Properties of the Riemann integral

By definition, the definite integral is essentially the signed area. From this fact, we can derive a series of properties of the Riemann integral. We also see that continuous functions are Riemann-integrable and prove the Fundamental Theorem of Calculus.

For $a<b<c$, we have

$$\begin{array}{}\text{(eq:intsum)}& {\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx.\end{array}$$

This means the sum of two areas is equal to the area of the combined region.

If $a<b$, we adopt the following convention:

$${\int }_b^{a}f(x)dx=-{\int }_a^{b}f(x)dx.$$

Then Eq. (eq:intsum) holds irrespective of the order of $a,b$ and $c$.

Furthermore, the following proposition should be trivial from the definition:

Theorem (Linearity of integral)

Let $f(x)$ and $g(x)$ be functions that are integrable on $[a,b]$. Let $k,l\in \mathbb{R}$ be constants. Then $kf(x)+lg(x)$ is also integrable on $[a,b]$ and

$${\int }_a^b(kf(x)+lg(x))dx=k{\int }_a^{b}f(x)dx+l{\int }_a^{b}g(x)dx.$$

In other words, the integral operation is linear.

Remark. Recalling that the Riemann integral is the limit of the sum of small rectangular areas, the above property is essentially the same as

$$\sum _i(k{a}_i+l{b}_i)=k\sum _i{a}_i+l\sum _i{b}_i.$$

□

As we have seen above, not all functions are integrable. However, continuous functions are always integrable.

Theorem (Continuous functions are integrable)

A function that is continuous on $[a,b]$ is integrable on $[a,b]$.

The proof is given later.

Theorem (Fundamental Theorem of Calculus)

Let $f(x)$ be a continuous function on an open interval $I$. Let $a\in I$ be a constant. For any $x\in I$, $f(x)$ is integrable either on $[a,x]$ (if $a<x$) or on $[x,a]$ (if $a>x$). Let us define the function $F(x)$ by

$$F(x)={\int }_a^{x}f(t)dt.$$

Then $F(x)$ is a differentiable function on $I$ and $F^{\prime }(x)=f(x)$.

Proof. It suffices to show the following equation

$$\begin{array}{}\text{(eq:fundcal)}& \underset{h\to 0}{lim}\frac{F(x+h)-F(x)}{h}=f(x).\end{array}$$

Note that

$$F(x+h)-F(x)={\int }_a^{x+h}f(t)dt-{\int }_a^{x}f(t)dt={\int }_x^{x+h}f(t)dt.$$

Let us define

$$\begin{array}{rcl}m& =& min\{f(t)\mid x\le t\le x+h\},\\ M& =& max\{f(t)\mid x\le t\le x+h\},\end{array}$$

and choose $S,s\in [x,x+h]$ such that

$$\begin{array}{rcl}f(s)& =& m,\\ f(S)& =& M,\end{array}$$

which is possible due to the Extreme Value Theorem. Then we have

$$mh\le {\int }_x^{x+h}f(t)dt\le Mh$$

so

$$f(s)\le \frac{F(x+h)-F(x)}{h}\le f(S).$$

As $h\to 0$, we have $s\to x$ and $S\to x$. Since $f(x)$ is continuous, $f(s)\to f(x)$ and $f(S)\to f(x)$ as $h\to 0$. By the Squeeze Theorem, Eq. (eq:fundcal) holds. ■