Integration : Find the Arc Length over an Interval
October 6th, 2022
Find the arc length of the graph of the function over the indicated interval:
y=1/6 x^3 + 1/(2x^2), [1,3]
I know S = Intergral( sqr( 1 + [f'(x)]^2 )) dx from 1 to 3.
I get y’ = [ 1/9 x^4 – 1/3 + 1/(4x^4) ] dx
Therefore,
S = Intergal( sqr( 1 + 1/9 x^4 – 1/3 + 1/(4x^4) )) dx from 1 to 3
= Intergal( sqr( 2/3 + 1/9 x^4 + 1/(4x^4) )) dx from 1 to 3
= Intergal( sqr( 2/3 ) + 1/3 x^2 + 1/(2x^2) ) dx from 1 to 3
Taking the Intergral:
S = [ sqr(2/3) x + 1/9 x^3 + ???.