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 .

$f\left(x\right)=x^3,\left[a,b\right]=\left[-2,2\right],n=4$

Short answer

The arc length is$2\left(\sqrt{50}+\sqrt{2}\right)$.

Step-by-step solution

Step 1. Given information .

Consider the given function$f\left(x\right)=x^3$.

Step 2. Formula used .

The formula used to find the arc length of $f\left(x\right)$from $x=a$to $x=b$is ,

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

where$\delta x=\frac{b-a}{n},x_k=a+k\delta x,\delta y_k=f\left(x_k\right)-f\left(x_{k-1}\right)$for all$k=0,1,2.........n$.

Step 3. Find the arc .

$$\begin{gathered} f\left(x\right)=\underset{n\to \infty }{\lim }\sum _{k=1}^4\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x \\ \delta x=\frac{2+2}{4}=1 \\ {x}_k=a+k\delta x \\ {x}_1=-2+1=-1 \\ {x}_2=-2+2=0 \\ {x}_3=-2+3=1 \\ {x}_4=-2+4=2 \end{gathered}$$

Further simplify .

$$\begin{gathered} \delta y_1=f\left(x_1\right)-f\left(x_0\right)=-1+2=1 \\ \delta y_2=f\left(x_2\right)-f\left(x_1\right)=0-\left(-1\right)=1 \\ \delta y_3=f\left(x_3\right)-f\left(x_2\right)=1-0=1 \\ \delta y_4=f\left(x_4\right)-f\left(x_3\right)=2-1=1 \end{gathered}$$

Substitute the values in formula .

localid="1649238741200" role="math" $$\begin{gathered} Arclength=\sum _{k=1}^4\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x \\ =\sum _{k=1}^4\sqrt{1+{\left(\frac{\delta y_1}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_2}{\delta x}\right)}^2·}\delta x+\sqrt{1+{\left(\frac{\delta y_3}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_4}{\delta x}\right)}^2}·\delta x \\ =\sqrt{1+{\left(7\right)}^2}+\sqrt{1+1}+\sqrt{1+1}+\sqrt{1+{\left(7\right)}^2} \\ =\sqrt{50}+\sqrt{2}+\sqrt{2}+\sqrt{50} \\ =2\left(\sqrt{50}+\sqrt{2}\right) \end{gathered}$$

Step 4. Sketch the graph of function fx=x3 .

The sketch of the given function is shown below .