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.

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

Short answer

The arc length is $20.6825$.

The graph is sketched as below:

Step-by-step solution

Step 1. Given information .

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

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 .

$$\begin{gathered} f\left(x\right)=e^x,a=-1,b=3,n=4 \\ \partial x=\frac{b-a}{n}=\frac{3+1}{4}=1 \\ {x}_k=a+k\partial x \\ {x}_0=-1+0=-1,x_1=-1+1=0 \\ {x}_2=-1+2=1,x_3=-1+3=2 \\ {x}_4=-1+4=3 \\ \partial y_1=f\left(0\right)-f\left(1\right)=e^0-e^{-1}=1-0.368=0.632 \\ \partial y_2=f\left(1\right)-f\left(0\right)=e-1=2.718-1=1.718 \\ \partial y_3=f\left(2\right)-f\left(1\right)=e^2-e=4.6707 \\ \partial y_4=f\left(3\right)-f\left(2\right)=e^3-e^2=12.6964 \end{gathered}$$

$$\begin{gathered} Arclength=\left[\sqrt{1+0.399}+\sqrt{1+2.951}+\sqrt{1+21.8154}+\sqrt{1+161.198}\right]\times 1 \\ =1.1827+1.9877+4.7765+12.7856 \\ =20.6825 \end{gathered}$$

Step 4. Sketch of the function fx=ex .

The sketch of the given function is shown below .