Calc Arc Length 0 to 8, sqrt(1+x)

[MathWarrior]

Find the arc length of [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP595419ebfac4cg3g1925000036iede4e50653655?MSPStoreType=image/gif&s=4&w=65&h=36 from 0 to 8Formula for Arc Length:
integral from a to b of sqrt(1+[f(x)]^2)
Attempt:

f'(x) = 2/3 * 3/2 * x ^(1/2)
f'(x) = x^(1/2)

integral from 0 to 8 of sqrt(1+[x^(1/2)]^2)

integral from 0 to 8 of sqrt(1+x)


After this point I find myself lost.

I know from searching online and such the next step should read:

2/3*(1+x)^(3/2) from 0 to 8

but I am unsure of how you get to this step? Keep in mind we have yet to learn trig substitution.

 

[lanedance]

Formula for Arc Length:
integral from a to b of sqrt(1+[f(x)]^2)

you mean
integral from a to b of sqrt(1+[f'(x)]^2)

 

[lanedance]

or
\int_b^a \sqrt(1+[f'(x)]^2) dx

 

[lanedance]

so reapeating your steps to check
f(x) = \frac{2}{3}x^{\frac{3}{2}}+1
f(x) = x^{\frac{1}{2}}

arc length is then
\int_b^a \sqrt(1+[fsingle-quote(x)]^2) dx
\int_0^8 \sqrt(1+(x^{\frac{1}{2}})^2) dx
\int_0^8 \sqrt(1+x) dx

which all looks good

so how about trying a variable change (translation) to integrate this?

 

[MathWarrior]

variable change (translation) to integrate this?

Would that be when you substitute?

u = 1+x
du/dx = 1

dx = du ?

integral from 0 to 8 of u^1/2 du?

then you get...

(u^3/2)/(3/2) du => 2/3*u^(3/2) => 2/3*(1+x)^(3/2)from 0 to 8

then you get

2/3*(1+8)^(3/2) =

2/3*9^(3/2) then the answer comes out to 18 which isn't right it supposedly 52/3 So how did i mess up?

 

[vela]

You forgot to use the lower limit.

 

[MathWarrior]

ty ^_^

 

[lanedance]

so you need to chnage the limits for the variable change