A certain length of the curve is known as the arc length. Basically, it is the distance between two points that exists on the curve. You can determine the arc length by using the technique of integration. As integration is itself a tricky procedure to carry out, an online arc length calculator by is there for you people. This free length of arc calculator not only determines the arc length but also many other geometrical parameters. 


Here in this article, I will be discussing how you can enumerate the arc length of a function.

Let’s dive in!

Proper Way To Find Arc Length Of A Function:

Now suppose you have the following function:


$$ y = f\left(x\right) $$


Now what you want here is to find the arc length of this function within the interval [a, b]. Is it? Okay let’s move ahead!

Determine The Length Of The Curve:

First of all, you need to find the length of the curve. For this particular purpose, you must divide the curve into equal subintervals, each one consisting of length \(\deltax\). Here, the length of each segment is given as follows:


$$ \left| {{P_{i – 1}}\,\,{P_i}} \right| $$

And the length of the curve will be:


$$ L \approx \sum\limits_{i = 1}^n {\left| {{P_{i – 1}}\,\,{P_i}} \right|} $$


If we take larger and larger value of n, then you will get the exact length which can also be determined with the help of arc length calculator and is as below:


$$ L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left| {{P_{i – 1}}\,\,{P_i}} \right|} $$


This is the exact length.

Get A Better Grip On Line Segments:

Here you need to define the each segment by the following equation:


$$ \Delta {y_i} = {y_i} – {y_{i – 1}} = f\left( {{x_i}} \right) – f\left( {{x_{i – 1}}} \right) $$


From this, you can directly compute the values of lines as follows:


$$ \left| {{P_{i – 1}}\,\,{P_i}} \right| = \sqrt {{{\left( {{x_i} – {x_{i – 1}}} \right)}^2} + {{\left( {{y_i} – {y_{i – 1}}} \right)}^2}} $$

$$ = \sqrt {\Delta {x^2} + \Delta y_i^2} $$

Apply Mean Value Theorem:

By mean value theorem, we have:


$$ \begin{align*}f\left( {{x_i}} \right) – f\left( {{x_{i – 1}}} \right) & = f’\left( {x_i^*} \right)\left( {{x_i} – {x_{i – 1}}} \right)\\ & \Delta {y_i} = f’\left( {x_i^*} \right)\Delta x\end{align*} $$


Now the length can be written as below as well:


$$ \begin{align*}\left| {{P_{i – 1}}\,\,{P_i}} \right| & = \sqrt {{{\left( {{x_i} – {x_{i – 1}}} \right)}^2} + {{\left( {{y_i} – {y_{i – 1}}} \right)}^2}} \\ &  = \sqrt {\Delta {x^2} + {{\left[ {f’\left( {x_i^*} \right)} \right]}^2}\Delta {x^2}} \\ &  = \sqrt {1 + {{\left[ {f’\left( {x_i^*} \right)} \right]}^2}} \,\,\,\Delta x\end{align*} $$


In case of any difficulty, you can su=bject to a free arc length calculator for accurate outputs.

Using Definite Integrals:

Now the use of a definite integral makes it nothing but;


$$ L = \int_{{\,a}}^{{\,b}}{{\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx}} $$


$$ L = \int_{{\,c}}^{{\,d}}{{\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy}} $$


The free arc length formula calculator automatically determines these particular parameters without subjecting you towards any difficulty.

Final Formulas:

The arc length formulas for any curve represented by a certain function is as follows:


$$ L = \int{{ds}} $$




$$ \begin{array}{*{20}{l}}\begin{align*}ds & = \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx\,\hspace{0.25in}{\mbox{if }}y = f\left( x \right),\,\,a \le x \le b\\ ds & = \sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy\,\hspace{0.25in}{\mbox{if }}x = h\left( y \right),\,\,c \le y \le d\end{align*}\end{array} $$

It seems a little bit difficult but you can make it happen for you with the help of a free arclength calculator.


Find the length of the function below:


$$ y = \ln \left( {\sec x} \right) \hspace{0.25in} 0 \le x \le \frac{\pi }{4} $$


Well, a fast way to resolve the arc length of this function is the use of an arc length calculator. But you also need to have hands on grip on manual calculations.

As the given function is in the form of f(x), you need to find its derivative first:


$$ \frac{{dy}}{{dx}} = \frac{{\sec x\tan x}}{{\sec x}} = \tan x $$ 


$$ {\left( {\frac{{dy}}{{dx}}} \right)^2} = {\tan ^2}x $$


Now we have:


$$ \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} $$

$$ = \sqrt {1 + {{\tan }^2}x} $$

$$ = \sqrt {{{\sec }^2}x} $$

$$ = \left| {\sec x} \right| $$

$$ = \sec x $$


The arc length is given as:


$$ \begin{align*}L & = \int_{{\,0}}^{{\,\frac{\pi }{4}}}{{\sec x\,dx}}\\ &  = \left. {\ln \left| {\sec x + \tan x} \right|} \right|_0^{\frac{\pi }{4}}\\ &  = \ln \left( {\sqrt 2  + 1} \right)\end{align*} $$


In this article, I thoroughly discussed how to find the arc length of a function and the importance of a free arc length calculator in this regard.

By Adam

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