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

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

Short answer

The arc length is$6+\frac{1}{e^2}-\frac{2}{e}+2e^2-2e-2e^3+e^4$.

Step-by-step solution

Step 1. Given information .

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

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 .

$f\left(x\right)=\sum _{k=1}^4\sqrt{1+{\left(\frac{\delta y_k}{\delta x}\right)}^2}·\delta x$

$$\begin{gathered} \delta x=\frac{b-a}{n}=\frac{4}{4}=1 \\ {x}_0=a+k·\delta x=-1 \\ {x}_1=-1+1=0,x_2=-1+2=1 \\ {x}_3=-1+3=2,x_4=-1+4=3 \\ \delta y_k=f\left(x_k\right)-f\left(x_{k-1}\right) \\ \delta y_1=1-\frac{1}{e},\delta y_2=e-1 \\ \delta y_3=e^2-e,\delta y_4=e^3-e^2 \end{gathered}$$

$$\begin{gathered} Arclength=\sum _{k=1}^n\sqrt{1+{\left(\frac{\delta y_1}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_2}{\delta x}\right)}^2}·\delta x+\sqrt{1+{\left(\frac{\delta y_3}{\delta x}\right)}^2}\delta x+\sqrt{1+{\left(\frac{\delta y_4}{\delta x}\right)}^2}·\delta x \\ =\sqrt{1+{\left(1-\frac{1}{e}\right)}^2}·1+{\sqrt{1+\left(e-1\right)}}^2·1+\sqrt{1+{\left(e^2-e\right)}^2}·1+\sqrt{1+{\left(e^3-e^2\right)}^2}·1 \\ =6+\frac{1}{e^2}-\frac{2}{e}+2e^2-2e-2e^3+e^4 \end{gathered}$$