I have little problems to solve this it would be great if some body could help me.
There are two parts to this proof
Part 1,
$\displaystyle \int [ f(x) + g(x) ] dx = \int f(x)dx + \int g(x) dx $
Part 2,
$\displaystyle \int af(x) dx = \int a f(x) dx $
Let us compute part 1 first,
$\displaystyle \int [ af + bg](x) dx = \int [af(x) + bg(x)] dx $
The above doesn't need to be proved, it is simple expansion.
Let $\displaystyle F(x) $ be the anti-derivative of $\displaystyle f(x) $ and let $\displaystyle G(x) $ be the anti-derivative of $\displaystyle g(x) $
In other words,
$\displaystyle F`(x) = f(x) $ and $\displaystyle G`(x) = g(x) $
Noting that $\displaystyle [F(x) + G(x)]` = F`(x) + G`(x) $
This is produced by both of the following,
$\displaystyle \int f(x)dx + \int g(x)dx = [F(x) + G(x)]` + c = F`(x) + G`(x) + c $
and
$\displaystyle \int [ f(x) + g(x) ]dx = F`(x) + G`(x) + c $
Thus,
$\displaystyle \int [ f(x) + g(x) ] = \int f(x)dx + \int g(x)dx $
But I didn't include the constants, here. So what we are left with from the question is
$\displaystyle \int af(x)dx + \int bg(x)dx $
Let $\displaystyle F(x) $ be the anti-derivative of $\displaystyle f(x) $
Then,
$\displaystyle [aF(x)]` = aF`(x) = af(x) $
So,
$\displaystyle \int af(x) dx= aF`(x) + c = a \int f(x) dx $