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Simplifying Integration Question

Hey. :smile:

Using I to stand for the integral sign:

I s dx + I t dx = I (s+t) dx (where s and t are functions of x).

Why is the above statement true?
Reply 1
Avatar for nuodai
nuodai
17
What level are you working at? If it's A-level then just accept the answer "it just is"; if it's not, then you need to consider the definition of Riemann integrability. If L(f,D)L(f, D) and U(f,D)U(f, D) denote the lower and upper sums of ff on a partition DD respectively, then note that:

L(s,D)+L(t,D)≤L(s+t,D)≤U(s+t,D)≤U(s,D)+U(t,D)L(s, D) + L(t, D) \le L(s+t, D) \le U(s+t, D) \le U(s, D) + U(t, D)
(edited 13 years ago)
Reply 2
Avatar for Magu1re
Magu1re
OP
3
Original post by nuodai
What level are you working at? If it's A-level then just accept the answer "it just is"; if it's not, then you need to consider the definition of Riemann integrability. If L(f,D)L(f, D) and U(f,D)U(f, D) denote the lower and upper sums of ff on a partition DD respectively, then note that:

L(s,D)+L(t,D)≤L(s+t,D)≤U(s+t,D)≤U(s,D)+U(t,D)L(s, D) + L(t, D) \le L(s+t, D) \le U(s+t, D) \le U(s, D) + U(t, D)


A Level. It has come up in a STEP question. Working the othet way then I would say that would be acceptable, i.e. I (s+t) dx = I s dx + I t dx, because that is what I do when integrating. Integrate each term seperately just like I would do for differentiation. I suppose that tells me that the two must be equal but I was looking for a better explaination. However, it looks as though the proper explaination is too complex for me! haha
Reply 3
Avatar for DFranklin
DFranklin
18
At A-level, you should just assume that ∫λf(x)+μg(x) dx=λ∫f(x) dx+μ∫g(x) dx\int \lambda f(x) + \mu g(x) \,dx = \lambda \int f(x) \,dx + \mu \int g(x) \,dx (for any real numbers λ,μ\lambda, \mu).

Edit: Well, actually at *any* level, unless you're specifically asked to prove it.
(edited 13 years ago)

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