00:02
Okay, so we have y to be equal to 2 over 3 x over 2 plus 1.
00:06
So the first thing you can as usual is to find the first derivative of the function, which in this case instead of f of x prime, i'm just representing it as of y prime.
00:15
So if you do that you end up with y prime to be equal to 2 over 3 times 3 over 2 times x to the power 3 over 2 minus 1.
00:24
And then when you simplify, you end up with square root of x.
00:27
Now the square of that function is what x.
00:30
So the arc length, s, is giving us x equals integral from 0 to 1, square root of 1 plus x d x.
00:40
As it is the easiest way to integrate is to use substitution.
00:44
So what we do is we let u to be equal to 1 plus x.
00:48
Now the u, when you find the derivative of u with respect to x, we end up with the u dx to be equal to 1, which implies that the x is a same...
