Length of a curve || Deriving formula

Question

I'm currently watching a video, which is deriving the formula to compute the length of a curve. $$ \lim_{n\to\infty} \sum_{k=1}^n \sqrt{1+(f'(x_k))^2} \delta x $$ $$ \int_a^b \sqrt{1+(f'(x))^2} \ dx $$

The last two steps of the proof, are the steps I don't quite understand. The thing which has my head spinning is how it magically turned into a integral? Why was this done?

Answer 1

A definite integral $\int^b_a{f(x)\,dx}$ can be interpreted as the area under the curve $f(x)$ bounded by $x=a$ and $x=b$, as shown by the shaded area in the following diagram:

One way to approximate the area (and therefore the integral) is to split the area under the curve into $n$ thin rectangles of equal width $\delta x = \frac{b-a}{n}$, and sum up the area of them.

The height of the $k$th rectangle will have a height $x_k$, where (e.g.) $x_k=(k-1)\delta x$ (I've assumed that the top left corner of each rectangle lies on the curve. This is not important if $n$ is infinite.) Therefore, the area of the $k$th rectangle is $f(x_k)\,\delta x$. The sum of the area under the curve will be the sum of all the rectangular areas, i.e.:

$$\sum_{k=1}^n {f(x_k)\,\delta x}$$

This is only an approximation of the area under the curve. However, the approximation improves as the number of rectangles increase. In fact, it becomes exact as $n\rightarrow \infty$. Noting that the area under the curve is equal to the definite integral, it can be seen that:

$$\int^b_a{f(x)\,dx}=\lim_{x\rightarrow\infty}\sum_{k=1}^n {f(x_k)\,\delta x}$$

Substitute my $f(x)$ for your $\sqrt{1+\left(f'(x)\right)^2}$ and that should show how the last two steps were made.

Answer 2

You need to understand the definition of Riemann integral $$\int_{a}^{b} f(x)\,dx$$ properly. Without going into too many details, here is a brief exposition.

Let $f$ be defined and bounded on interval $[a, b] $. A set $P$ of the form $$P=\{x_{0},x_{1},x_{2},\dots,x_{n}\}$$ is called a partition of $[a, b]$ if $$a=x_{0}Riemann sum for $f$ over partition $P$. The difference $(x_{i} - x_{i-1})$ is also denote by $\Delta x_{i} $ or $\delta x_{i} $. Informally, if all Riemann sums of type $$S(f, P) =\sum_{i=1}^{n}f(t_{i})\Delta x_{i} $$ tend to the same limit $I$ as $n\to\infty$ then we say that Riemann integral $$\int_{a} ^{b} f(x) \, dx$$ exists and is equal to $I$.

A sufficient condition for existence of the Riemann integral is that the function $f$ under consideration be continuous on the interval $[a, b] $ under consideration. By definition we can see that if the Riemann integral $$\int_{a} ^{b} f(x) \, dx$$ exists then any specific Riemann sum tends to the value of the integral as $n\to\infty$.

In the current scenario we can see that the given sum is a Riemann sum for function $\sqrt{1+\{f'(x)\}^{2}}$ and hence it tends to the given integral as $n\to\infty$ provided the integral exists. Thus the sudden transformation from a sum to an integral is a direct consequence of the definition of integral and there is no magic here. Probably you don't see it that way because most introductory textbooks of calculus don't provide the definition of Riemann integral properly.