find the arc length of the graph of the function over the interval [1, 3]?

Question

y= [(1/6)x^3] + (1/2x)

Answer 1

The previous poster got it wrong. You need to take a derivative before you square.

For arc length, you integrate
sqrt[1+(y')^2]
between the x-values given.
In this case, the derivative is
(1/2)x^2 -(1/2)x^(-2).
If you square this and add one, you will find the
result to be a perfect square of
(1/2)x^2 +(1/2)x^(-2).
So you can take the square root and integrate the result from 1 to 3.
The anti-derivative is
(1/6)x^3 - 1/(2x),
so you evaluate this at 3 and subtract what you get when
you evaluate it at 1.

Answer 2

Arc-length = int [1+(1/6 x^3+1/2x)^2]^(1/2) [integral is evaluated from 1 to 3]
Complete the square and then integrate.

Answer 3

You may use the formula int(sqr(1+(y')^2)) from values 1 to 3 where y' is 1/2(x^2+1) .

Answer 4

Is that something that is really too hard for you to do yourself? Don't you have the formula?

Answer 5

L = ?sqrt(one million+(dx/dy)²)*dy dx/dy = one million/3 * (one million*?y + (y-3)/(2?y)) = one million/3 * (3y-3)/(2?y) = (y-one million)/(2?y) one million+(dx/dy)² = one million+(y²-2y+one million)/(4y) = (y²+2y+one million)/4y sqrt(one million+(dx/dy)²) = (y+one million)/(2?y) L = one million/2*?(?y + one million/?y)*dy = one million/2*[2/3*(a hundred and forty four^one million.5-one million) + 2*(?a hundred and forty four-one million)] = one million/2*[2/3*(12³-one million)+22] = (12³-one million)/3+11 = 1760/3