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Given is the function $\Bbb{1}_{\{1\}}(x)$. Prove that this funtion has no antiderivative on [0,2].
You may use that any differentiable function is continuous and that all antiderivatives of the zero-function are constant.

It has been a while for me doing this kind of mathematics. Can someone help me prove the statement?

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    Suppose it has an antiderivative $F$. In particular $F$ must be continuous. Moreover it must be constant on $[0,1)$ and on $(1, 2]$ ($F'=0$ there). putting together these two things, necessarily $F$ must be constant. But then $F'=0 \neq 1_{\{ 1 \}}$. – 2017-02-05
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    what about an antiderivative of the following form: $F = c,$ if $\; x\in [0,1) \cup (1, 2]$ and $F = x\; $if$ x = 1$. With c = 1. Or does that make no sense at all? – 2017-02-05
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    Wouldn't my example of an $F$ be a horizontal line, which is continuous? – 2017-02-05
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    But taking c = 1, => F(x) = 1 for x != 1, and when x = 1, F(x) = 1 because F(x) = x at that point. Hence, for all values of x, F(x) has a value of 1. Right? – 2017-02-05
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    Sorry, I misread your functino. Yes, your function is simply the constant $1$, whose derivative is indeed $0$ everywhere. – 2017-02-05

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