# Proving linearity property of the integral.

#### chirkowatson

I have little problems to solve this it would be great if some body could help me.

Last edited:

#### HallsofIvy

MHF Helper
I have little problems to solve this it would be great if some body could help me.
What do you have to work with? Do you have any "rules" of integration or are you to prove this from the Riemann sums definition?

#### AllanCuz

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,

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