# Thread: Help on a Calc Problem? Find the length of the curve?

1. ## Help on a Calc Problem? Find the length of the curve?

Find the length of the curve, L.
y^2= 16(x+1)^3 where 0<=x<=2 and y>0
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Ok, so first I solved the equation for y
y= 4(x+1)^3/2

I know that the equation for the length of a curve is:
s= int (a to b) sqrt(1+ (f'(x))^2)

I then took the derivative of y= 4(x+1)^3/2 with respect to x and came up with:
6(x+1)^(1/2)

Then, I plugged what I knew into the formula...
s= int (0 to 2) sqrt( 1+ (6(x+1)^1/2)^2) dx and simplified...
s= int (0 to 2( sqrt( 1+36(x+1)) dx

Now I am stuck! If I tried distributing the 36 and combining it with the 1 but I get 36x+37 under the square root which I don't know how to simply to find its AntiD.

(The only problems like this I know how to work are the ones where you get the + or - 1/2 in the derivative and replace it with the difference/sum of the derivative squared. This one is different, unfortunately!) Oh, and Thanks very much in advance.

2. You are correct so far. Now all we need to do is perform a simple substitution:

$\int_0^2\sqrt{36x+37}\,dx=\frac{1}{36}\int_0^2 36\sqrt{36x+37}\,dx.$

Now we let $u=36x+37$ and the antiderivative of $36\sqrt{36x+37}$ follows at once.

Hope this helps!