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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.

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.

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:

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.

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 \(\lvert f(x)\rvert=f_+(x)+f_-(x)\text{.}\)

This follows from the Cauchy Criterion and the fact that for any interval, \(\displaystyle \sup_I f_+ - \inf_I f_+\leq \sup_I f-\inf_I f\text{.}\)

Now because \(f_+\) and \(f_-\) are integrable, their sum \(\lvert f\rvert\) is also.

The claimed inequality follows by noting that \(f_-\) is nonnegative.

What result about the real numbers is the inequality in Theorem 7.2.5 a version of?