You are correct so far. Now all we need to do is perform a simple substitution:
Now we let and the antiderivative of follows at once.
Hope this helps!
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.