Q. 5 - Calculus

Question

How is the Mean Value Theorem involved in proving that the arc length of a function on an interval can be represented by a definite integral? 

Step-by-Step Solution

Answer

Arc length is $\underset{n\to \infty }{\lim }\sum _{k=1}^n\sqrt{1+{\left(f\text{'}\left(x_k^{*}\right)\right)}^2}∆x={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx$

Step 1. Given Information.

The mean value theorem states that,

If $f\left(x\right)$ is differential and continuous at $\left[a,b\right]$, then there is c such as $a<c<b$and

$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Step 2. Explanation.

To find arc length using Mean value theorem. 

If $f\left(x\right)$ is a differentiable function, then the Mean Value Theorem applies to the function on each subinterval $[x_{k-1}, x_k]$ then, there exists some point $x_k^{*}\in \left(x_{k-1},x_k\right)$

such that,

$f\text{'}\left(x_k^{*}\right)=\frac{f\left(x_k\right)-f\left(x_{k-1}\right)}{x_k-x_{k-1}}=\frac{∆y_k}{∆x}$

so, 

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\sqrt{1+{\left(\frac{∆y_k}{∆x}\right)}^2}∆x=\underset{n\to \infty }{\lim }\sum _{k=1}^n\sqrt{1+{\left(f\text{'}\left(x_k^{*}\right)\right)}^2}∆x$

As the derivative $f\text{'}\left(x\right)$ is assumed to be continuous, then  function is also continue.

therefore, the limit of sums represents the definite integral of $\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}$ on interval $\left[a,b\right]$.

so that,

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\sqrt{1+{\left(f\text{'}\left(x_k^{*}\right)\right)}^2}∆x={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx$