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,

\(\displaystyle \int af(x) dx= aF`(x) + c = a \int f(x) dx \)