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**Definition 1**. *Let \(I\) be an interval and \(f:I\to\mathbb{R}\) be continuous. Then a
differentiable function \(F:I
\to\mathbb{R}\) is called an **antiderivative of \(f\)* if \(F'=f\).

**Theorem 2**. *Let \(I\) be an interval, \(f:I\to\mathbb{R}\) be continuous and \(a\in I\). For \(x\in I\) define \[F(x)=\int_a^xf(\xi)d\xi.\] Then \(F\) is differentiable and an antiderivative
of \(f\).*

*Proof:* Let \(x\in
I\) and \(h\neq0\) such that
\(x+h\in I\). Then, by using the mean
value theorem of integration we obtain \[\begin{aligned}
\frac1h
\left(F(x+h)-F(x)\right)=&\;\frac1h\left(\int_a^{x+h}f(\xi)d\xi-\int_a^xf(\xi)d\xi\right)\\
=&\frac1h\int_x^{x+h}f(\xi)d\xi=\frac1h\cdot h f(\hat{x})=f(\hat{x})
\end{aligned}\] for some \(\hat{x}\) between \(x\) and \(x+h\). If \(h\) tends to \(0\) then \(\hat{x}\to x\) and thus \[\lim_{h\to0}\frac1h
\left(F(x+h)-F(x)\right)=f(x).\] This shows the desired
result.\(\Box\)

Now we consider how two antiderivatives of a given continuous \(f:I\to\mathbb{R}\) differ.

**Theorem 3**. *Let \(I\) be an interval, \(f:I\to\mathbb{R}\) be given and let \(F:I\to\mathbb{R}\) be an antiderivative of
\(f\), i.e., \(F'=f\). Then \(G:I\to\mathbb{R}\) is an antiderivative of
\(f\) if and only if \(F-G\) is constant.*

*Proof:* “\(\Rightarrow\)”: Let \(G:I\to\mathbb{R}\) be an antiderivative of
\(f\). Then \[(F-G)'=F'-G'=f-f=0\] and
hence, \(F-G\) is constant due to the
mean value theorem of differentiation.

“\(\Leftarrow\)”: If \(F-G\) is constant, i.e., \(F(x)-G(x)=c\) for some \(c\in\mathbb{R}\) and all \(x\in I\), then \(0=(F-G)'=F'-G'\) and thus \(f=F'=G'\). \(\Box\)

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