Find the arc length of the given curve between the specified points.

Question

Y= x^3/12 + 1/x from *1, 13/12) to (4, 67/12)

x = t x' = 1

y = t^3/12 + 1/t y'= x^2/4 - 1/x^2

integral sign upper bound 4 lower bound 1 sqrt( 1 + (x^2/4 - 1/x^2)^2

integral sign upper bound 4 lower bound 1 x^2/4 - 1/x^2 and the antiderivative of that is x^3/12 + 1/x

but that does not give the correct answer when evaluated out it gives 4.5 when the correct answer is 6

so why does the negative stay when the correct antiderivative does not have a negative?

Answer 1

∫₁⁴ √(1+(x²/4 - x^-2)²)dx

What do you get when you square the binomial under the radical?

x⁴/16 - 1/2 + x^-4

So under the radical, because 1 - 1/2 = 1/2):

1/2 + x⁴/16 +1/x⁴

At this point get a common denominator (16x⁴) and combine the numerators:

(x⁸+8x⁴+16)/16x⁴

The numerator can be written as a binomial squared: (x⁴+4)².

So after taking the square root:

∫₁⁴[(x⁴+4)/4x²]dx

So you are finding the antiderivative of: (1/4)x²+x^-2

F(x) =(1/12)x³- 1/x

F(4) = 64/12 - 1/4 = 64/12 - 3/12 = 61/12

F(1) = 1/12 - 12/12 = -11/12

F(4) - F(1) = 61/12 - (-11/12) = 6

Double check here:

desmos.com/calculator/tad7vyrcb2