1. Arc length problem

L= definite Integral a - b [sqrt(1+(dy/dx)^2)]dx

y - 2 = (x+2)^3/2 points (-2,2) to (2,10)
y=(x+2)^3/2 +2
y' = 3/2 (x+2)^1/2
1 + (y')^2 = 9x/4 +9/2 + 1 = (9/4)x +11/2

I am getting stuck when it comes to setting up the integral, i forgot the law is it correct to distribute the integral when adding ?

Def. Int. (-2 --> 2) of [sqrt((9/4)x +11/2)]dx =
[(3/2)x^3/2 +2.345x + C]-2 --> 2

but I really don't like that last term 2.345x, i think i did it wrong. If you can point me in the right direction it would be greatly appreciated.

2. Originally Posted by bgonzal8
L= definite Integral a - b [sqrt(1+(dy/dx)^2)]dx

y - 2 = (x+2)^3/2 points (-2,2) to (2,10)
y=(x+2)^3/2 +2
y' = 3/2 (x+2)^1/2
1 + (y')^2 = 9x/4 +9/2 + 1 = (9/4)x +11/2

I am getting stuck when it comes to setting up the integral, i forgot the law is it correct to distribute the integral when adding ? This is true, but you can't forget about the radical. It is Not OK to distribute the integral through that.

Def. Int. (-2 --> 2) of [sqrt((9/4)x +11/2)]dx =
[(3/2)x^3/2 +2.345x + C]-2 --> 2

but I really don't like that last term 2.345x, i think i did it wrong. If you can point me in the right direction it would be greatly appreciated.
Here's the set up...

$s=\int_{-2}^2\sqrt{1+\frac{9}{4}(x+2)}dx$

$=\int_{-2}^2\sqrt{\frac{9}{4}x+\frac{11}{2}}dx$

$du=\frac{9}{4}dx\Rightarrow{\frac{4}{9}{du}}=dx$

So we now have

$s=\frac{4}{9}\int_{1}^{10}\sqrt{u}du$.

Note the new limits of integration.

3. Thank you.