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 $y_k=0,1,2......n$.

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

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

Step 3. Find the arc length .

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

$$\begin{gathered} \delta x=\frac{2+2}{4}=1 \\ {x}_k=a+k·\delta x \\ {x}_0=-2+0·1=-2,x_1=-2+1=-1 \\ {x}_2=-2+2=0,x_4=-2+4=2 \\ \delta y_k=\left({\delta }_k-{\delta }_{k-1}\right) \\ \delta y_1=f\left(x_1\right)-f\left(x_0\right)=-1+8=7 \\ \delta y_2=f\left(x_2\right)-f\left(x_1\right)=0+1=1 \\ \delta y_3=f\left(x_3\right)-f\left(x_2\right)=1-0=1 \\ \delta y_4=f\left(x_4\right)-f\left(x_3\right)=8-1=7 \end{gathered}$$

Arc length $$\begin{gathered} ={\sqrt{1+\left(\frac{\delta y_1}{\delta x}\right)}}^2·\delta x+\sqrt{1+{\left(\frac{\delta y_2}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_3}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_4}{\delta x}\right)}^2}·\delta x \\ =\sqrt{1+{\left(7\right)}^2}+\sqrt{2}+\sqrt{2}+\sqrt{50} \\ =2\sqrt{50}+2\sqrt{2} \\ =2\left(\sqrt{50}+\sqrt{2}\right) \end{gathered}$$