Extension: Formal Definition of Riemann Integral

Just a recap, we define

$$\int_a^bf(x)\,\text{d}x$$

to be the signed area bounded by the curve $f(x)$ and the $x$-axis on the interval $x\in[a,b]$.

As we've seen in the earlier articles involving Riemann sums, one way to approximate the area under a curve $f(x)$ on the interval $[a,b]$ is the left Riemann sum:

$$\int_a^bf(x)\,\text{d}x\approx\sum_{i=1}^n\left[\frac{b-a}{n}f\left(a+\frac{b-a}ni\right)\right]$$

Let's break this down without all the fancy sigmas and fractions. This approximation is just summing the area, $\text{height}\times\text{width}$, of $n$ rectangles of equal width on the interval $[a,b]$. Since there are $n$ rectangles of equal width on an interval of length $b-a$, the width of each rectangle is $\frac{b-a}n$. The first rectangle starts at $x=a$, the second rectangle is one width over, or $x=a+\frac{b-a}n$, the third is two widths over, or $x=a+2\frac{b-a}n$, etc, to the last rectangle being $n-1$ widths over, or $x=a+(n-1)\frac{b-a}n$. Since the height of each rectangle is just $f(x)$, we just sum up

$$\frac{b-a}nf(a)+\frac{b-a}nf\left(a+\frac{b-a}n\right)+\frac{b-a}nf\left(a+2\frac{b-a}n\right)+\dots\\\dots+\frac{b-a}nf\left(a+(n-1)\frac{b-a}n\right)$$

which is what the sum expresses.

Notice that as we pack more rectangles, the approximation becomes better and better. In fact, if we pack infinitely many rectangles, the approximation is exact. Therefore,

$$\int_a^bf(x)\,\text{d}x=\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\Delta x$$

where $x_i=a+\frac{b-a}ni$ and $\Delta x=\frac{b-a}n$.

As an example, to find the area under the curve $f(x)=x^2$ on the interval $[0,1]$, we can take $x_i=0+\frac{1-0}ni=\frac{i}n$, $\Delta x=\frac{1-0}n=\frac1n$, and

\begin{align*} \int_0^1x^2,\text{d}x&=\lim_{n\to\infty}\sum_{i=1}^n(x_i)^2\Delta x \ &=\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{i}n\right)^2\frac1n \ &=\lim_{n\to\infty}\frac1{n^3}\sum_{i=1}^ni^2 \ &=\lim_{n\to\infty}\frac1{n^3}\left(\frac{n(n+1)(2n+1)}6\right) \ &=\lim_{n\to\infty}\frac16\left(\frac{n}n\right)\left(\frac{n+1}n\right)\left(\frac{2n+1}n\right) \ &=\lim_{n\to\infty}\frac16\left(1+\frac1n\right)\left(2+\frac1n\right) \ &=\frac16\cdot1\cdot2=\frac13 \end{align*}

This may get the answer, but it's an awfully long solution. Try finding an easy method to integrate functions. How is it related to the derivative?

The answer lies in the fundamental theorem of calculus. We'll see more of that later.