What is the arc length of #f(x)= x ^ 3 / 6 + 1 / (2x) # on #x in [1,3]#?

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What is the arc length of #f(x)= x ^ 3 / 6 + 1 / (2x) # on #x in [1,3]#?

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Answer 1 · 2015-11-11T21:21:29

#14/3 approx 4.667#

Explanation:

If #f : x mapsto x^3/6 + 1/(2x)#, the length is
#L=int_1^3 sqrt(1+f'(x)^2) dx#.

Because #f'(x) = x^2/2 - 1/(2x^2) = (x^4-1)/(2x^2)#, you have
#f'(x)^2+1 = (x^8-2x^4+1)/(4x^4) + 1 = (x^8+2x^4+1)/(4x^4) = ((x^4+1)/(2x^2))^2#,
so, you can write
#L=int_1^3 (x^4+1)/(2x^2) dx = int_1^3 (x^2/2 + 1/(2x^2)) dx#
#L = [x^3/6 - 1/(2x)]_1^3 = 14/3#

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What is the arc length of #f(x)= x ^ 3 / 6 + 1 / (2x) # on #x in [1,3]#?

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