How to Prove a Function is Riemann Integrable and Compute its Integral

Proving that a function is Riemann integrable on a specific interval involves several steps. Here, we will go through the process with a polynomial function example and another more complex example involving rational and irrational numbers. Let's start with a polynomial function $f(x) 2x$ on the interval $[0, 1]$.

Proving the Function is Riemann Integrable

Step 1: Show that the function is continuous. Polynomial functions, like $f(x) 2x$, are continuous everywhere, implying that $f(x)$ is continuous on the closed interval $[0, 1]$. Therefore, by the definition of Riemann integrability, $f(x)$ is Riemann integrable on $[0, 1]$.

Step 2: Use the definition of the Riemann integral. The Riemann integral of a function $f$ over the interval $[a, b]$ is expressed as the limit of Riemann sums:

$$int_{a}^{b} f(x) dx lim_{n to infty} sum_{i1}^{n} f(x_i^*) Delta x$$

where $Delta x frac{b-a}{n}$ and $x_i^*$ is a sample point in the $i$-th subinterval $[x_{i-1}, x_i]$.

Step 3: Compute the integral using the Fundamental Theorem of Calculus. We find the antiderivative of $f(x) 2x$:

$$F(x) int 2x dx x^2 C$$

For the definite integral from $0$ to $1$:

$$int_{0}^{1} 2x dx F(1) - F(0)$$

Calculating $F(1)$ and $F(0)$:

$$F(1) 1^2 1, F(0) 0^2 0$$

Thus, we have:

$$int_{0}^{1} 2x dx 1 - 0 1$$

Proving the Riemann Integrability and Integral of a More Complex Function

Consider the function $f(x)$ defined as $f_0 f_1 1$ and $f(m/n) frac{1}{n}$ where $m/n$ is a rational number with $m, n geq 1$ and $f(x) 0$ if $x$ is an irrational number. Let the interval be $[a, b]$ as any closed interval.

We show that $f$ is integrable on $[a, b]$ with zero integral.

Let $epsilon > 0$. Choose $n$ such that $frac{1}{n} . Suppose the set $A { k/n : k in mathbb{Z}, 1 leq n, k, n }$ has $N$ elements. Choose $delta > 0$ such that $Ndelta and choose a partition $P { x_0, x_1, x_2, ldots, x_m }$ of $[a, b]$ with $|P| . Let $C_i in [x_{i-1}, x_i]$ for $i 1, 2, 3, ldots, m$. Note that $f(C_i) frac{1}{n}$ if $[x_{i-1}, x_i]$ does not intersect $A$ and $f(C_i) 1$ otherwise. It may happen that $C_i C_{i-1} x_i$ or $C_i C_{i 1} x_{i 1}$, thus there are at most $2N$ sub-intervals of $P$ containing elements of $A$.

Consider the Riemann sum $S sum_{i1}^{m} f(C_i) Delta x_i$. We divide the Riemann sum into two groups: one where the intervals contain $A$ and the other where intervals do not intersect $A$. We have:

$$S sum_{i1}^{m} f(C_i) Delta x_i sum_{[x_{i-1}, x_i] text{ contains } A} f(C_i) Delta x_i sum_{[x_{i-1}, x_i] text{ does not intersect } A} f(C_i) Delta x_i$$

The first sum satisfies:

$$sum_{[x_{i-1}, x_i] text{ contains } A} f(C_i) Delta x_i leq N cdot frac{1}{n} cdot frac{b-a}{n}

The second sum satisfies:

$$sum_{[x_{i-1}, x_i] text{ does not intersect } A} f(C_i) Delta x_i leq 2N cdot 1 cdot delta

Therefore:

$$S

Hence, $f$ is integrable with integral 0.