Second Fundamental Theorem of Calculus
Characterization of all antiderivatives
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Study Second Fundamental Theorem of Calculus
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 \(f=F'=G'\). \(\Box\)
Remark 4. It is very important to note that the statement of Theorem 3 is only valid for functions defined on intervals. For instance, consider the function \(f:\mathbb{R}\backslash\{0\}\to\mathbb{R}\) with \(f(x)=\frac1x\). An antiderivative is given by \(F:\mathbb{R}\backslash\{0\}\to\mathbb{R}\) with \[F(x)=\begin{cases}\log(x)&:x>0,\\\log(-x)&:x<0\end{cases}=\log(|x|).\] Another antiderivative is given by \[G(x)=\begin{cases}\log(x)&:x>0,\\\log(-x)+1&:x<0\end{cases}.\] The difference between \(G\) and \(F\) is given by \[G(x)-F(x)=\begin{cases}0&:x>0,\\1&:x<0\end{cases}\] and therefore not constant.
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