How do you find the length of a curve in calculus?

In Cartesian coordinates for y = f(x) defined on interval [a,b] the length of the curve is
${\displaystyle \Rightarrow L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}$
In general, we could just write:
${\displaystyle \Rightarrow L={\int }_a^{b}ds}$Solution:
Let's use Cartesian coordinates for this explanation.
If we consider an arbitrary curve defined as ${\displaystyle y=f\left(x\right)}$ and are interested in the interval ${\displaystyle x\in [a,b]}$, we can approximate the length of the curve using very tiny line segments.
Consider a point on the curve ${\displaystyle P_i}$. We can compute the distance of a line segment by finding the difference between two consecutive points on the line ${\displaystyle |P_i-P_{i-1}|}$ for ${\displaystyle i\in [1,n]}$ where n is the number of points we've defined on the curve.
This means that the approximate total length of curve is simply a sum of all of these line segments:
${\displaystyle L\approx \sum _{i=1}^n|P_i-P_{i-1}|}$
If we want the exact length of the curve, then we can make the assumption that all of the points are infinitesimally separated. We now take the limit of our sum as ${\displaystyle n\to \infty }$.
${\displaystyle L=\underset{n\to \infty }{lim}\sum _{i=1}^n|P_i-P_{i-1}|}$
Since we are working in the xy-plane, we can redefine our distance between points to take on the typical definition of Euclidean distance.
${\displaystyle |P_i-P_{i-1}|=\sqrt{{(y_i-y_{i-1})}^2+{(x_i-x_{i-1})}^2}=\sqrt{\delta y^2+\delta x^2}}$
We can now apply the Mean Value Theorem, which states there exists a point ${\displaystyle x_i^{\prime }}$ lying in the interval ${\displaystyle [x_{i-1},x_i]}$ such that
${\displaystyle \Rightarrow f\left(x_i\right)-f\left(x_{i-1}\right)=f\prime \left(x_i^{\prime }\right)(x_i-x_{i-1})}$
which we could also write (in the same notation) as
${\displaystyle \Rightarrow \delta y=f\prime \left(x_i^{\prime }\right)\delta x}$
This means that we now have
${\displaystyle |P_i-P_{i-1}|=\sqrt{{[f\prime \left(x_i^{\prime }\right)\delta x]}^2+\delta x^2}}$
Simplifying this expression a bit gives us
${\displaystyle |P_i-P_{i-1}|=\sqrt{{[f\prime \left(x_i^{\prime }\right)]}^2\delta x^2+\delta x^2}}$
${\displaystyle |P_i-P_{i-1}|=\sqrt{({[f\prime \left(x_i^{\prime }\right)]}^2+1)\delta x^2}}$
${\displaystyle |P_i-P_{i-1}|=\sqrt{(1+{[f\prime \left(x_i^{\prime }\right)]}^2)}\delta x}$
We can now use this new distance definition in our summation for our points.
${\displaystyle L=\underset{n\to \infty }{lim}\sum _{i=1}^n\sqrt{(1+{[f\prime \left(x_i^{\prime }\right)]}^2)}\delta x}$
Sums are nice, but integrals are better in continuous situations! Because integrals and sums are both "summation" tools, it's simple to write this as a definite integral. We can also remove our sum index from the integral.
${\displaystyle L={\int }_a^b\sqrt{(1+{[f\prime \left(x\right)]}^2)}\delta x}$
Writing this a little bit more typically yields
${\displaystyle L={\int }_a^b\sqrt{(1+{\left(\frac{dy}{dx}\right)}^2)}dx}$
We have arrived at our result! In general, the length is usually defined for a differential of arclength ds
${\displaystyle L={\int }_a^{b}ds}$
where ds is defined accordingly for whatever type of coordinate system you are working in. However, I wanted the explanation to be clearer, so I just chose Cartesian ones for simplicity. You could also use polar or spherical coordinates by simply making the necessary substitutions.