Calculus II - Arc Length

Just wanted to verify my work and see if anything could be done more easily.

Find the exact arc length of the portion of the graph of $$y = (1/6)x^3 + 1/(2x) $$ from x = 1 to x=2

dy/dx = $$(x^2/2) - (1/(2x^2))$$

$$(dy/dx)^2 = x^4/4 + (1)/(4x^4) - 1/2 $$

$$(dy/dx)^2+1 = x^4/4 + (1)/(4x^4) + 1/2 $$

$$\int\limits_{1}^{2}\sqrt{x^4/4 + 1/(4x^4) + 1/2}$$

Arc Length = 17/12

Question

Calculus II - Arc Length

Just wanted to verify my work and see if anything could be done more easily.

Find the exact arc length of the portion of the graph of $$y = (1/6)x^3 + 1/(2x)  $$ from x = 1 to x=2

dy/dx = $$(x^2/2) - (1/(2x^2))$$

$$(dy/dx)^2 = x^4/4 + (1)/(4x^4) - 1/2 $$

$$(dy/dx)^2+1 = x^4/4 + (1)/(4x^4) + 1/2 $$

$$\int\limits_{1}^{2}\sqrt{x^4/4 + 1/(4x^4) + 1/2}$$

Arc Length = 17/12

anonymous · Answer

My biggest question is 1) Whether the answer is right and 2) Whether it is possible to get a perfect square from (dy/dx)^2 + 1, since the integral of that was an absolute nightmare.