Q. 75 - Calculus

Question

Consider that f(x) is a continuous function on an interval [a,b] and n is a positive integer. If the interval [a, b] is divided into n subintervals of equal width with the points of division represented by $$\begin{gathered} x_0=a,x_1,x_{2........}{x}_n=b and if △x=\frac{b-a}{n}, then prove that \\ \sum _{k=1}^{\infty }\sqrt{{\left(x_k-x_{k-1}\right)}^2+\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2}=\sum _{k=1}^n\sqrt{1+{\left(\frac{△y_k}{△x}\right)}^2△x} \end{gathered}$$

Step-by-Step Solution

Answer

$\sum _{k=1}^{\infty }\sqrt{{\left(x_k-x_{k-1}\right)}^2+\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2}=\sum _{k=1}^n\sqrt{1+{\left(\frac{△y_k}{△x}\right)}^2△x}$

Hence proved .

Step 1. Given

Consider that f(x) is a continuous function on an interval [a,b] and n is a positive integer. 

Step 2. Explanation

The function y = f(x) has a continuous graph on the interval [a, b] , which has been divided into subintervals with k ^th subdivision point as x_k for k=0,1,2,...,) . Thus the length triangle $△x$ of each subinterval is defined by$△x=\frac{b-a}{n}$  Now, if $x_{k-1} and x_k$are the end points of the k^th interval, then $△x=x_{k+1}-x_k$ and the corresponding values of y are f(x_k-1 ) and f(x₂) . So, the length of the line segment joining these points on the graph of the curve is given by the distance formula as

$\sqrt{{\left(x_k-x_{k-1}\right)}^2+\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2}$

The arc length of the function is then approximated by the sum of all such line segments. That is the arc length is approximated by the expression

$\sqrt{{\left(x_k-x_{k-1}\right)}^2+\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2}$

Replacing $x_{k+1}-x_k$the width of the k ^th subinterval triangle x and $\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2$ by the ordinate $△y_k$of the k^th subinterval, the above expression equals

$$\begin{gathered} \sum _{k=1}^{\infty }\sqrt{{\left(△x\right)}^2+{\left(△y_k\right)}^2} \\ S\mathrm{ince} \mathrm{triangle} △\mathrm{x} \mathrm{is} \mathrm{a} \mathrm{positive} \mathrm{number}, \mathrm{write} \mathrm{the} \mathrm{last} \mathrm{expression} \mathrm{as} \\ \sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{1+{\left(\frac{△{\mathrm{y}}_{\mathrm{k}}}{△\mathrm{x}}\right)}^2△\mathrm{x}} \\ \mathrm{Hence} \mathrm{Proved} \sum _{\mathrm{k}=1}^{\infty }\sqrt{{\left({\mathrm{x}}_{\mathrm{k}}-{\mathrm{x}}_{\mathrm{k}-1}\right)}^2+\left(\mathrm{f}({\mathrm{x}}_{\mathrm{k}}\right)-\mathrm{f}{\left({\mathrm{x}}_{\mathrm{k}-1}\right)}^2}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{1+{\left(\frac{△{\mathrm{y}}_{\mathrm{k}}}{△\mathrm{x}}\right)}^2△\mathrm{x}} \end{gathered}$$