00:01
In order to solve this problem, i'm going to first derive a general result and then apply it to the problem.
00:10
So we're first told that if we have something that looks like this, so if we have a derivative of some function, and this integral here will just be a function of x, because we're integrating over the whole t variable, if we have something like this, this will be equal to assuming some right continuity conditions, which we can assume here.
00:49
We can move the derivative inside, but we have to make it into a partial derivative.
00:58
But this is only true if a and b are constants.
01:05
So for the more general case, imagine that we have something like f of x, is equal to the integral from v of x, so a function of x, to u of x, g of x and t, d t.
01:35
So this integral is still just a function of x because we're integrating over the t variable, but we have these, rather than just having constants, we have these functions as the limits of our integration.
01:46
So if we want to find, so if we want to find d by d x of f of x, is this equal to? well, the problem becomes much easier if we have something that resembles constants.
02:09
So we're going to make just a little substitution.
02:13
And what we're going to say is we're going to define a function that's g of, let's just call it u, v and x and this is just equal to v u and then our function so to make this a little bit clearer um maybe here we can change this to a of x and b of x so we don't get our u's and v is confused so b of x a of x so we've replaced we've replaced that uh u is equal to b of x and v is equal to a of x.
03:12
But in the scheme of this function here, u and v are just constants.
03:17
They're parameters that we put in to the function in order to work out the integral.
03:25
And then we should keep in mind that if we have g evaluated at a of x, b of x, and x, this is just equal to f of x.
03:43
So we're going to find the derivative of this with respect to x, and then we're going to see how it will reduce to finding this derivative.
03:55
So if we have d by dx of g, well, we're going to have to use the chain rule here.
04:06
So because g is a function of three variables, u, v, and x.
04:11
So this will be g of u, which just means the partial derivative of g with respect to u times d -u -d -x, and then minus g of v, d -v -d -x.
04:33
So i plus, so the partial of g with respect to v times dv -d -x, and then just plus the partial of g with respect to v times d -v -d -x, and then just plus the normal total derivative.
04:51
So then we need to work these things out.
04:59
So if we just look at this first term right here, because the other ones will follow in the same way.
05:08
So if we have d by d u of g, but g is just equal to v, u, gxt, d, d, t, then from just, calculus one, we know that this is simply equal to g evaluated at x, and then whatever the variable is here that we're integrating over, replace it with this variable.
05:46
So gxu.
05:53
So if we apply that to all three terms, we will end up with an expression that looks like this.
06:06
So so first it's g evaluated at x, u, and then d u, d x, and then in this term we're going to get a minus because the v is in the bottom of the integral, but for the rule that we just used to get this term, the u is in the top of the integral, but you can always exchange these to make this a a v up here and a u down here by just sticking a minus sign out in front.
06:45
So this is similarly gx, v, dv, dx.
06:54
And then, since u and v are acting like their constants, for this term here, it's just going to be like this rule up here.
07:05
So we can exchange the total derivative for a partial derivative and move it through the integral.
07:10
So it's going to be plus the integral from v to you of the partial of little g with respect to x, x, x, t, dt.
07:29
And so this is equal to our total derivative of big g.
07:38
And so this is an expression that we're going to use over and over again throughout this problem.
07:50
So in this problem, we're told that f of x is equal to the integral from x squared to 1 of the square root of t cubed plus x squared d t.
08:13
So this is our g of x t.
08:21
But notice that whereas we had functions, ax and bx, here we just have an x squared, which is a function, and then a constant, which is constant 1.
08:46
So for our u term that involves the x, this will be zero.
08:52
So this dg, d, u term is irrelevant.
08:58
So we only need to assign a function, a new function, big g, of x and v, not x, u and v, which makes our lives a little bit easier.
09:08
So we can define a function...