Question

Approximate the arc length of f (x) on [a, b], using the approximation $\sum _{k=1}^n\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$with the given value of n. In each problem list the values of $\delta y_k$for role="math" localid="1649228183425" $k=0,1,2,3.....n$.

$f\left(x\right)=Inx,\left[a,b\right]=\left[1,3\right],n=4$

Short answer

The arc length is$\frac{\sqrt{13}+3\sqrt{5}}{4}$.

Step-by-step solution

Step 1. Given information .

Consider the given function$f\left(x\right)=Inx,\left[a,b\right]=\left[1,3\right],n=4$.

Step 2. Formula used .

Arc length of$f\left(x\right)=\sum _{k=1}^n\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$.

Step 3. Find the arc length .

$f\left(x\right)=\sum _{k=1}^n\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$

$$\begin{gathered} \delta x=\frac{b-a}{n}=\frac{1}{2} \\ {x}_0=a+k·\delta x=1,x_1=\frac{3}{2},x_2=2 \\ {x}_3=\frac{5}{2},x_4=3 \\ \delta y_1=f\left(x_1\right)-f\left(x_0\right)=\frac{3}{2} \\ \delta y_2=f\left(x_2\right)-f\left(x_1\right)=\frac{1}{2} \\ \delta y_3=f\left(x_3\right)-f\left(x_2\right)=\frac{1}{2} \\ \delta y_4=f\left(x_4\right)-f\left(x_3\right)=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} Arclength=\sqrt{1+{\left(\frac{\delta y_1}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_2}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_3}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_4}{\delta x}\right)}^2}·\delta x \\ =\frac{\sqrt{13}}{4}+\frac{\sqrt{5}}{4}+\frac{\sqrt{5}}{4}+\frac{\sqrt{5}}{4} \\ =\frac{\sqrt{13}+3\sqrt{5}}{4} \end{gathered}$$