Riemann Integrability, Linear Transformations

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Homework Statement



If f,g are Riemann integrable on [a,b], then for c,d real numbers,

(let I denote the integral from a to b)

I (cf + dg) = c I (f) + d I (g)


Homework Equations





The Attempt at a Solution



I have the proofs for

c I(f) = I (cf)

and

I (f+g) = I (f) + I(g)

I'm wondering how would I put them together to prove the statement?
Can I just say that because I know these 2 propositions are true, then any linear combination would also work?
 
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Should be easy to put them together.

I(cf + cg) = I(cf) + I(dg) (since you know I(f+g) = I(f) + I(g), just replace f with cf and g with dg)
= cI(f) + dI(g) (since you know I(cf) = I(f))

I think the only problem with this is that you have to prove that if f is Riemann integrable then so is cf, which shouldn't be hard.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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