00:01
Here, we're given some values of f and its derivatives, and we know that it's continuous, and we want to find out the integral from 1 to 4 of x times f double prime of x.
00:14
Now, this looks to me like a prime opportunity to do differentiation by parts.
00:20
So we need to find a u and a v.
00:22
So i like to write it out like this, u, du, and dv.
00:28
So we need to take out our integral and break it up in terms of a u and a dv.
00:33
U needs to be something we can take the derivative, and v, dv needs to be something we can take the integral.
00:40
So really, we have no choice here because we can't take the derivative of f.
00:43
We know nothing about the third derivative.
00:46
So let's just say dv is f double prime of x, dx, which means u has to be x.
00:51
And that actually works out quite well because then du is one.
00:55
Now we take the integral of v and we get f prime of x.
00:59
So this integral will be u, v, evaluated at.
01:05
1 to 4 minus the integral of vdu, which is f prime of x d x.
01:17
And sorry, i should have written s is 1dx here...