Find the length of the specified arc of the given curve

[gigi9]

Calc help please!

Plz help me out w/ the problem below. Thanks.
Find the length of the specified arc of the given curve:
y=1/3sqrt(x)*(3-x), 0<=x <=3
Formular to find arc length is integral from a to b of sqrt(1+f'(x)^2)dx
I got to these steps below and not sure exactly wat to do next
***y'=1/6x^(-1/2)(3-3x)=1/2x^(-1/2)(1-x)
***V= integral from 0 to 3 of sqrt{1+[1/2x^(-1/2)*(1-x)])2} dx
=integral 0 to 3 of sqrt[1/2+1/4*x^-1+1/4*x] dx

 

[KLscilevothma]

V= integral from 0 to 3 of sqrt{1+[1/2x^(-1/2)*(1-x)])2} dx
=integral 0 to 3 of sqrt[1/2+1/4*x^-1+1/4*x] dx

1/(4x) + 1/2 + x/4 can be factorized to (1/2x^-1/2 + 1/2^1/2)²

 

[M98Ranger]

To find the length of the specified arc, we can use the formula given in the problem: integral from a to b of sqrt(1+f'(x)^2)dx. In this case, a=0 and b=3, so we have:

L = integral from 0 to 3 of sqrt(1+(1/2x^(-1/2)(1-x))^2)dx

Next, we need to simplify the expression inside the square root:

1+(1/2x^(-1/2)(1-x))^2 = 1+1/4x^-1+1/4x

Substituting this into the original equation, we have:

L = integral from 0 to 3 of sqrt(1/2+1/4x^-1+1/4x) dx

Now, we can use the power rule for integration to solve this integral:

L = integral from 0 to 3 of (1/2+1/4x^-1+1/4x)^(1/2) dx

= (1/2+1/4x^-1+1/4x)^(3/2) / (3/2) from 0 to 3

= (1/2+1/4*3^-1+1/4*3)^(3/2) / (3/2) - (1/2+1/4*0^-1+1/4*0)^(3/2) / (3/2)

= (1/2+1/4+3/4)^(3/2) / (3/2) - (1/2+0+0)^(3/2) / (3/2)

= (2)^(3/2) / (3/2)

= 2^(3/2) / 3

Therefore, the length of the specified arc is 2^(3/2) / 3 units.