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    Hey.

    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?
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    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) and U(f, D) denote the lower and upper sums of f on a partition D respectively, then note that:

    L(s, D) + L(t, D) \le L(s+t, D) \le U(s+t, D) \le U(s, D) + U(t, D)
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    (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) and U(f, D) denote the lower and upper sums of f on a partition D respectively, then note that:

    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
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    At A-level, you should just assume that \int \lambda f(x) + \mu g(x) \,dx = \lambda \int f(x) \,dx + \mu \int g(x) \,dx (for any real numbers \lambda, \mu).

    Edit: Well, actually at *any* level, unless you're specifically asked to prove it.
 
 
 
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