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I have in general issues understanding the transition of sums from in to out( or from out to in) of integrals. Like in this example:

$$\sum^{n}_{k=0}\int^{1}_{0}(-t)^{k}f(t)dt =\int^{1}_{0}\sum^{n}_{k=0}(-t)^{k}f(t)dt $$

Why is this possible? It does seem to me that it fundamentally changes the part that needs to be integrated, yet it still holds true.

And when can't I do this? What's the general rule for such manipulations?

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    Because the integral is linear. :-) – 2017-01-02
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    In this case the reason you can interchange the order of summation and integration is there are finite terms and you can apply the linear property of integration. But if it is an infinite sum, then you need to be careful. – 2017-01-02
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    @HengrongDu Oh I see, now it got clearer to me. Post that as an answer and I'll accept it. – 2017-01-02

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In your case we have a finite number of terms in the summation. This means that the integral can distribute across the sum in the sense that the integral operator is linear. For an operator to be linear two properties must hold (where $c$ is constant and $a,b$ are not necessarily constant): $$f(a+b) = f(a) + f(b)\\ f(cx)=cf(x)$$ That first property is the one we want - clearly we can can write $$f(a+b+c) =f((a+b) +c) = f(a+b) +f(c) = f(a) + f(b) + f(c) $$ We can do this process on as many finitely summed terms as we want. This is what is happening when you interchange an integral and a finite summation.

If we have an infinite sum, then we can apply Tonelli's Theorem . Tonelli's theorem says if $f_n(x) \ge 0$ for all $n,x$, then $$\sum \int f_n(x) dx = \int \sum f_n(x) dx$$ without any further conditions needed (where the integral is a Lebesgue Integral)