Question

Approximate the arc length of f (x) on [a, b], using n line segments and the distance formula. Include a sketch of f (x) and the line segments .

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

Short answer

The arc length is 2.3 .

The graph is sketched as below:

Step-by-step solution

Step 1. Given information .

Consider the given function$f\left(x\right)=Inx$.

Step 2. Formula used .

Arc length$=\underset{n\to \infty }{\lim }\sum _{k=1}^n\sqrt{1+{\left(\frac{\partial y_k}{\partial x}\right)}^2}·\partial x$

Step 3. Find arc length .

$$\begin{gathered} f\left(x\right)=Inx,a=1,b=3 \\ \partial x=\frac{b-a}{n}=\frac{3-1}{4}=\frac{1}{2} \\ {x}_k=a+k\partial x \\ {x}_0=1,x_1=1+\frac{1}{2}=\frac{3}{2} \\ {x}_3=1+2\times \frac{1}{2}=2,x_4=1+4\times \frac{1}{2}=3 \\ \partial y_1=f\left(x_1\right)-f\left(x_0\right)=\ln \left(\frac{3}{2}\right)-\ln \left(1\right)=0·4055-0=0·4055 \\ \partial y_2=f\left(x_2\right)-f\left(x_1\right)=\ln \left(2\right)-\ln \left(\frac{3}{2}\right)=0·28768 \\ \partial y_3=f\left(x_3\right)-f\left(x_2\right)=\ln \left(\frac{5}{2}\right)-\ln \left(2\right)=0·22314 \\ \partial y_4=f\left(x_4\right)-f\left(x_3\right)=\ln \left(3\right)-\ln \left(\frac{5}{2}\right)=0·1823 \end{gathered}$$

Arc length$$\begin{gathered} =\left[\sqrt{1+{\left(\frac{0·405}{0·5}\right)}^2}+\sqrt{1+{\left(\frac{0·2876}{0·5}\right)}^2}+\sqrt{1+{\left(\frac{0·22314}{0·5}\right)}^2}+\sqrt{1+{\left(\frac{0·182}{0·5}\right)}^2}\right]\times 0·5 \\ =\left[\sqrt{1.6561}+\sqrt{1·3308}+\sqrt{1.199}+\sqrt{1.132}\right]\times 0·5 \\ =4.59\times 0.5 \\ =2.3 \end{gathered}$$

Step 4. Sketch of the function fx=In x .

The sketch of the given function is shown below .