00:01
For this problem, we are to find the arc length of the curve y equals 4x squared minus 2 on the closed interval 0 to 3.
00:07
Now we have to recall that our arc length, l, is equal to the integral from a to b, of the square of 1 plus the square of f prime of x, dx.
00:18
Now, if our f of x equals 4x squared minus 2, then our f prime of x would be 8x, and the square of that, is equal to 64x squared.
00:35
So our arc length l is equal to the integral from 0 to 3 of this square root of 1 plus 64 x squared dx.
00:44
And then from here we have to do trigonometric substitution.
00:47
We want to let 8x equal to tangent theta so that 8 dx is equal to secant squared theta d theta.
01:00
So dx will be 1 8 secan squared theta d theta.
01:07
Now substituting what do we have here, we have l equal to the integral from 0 to 3 of the square root of 1 plus 64 times 1 over 8 tangent theta and then squared times dx which is 1 over 8 secant squared theta d theta.
01:27
That's integral from 0 to 3, the square root of 1 plus 2.
01:31
Tangent squared theta times 1 over 8, secrets squared theta d theta.
01:38
And because 1 plus tangent squared theta is secund squared theta, then you have 1 over 8 integral from 0 to 3 of secant theta times secant squared theta d theta.
01:49
Now we will have to use integration by parts here.
01:53
We want to let u equal to secant theta, dv equal to secan squared theta d theta.
02:01
So our du is equal to secant theta, tangent theta, d theta, while v is equal to tangent theta.
02:12
So from here we have this equal to 1 over 8 times u times v, that secant theta tangent theta, minus the interval of v times d u, that secant theta times tangent squared theta d theta...