Expert: Alon Mandes - 10/9/2009

Question
Dear My Alon
Im sorry to disturb you but I had a calculus question that I needed a bit of help with. This is the question:
Calculate the arc-length for both of your two functions correct to three decimal places using the arc length formula (function 1: f(x)=6x(1-x)and function 2:f(x)=π/2(sin(xπ))

 
L= ∫ (1 + [[f'(x)]squared]])square root {sorry im not sure if you understand the formula ive put there but im not sure how to retype it)-the limits are 1 and 0

For your functions you must calculate the resulting integrals are far as you possibly can by using the exact integration techniques presented in this unit. If necessary, to obtain a decimal approximation, you may then use a technique such as Simpson’s Rule or the Trapezoidal Rule.

Thanks so much for your time=)
Sincerely Dil

Answer
Hello Dil,
The definition of the arc length is :
1
∫√[1+f'(x)²] dx . In our case f(x)=6x(1-x), therefore f'(x)=6-12x . Hence,
0          1
ArcLength=∫√[1+(6-12x)²] dx . This integral is not immediate, so wee need to calculate it
        0
using Numerical method such as "Trapezoidal Rule" . In this method we divide our integration
into n subsection, gaining Trapezoidal shape of the area . Let's suppose n=5, this gives us :
xo=0 , x1=1/5 , x2=2/5 , x3=3/5 , x4=4/5 , x5=1 . So, the area under the graph will be :
A=[(1-0)/(5*2)]*[f(xo)+2f(x1)+2f(x2)+2f(x3)+2f(x4)+f(x5) . ("Note that the area will gives us
the arclength"). So, let's get to work : In our case f(x)=√[1+(6-12x)²].
1
∫√[1+(6-12x)²] dx = [(1-0)/(5*2)]*[f(xo)+2f(x1)+2f(x2)+2f(x3)+2f(x4)+f(x5)]
0

=(1/10)*[f(0)+2f(1/5)+2f(2/5)+2f(3/5)+2f(4/5)+f(1)]
=(1/10)*[√{1+(6-12*0)²}+2√{1+(6-12/5)²}+2√{1+(6-24/5)²}+2√{1+(6-36/5)²}+2√{1+(6-48/5)²}+
 2√{1+(6-12)²}]
=(1/10)*[√{1+36}+2√{1+(18/5)²}+2√{1+(6/5)²}+2√{1+(-6/5)²}+2√{1+(-18/5)²}+2√{1+36}]
=(1/10)[2√37 + 4√{1+(18/5)²} + 4√{1+(36/25)}]
=(1/10)[2√{37} + 4√{43/25} + 4√{61/25}]
=(1/10)[2√{37} + (4/5)√{43} + (4/5)√{61}]
=(1/10)[2√{37} + (4/5)(√{43}+√{61})]
=2.365

I will leave to function 2 , to perform in the same manner as I did .
Best wishes,
Alon.