theorem 3 in the following page.

Properties of Limits
http://www.geocities.com/pkving4math2tor1/1_lim_and_cont/1_01_02_prop_of_lim.html


b. Again, we will use the definition of integral here:

integral(f+c) from a to b

= lim           Rsum(f(x)+c)
  n->infinity
= lim           Rsum(f(x)) + Rsum(c)
  n->infinity

Now, it's clear from the definition of the Riemann sum that Rsum(c) =
c*(b-a). To see this, first note that the Rsum(c) = c*Rsum(1)
(property of sum). But Rsum(1) is simply the sum of the partitions of
the segment [a,b], which has "length" equal to b-a. Therefore,
Rsum(1)=b-a, no matter how many partitions (n) of the segment there
are. Finally, using another property of the limit (proof of which can
be found at the link provided above),

= lim           Rsum(f(x)) +  lim          Rsum(c)
  n->infinity                 n->infinity

= int(f) + c*(b-a)

Since int(f) exists (because f is integrable) then int(f) + c*(b-a)
(which is a constant) also exists, and therefore, int(f+c) is also
integrable.


Google search strategy:
properties limits
properties integrals
Riemann sum


I hope this helps. If you have any doubt regarding my answer, please
don't hesitate to request a clarification before rating it. Otherwise,
I await your rating and final comments.


Best regards,
elmarto