Make a list of all the formal properties of area that you can think of. Here’s an example: area is translation-invariant, i.e., if you slide a region around the plane, its area does not change.
Section 7.2 Properties of the Definite Integral
Remembering that the integral is supposed to represent an area, let’s show that integrals act like areas.
Checkpoint 7.2.1.
Theorem 7.2.2.
If \(f,g:[a,b]\to \mathbb{R}\) are integrable, so is \(f+g\text{.}\) Moreover,
\begin{equation*}
\displaystyle\int_{[a,b]}f(x)+g(x)\ dx=\int_{[a,b]}f(x)\ dx + \int_{[a,b]}g(x)\ dx\ \ \ .
\end{equation*}
Proof.
Consider any partition \(P=\{x_0,\ldots,x_n\}\text{,}\) consider
\begin{equation*}
\displaystyle U(f+g;P)=\sum_{k=0}^{n-1}\sup_{[x_k,x_{k+1}]}(f+g) \left(x_{k+1}-x_k\right)\leq \sum_{k=0}^{n-1}\left(\sup_{[x_k,x_{k+1}]}f+\sup_{[x_k,x_{k+1}]}g\right) \left(x_{k+1}-x_k\right)=U(f;P)+U(g;P)\ \ \ .
\end{equation*}
Therefore, \(\displaystyle\overline{\int_{[a,b]}(f+g)(x)\ dx \leq U(f;P)+U(g;P)}\) for any partition.
Now, given \(\epsilon\gt 0\text{,}\) there are partitions \(P_1\) so that \(\displaystyle U(f;P_1)\lt \overline{\int_{[a,b]}f(x)\ dx}+\frac{\epsilon}{2}\text{,}\) and \(P_2\) so that \(\displaystyle U(g;P_1)\lt \overline{\int_{[a,b]}g(x)\ dx}+\frac{\epsilon}{2}\text{.}\) Setting \(P_\epsilon=P_1\cup P_2\text{,}\) we have that \(P_\epsilon\) refines both \(P_1\) and \(P_2\text{.}\) Therefore,
\begin{equation*}
\displaystyle\overline{\int_{[a,b]}(f+g)(x)\ dx}\leq U(f;P_\epsilon)+U(g;P_\epsilon)\leq U(f;P_1)+U(g;P_2)\lt \overline{\int_{[a,b]}f(x)\ dx}+\frac{\epsilon}{2}+\overline{\int_{[a,b]}g(x)\ dx} +\frac{\epsilon}{2}\ \ .
\end{equation*}
Since \(\epsilon\) is arbitrary, this means
\begin{equation*}
\overline{\int_{[a,b]}(f+g)(x)\ dx}\leq \overline{\int_{[a,b]}f(x)\ dx}+\overline{\int_{[a,b]}g(x)\ dx}\ \ \ .
\end{equation*}
A similar argument shows that
\begin{equation*}
\underline{\int_{[a,b]}(f+g)(x)\ dx}\geq \underline{\int_{[a,b]}f(x)\ dx}+\underline{\int_{[a,b]}g(x)\ dx}\ \ \ .
\end{equation*}
Now, because \(f\) and \(g\) are integrable, we can replace their integrals superior and inferior with their respective integrals to obtain
\begin{equation*}
\displaystyle\int_{[a,b]}f(x)\ dx + \int_{[a,b]}g(x)\ dx \leq \underline{\int_{[a,b]}(f+g)(x)\ dx}\leq \overline{\int_{[a,b]}g(x)\ dx}\leq \int_{[a,b]}f(x)\ dx+\int_{[a,b]}g(x)\ dx
\end{equation*}
which finishes the proof.
In the proof of Theorem 7.2.2, we brushed up against this result, which will be pretty handy going forward:
Proposition 7.2.3. Cauchy Criterion for Integrability.
\(f:[a,b]\to\mathbb{R}\) is integrable if and only if for each \(\epsilon\) there exists a partition \(P_\epsilon\) so that \(U(f;P_\epsilon)-L(f;P_\epsilon)\lt \epsilon\text{.}\)
Theorem 7.2.4.
If \(f\) is integrable on \([a,b]\text{,}\) and \(c\in(a,b)\) then \(f\) is integrable on \([a,c]\) and \([c,b]\text{;}\) moreover,
\begin{equation*}
\displaystyle\int_{[a,b]}f(x)\ dx=\int_{[a,c]}f(x)\ dx + \int_{[c,b]}f(x)\ dx \ \ \ .
\end{equation*}
Proof.
First, we’ll prove that \(f\) is integrable on \([a,c]\) and \([c,b]\text{.}\) Given \(\epsilon\gt 0\text{,}\) by the Cauchy Criterion, there is \(P_\epsilon\) so that \(U(f;P_\epsilon)-L(f;P_\epsilon)\lt \epsilon\text{.}\) Consider \(P=P_\epsilon\cup\{c\}\text{,}\) which naturally decomposes into a partition \(P'\) of \([a,c]\) and a partition \(P''\) of \([c,b]\text{.}\) We have
\begin{equation*}
U(f;P')-L(f;P')\leq U(f;P)-L(f;P)\leq U(f;P_\epsilon)-L(f;P_\epsilon)\lt\epsilon
\end{equation*}
and similarly,
\begin{equation*}
U(f;P'')-L(f;P'')\leq U(f;P)-L(f;P)\leq U(f;P_\epsilon)-L(f;P_\epsilon)\lt\epsilon\ \ ,
\end{equation*}
so that by the Cauchy Criterion, \(f\) is integrable on each of \([a,c]\) and \([c,b]\text{.}\)
Now let’s establish the claimed formula for \(\displaystyle\int_{[a,b]}f(x)\ dx\text{.}\)
Let \(\epsilon\gt 0\text{.}\) By the epsilon criterion, there are partitions \(P_{\epsilon,+}',P_{\epsilon,-}'\) of \([a,c]\) and \(P_{\epsilon,+}'',P_{\epsilon,-}''\) of \([c,b]\) so that
\begin{gather*}
\displaystyle\underline{\int_{[a,c]}f(x)\ dx}-\frac{\epsilon}{2}\lt L(f;P_{\epsilon,-}')\leq \underline{\int_{[a,c]}f(x)\ dx}\\
\displaystyle\overline{\int_{[a,c]}f(x)\ dx} \leq\displaystyle U(f;P_{\epsilon,+}')\lt \overline{\int_{[a,c]}f(x)\ dx}+\frac{\epsilon}{2}\\
\displaystyle\underline{\int_{[c,b]}f(x)\ dx}-\frac{\epsilon}{2}\lt L(f;P_{\epsilon,-}'')\leq \underline{\int_{[c,b]}f(x)\ dx}\\
\displaystyle\overline{\int_{[c,b]}f(x)\ dx} \leq\displaystyle U(f;P_{\epsilon,+}'')\lt \overline{\int_{[c,b]}f(x)\ dx}+\frac{\epsilon}{2}
\end{gather*}
Setting \(P_\epsilon'=P_{\epsilon,-}'\cup P_{\epsilon,+}'\) and \(P_\epsilon''=P_{\epsilon,-}''\cup P_{\epsilon,+}''\text{,}\) observe that \(P_\epsilon=P_\epsilon'\cup P_\epsilon''\) is a partition of \([a,b]\) so that \(U(f;P_\epsilon)=U(f;P_\epsilon')+U(f;P_\epsilon'')\) and \(L(f;P_\epsilon)=L(f;P_\epsilon')+L(f;P_\epsilon'')\text{.}\)
Therefore,
\begin{equation*}
\displaystyle L(f;P_\epsilon)-\epsilon \leq \int_{[a,c]}f(x)\ dx + \int_{[c,b]}f(x)\ dx \leq U(f;P_\epsilon)+\epsilon
\end{equation*}
which gives us the desired formula.
Theorem 7.2.5.
If \(f:[a,b]\to \mathbb{R}\) is integrable, then so is \(\lvert f\rvert:[a,b]\to \mathbb{R}\text{;}\) moreover,
\begin{equation*}
\displaystyle\left\lvert \int_{[a,b]}f(x)\ dx\right\rvert\leq \int_{[a,b]}\lvert f(x)\rvert\ dx\ \ .
\end{equation*}
Proof.
Define the positive part and negative part of \(f\) by
\begin{align*}
f_+(x)=\begin{cases}f(x)& \text{ if }f(x)\gt 0\\0&\text{ if }f(x)\leq 0\end{cases}\\
f_-(x)=\begin{cases}0& \text{ if }f(x)\gt 0\\-f(x)&\text{ if }f(x)\leq 0\end{cases}
\end{align*}
Notice that \(f(x)=f_+(x)-f_-(x)\) and \(\lvert f(x)\rvert=f_+(x)+f_-(x)\text{.}\)
Lemma 7.2.6.
Proof.
This follows from the Cauchy Criterion and the fact that for any interval, \(\displaystyle \sup_I f_+ - \inf_I f_+\leq \sup_I f-\inf_I f\text{.}\)
The claimed inequality follows by noting that \(f_-\) is nonnegative.
Checkpoint 7.2.7.
What result about the real numbers is the inequality in Theorem 7.2.5 a version of?
