Fundamental Theorem of Calculus: Part One

[in the rye]

I am a little confused over part 1 of the fundamental theorem of calculus. Part 2 makes perfect sense to me. I guess my confusion is if we have an integral g(x) defined from [a, b], and we are looking at point x, how do we know that g'(x) = f(x)? It makes sense in the idea that they are inverses. I understand that much. One "undoes" the other. I guess my confusion spawns from this idea that only the upper limit being a variable affects anything. It seems like the lower limit should be considered. Do we ignore it because it's a constant, so whenever we take the derivative it will just be 0? I don't really know where my confusion is. I think it's because the idea kind of seems so obvious that I'm over thinking it. That, and we covered Part 2 first, which seemed to kind of make Part 1 redundant since in Part 2 we were doing applications where if we had
G(x) = ∫ G'(x) dx, which seems to flow naturally that since an integral is an antiderivative, the inner function must be a derivative of the integral.

Also, in my book, they talk about using the chain rule if your upper limit is something like x². Is this because you have f(g(x)), where g(x) = x² so when you do part 1 of the fundamental theorem you have f'(g(x))g'(x)? The book doesn't explain this part. It just says to apply the chain rule and gives an anwser, but it's a tad confusing since we've never applied the chain rule to an integral before this.

 

[RUber]

By part one, are you referring to:
$g(x) = \int_a^x f(t) dt$
$g'(x) = f(x)$?

Let's call F the antiderivative of f, for now.
Then, I assume you know that
$\int_a^x f(t) dt = F(x)-F(a)$
using the properties of antiderivatives.
So, now bringing g(x) back,
$g(x)= F(x) - F(a)$
As you pointed out, when you take the derivative of this, the constant term will go to zero...
$g'(x) = F'(x)$ and you know that $F'(x) = f(x)$ .

As for if you have a function of x in the limits of integration...the process is the same.
$\int_a^{x^2} f(t) dt = F(x^2) - F(a)$
So its derivative would be (using the chain rule):
$F'(x^2)*2x = f(x^2)*2x$.

Does it make sense to look at it that way?

 

[in the rye]

By part one, are you referring to:
$g(x) = \int_a^x f(t) dt$
$g'(x) = f(x)$?

Let's call F the antiderivative of f, for now.
Then, I assume you know that
$\int_a^x f(t) dt = F(x)-F(a)$
using the properties of antiderivatives.
So, now bringing g(x) back,
$g(x)= F(x) - F(a)$
As you pointed out, when you take the derivative of this, the constant term will go to zero...
$g'(x) = F'(x)$ and you know that $F'(x) = f(x)$ .

As for if you have a function of x in the limits of integration...the process is the same.
$\int_a^{x^2} f(t) dt = F(x^2) - F(a)$
So its derivative would be (using the chain rule):
$F'(x^2)*2x = f(x^2)*2x$.

Does it make sense to look at it that way?

The second part does, and I follow you on the first part, at least symbolically. I guess I don't really see what's happening on the graph, though.

You have:
$g(x)= F(x) - F(a)$

So, presently g(x) is the distance between F(a) and F(x).

Then we take the derivative of that graph. So, we're looking at the slope of F(x) - F(a). Well, because 'a' is a constant, it drops out, and we're only concerned with the slope of F(x). Meaning that the slope of g'(x) = F'(x). Couldn't, for argument-sake, x = a? I guess it's confusing because if 'a' and 'x' are intervals on g'(x), why is the slope of 'a' always 0? I get that it's a constant, so-to-speak, but isn't it possible that our slope on the antiderivatives graph is greater than a? For example:
$g(x) = \int_a^x f(t) dt$
if f(t) = t², then g'(x) = x². This means g(x) = (1/3)x³ over the interval [a, x]. However, if we let a = 3, and x= 5, we know the slope at those points would be 9, and 25 by plugging them into g'(x), not 0.

I think this gets to the root of my confusion. It's obviously something I'm missing conceptually in thinking about it like this. Maybe because I'm thinking of 'a' as a variable is what's throwing me off... is that why?

 

[RUber]

I think I understand now.
The derivative of a constant (function) is 0 because the derivative is defined to be the instantaneous change in a function value caused by an infinitely small change in the variable.
It is sometimes written like: $\frac{d}{dx} f(x)$ indicating you are looking for the change in f caused by a change in x. If you were graphing y = f(x), this would be the same as the slope (change in y / change in x).
So...if you have f(a) as your function, you see it no longer depends on x at all, so changing x will not change the value of f(a), right?
That is not to say that the slope of f(x) at x = a is zero...it means that if you plotted y=f(a) for various x values, it would be a straight line.

In your example, $g(x) = \int_a^x t^2 dt$ you would get $g(a) = 0, g(x) = \frac{x^3}{3}$ just as you said. The slope of y=g(x) has absolutely no dependence on where a is, only where x is. You could have a = 0 or a = 2, but g'(3) is always going to be 9.

 

[RUber]

As an aside $\int_a^b f(x) dx$ is a constant function, since a and b are constants. So, the derivative of $\int_a^b f(t) dt$.is zero.

 

[RUber]

For example:
g(x)= \int_a^x f(t) dt
if f(t) = t², then g'(x) = x². This means g(x) = (1/3)x³ over the interval [a, x]. However, if we let a = 3, and x= 5, we know the slope at those points would be 9, and 25 by plugging them into g'(x), not 0.

I may not be relevant to the slope discussion, but this function would be: g(x) = (1/3)x³ - (1/3)a³. It is equivalent to the area under the curve f(t) = t^2 over the interval [a,x], but g(x), as a function should not be thought of in terms of an interval -- one x in gives one g(x) out.

 

[in the rye]

I think I understand now.
The derivative of a constant (function) is 0 because the derivative is defined to be the instantaneous change in a function value caused by an infinitely small change in the variable. It is sometimes written like: $\frac{d}{dx} f(x)$ indicating you are looking for the change in f caused by a change in x. If you were graphing y = f(x), this would be the same as the slope (change in y / change in x).

Right, I get this, but I think about it in terms of slope, which may be hurting me? I see it that since a constant is a horizontal line, the slope is then 0, making the derivative 0. So, I follow you here. I am just trying to bring the integral into the same elementary understanding so I get it conceptually.

So...if you have f(a) as your function, you see it no longer depends on x at all, so changing x will not change the value of f(a), right?

I see this symbolically since if I were to take the derivative of f(a), I would have f'(a), but then take the derivative of a, which would wipe out the derivative making it 0. I guess I don't see the second part where you say:

That is not to say that the slope of f(x) at x = a is zero...it means that if you plotted y=f(a) for various x values, it would be a straight line.

I don't understand why f(x) at x = a is different? While I understand that for various x values it wouldn't change, because as your x increase or decreases, 'a' is staying the same, but it seems to me that it still has SOME sort of slope based on my previous post.

In your example, $g(x) = \int_a^x t^2 dt$ you would get $g(a) = 0, g(x) = \frac{x^3}{3}$ just as you said. The slope of y=g(x) has absolutely no dependence on where a is, only where x is. You could have a = 0 or a = 2, but g'(3) is always going to be 9.

This is where I'm still lost. I don't see it. Because based on part 1 of the theorem g'(x) = x².Which means if x = a, you have g'(a) = x², so using a = 3, g'(3) = 9, not 0. Sorry, agh.

 

[PeroK]

Let's take an example. Let's look at the area under the curve $y = x$ from $x = 0$ to $x = t$

$A(t) = \frac{1}{2}t^2$ and $A'(t) = t$

Now, let's start somewhere else. Let's start at $x = 1$ (assuming $t > 1$)

$A(t) = \frac{1}{2}t^2 - \frac{1}{2}$ and $A'(t) = t$

Now, let's start at $x = 10$ (assuming t > 10)

$A(t) = \frac{1}{2}t^2 - 50$ and $A'(t) = t$

It makes no difference where you start. The rate of change of the area under the curve is the same at a given point: namely, the function value at that point.

 

[RUber]

I think this gets to the root of my confusion. It's obviously something I'm missing conceptually in thinking about it like this. Maybe because I'm thinking of 'a' as a variable is what's throwing me off... is that why?

Yes. a is not a variable.

This is where I'm still lost. I don't see it. Because based on part 1 of the theorem g'(x) = x².Which means if x = a, you have g'(a) = x², so using a = 3, g'(3) = 9, not 0. Sorry, agh.

Right. If you plot g(x)=x^2 , you would have a slope at x=a=3 of 9.
However, if you plotted g(a) over an interval [a,b], you would have a straight horizontal line because a is a constant, so g(a) is a constant.

I don't understand why f(x) at x = a is different? While I understand that for various x values it wouldn't change, because as your x increase or decreases, 'a' is staying the same, but it seems to me that it still has SOME sort of slope based on my previous post.

f(x) at x = a, means you are plotting a function that depends on x and looking at it at a specific point (x=a), this can be written as f(a).
The distinction is that a is constant, so if you plot f(a) for different values of x, it won't change. If you plot f(x) for different values of x, it will change. So f(x) can have a non-zero slope (as defined by $\frac{df}{dx}$ ) but f(a) cannot.

You might see something like, "find $f'(2) \text{if} f(x) = x^2$" This is asking you to find f', defined as $\frac{d}{dx}f(x)$ and evaluate it at 2, giving 4. It is not asking you to find $\frac{d}{dx}f(2)$ which would be zero.

 

[in the rye]

Yes. a is not a variable.

Right. If you plot g(x)=x^2 , you would have a slope at x=a=3 of 9.
However, if you plotted g(a) over an interval [a,b], you would have a straight horizontal line because a is a constant, so g(a) is a constant.f(x) at x = a, means you are plotting a function that depends on x and looking at it at a specific point (x=a), this can be written as f(a).
The distinction is that a is constant, so if you plot f(a) for different values of x, it won't change. If you plot f(x) for different values of x, it will change. So f(x) can have a non-zero slope (as defined by $\frac{df}{dx}$ ) but f(a) cannot.

You might see something like, "find $f'(2) \text{if} f(x) = x^2$" This is asking you to find f', defined as $\frac{d}{dx}f(x)$ and evaluate it at 2, giving 4. It is not asking you to find $\frac{d}{dx}f(2)$ which would be zero.

I think I understand. Essentially because 'a' is our starting value in the interval the area under the curve at 'a' is never changing, therefore must be 0. But, as we progress to 'x' is area under the curve is changing some rate which happens to be F'(x). It's just that whatever 'a' value you start at, the area will be changing at a rate of 0 because there is no change. If that's it, it couldn't have smacked me in the face any harder. It's one of those things that is so easy, it's hard. Ugh. It still feels a tad abstract, even though it's really not.

It's like were taking the integral from [a, a] and finding the rate of change with respect to x.

 

[RUber]

That's kind of right.
The area under the curve
$g(x) = \int_a^x f(t) dt$ is always going to be zero at x = a. Since you would have an interval width of 0.
And remember that
$g(x) = F(x) - F(a)$.
I think your confusion might have been in using the F' notation rather than d/dx, since F'(a) is often thought of as the slope of F'(x) at x =a. This, as we have discussed is not zero.
If you use d/dx, it is more clear.
$\frac{d}{dx} g(x) = \frac{d}{dx}F(x) - \frac{d}{dx}F(a)$.
In this case, it should be clear that F(a) does not play a role ... since it has no dependence on x.
$\frac{d}{dx}F(x) = f(x)$ and $\frac{d}{dx}F(a) = 0$.
So, using this definition, the slope of y=g(x) at x = a would be f(a).

 

[mathwonk]

given a (positive) graph y = f(x) and a fixed point a on the x axis, consider the "moving area function", i.e. A(x) = the portion of area under the graph of f and lying between the points a and x. As x moves to the right, this area increases, and the derivative A'(x) measures how fast it increases. Very roughly, this equals the amount of the increase in area when x moves one unit to the right.

Think of the case where the graph is a horizontal line, so we are taking the area of some rectangles/ But as x moves one unit to the right, the increase in area is that of a rectangle with base = one unit, and height = the height of the graph, namely the constant height f(x) for any x in the interval. Thus the derivative, or the area of that rectangle equals one tim es f(x) = f(x), the height.

If we have a more complicated graph, say increasing, the increase in area when the base increases one unit, from x to x+1, is not a rectangle, but sort of a triangle sloping up to the right. But it equals the area of some rectangle of base one, and whose height is that of a point on the graph somewhere maybe half way between x and x+1. Thus the increase in area would be the height of that rectngle, some number between f(x) and f(x+1).

Now to get the derivative we would make this measurement again for a smaller increase in the x value, say from x to x+1/2, and the derivative would be roughly the increase in area, divided by the base length which is 1/2. This area is again that of some rectangle with height between f(x) and f(x+1/2) multiplied by the base 1/2. so after we divide b y th base length, we just get the height again. So doing this over an over with shorter and shorter base lengths, we are always approximating the derivative as the heights of the graph at points getting closer to x. Thus in the limit the derivative is the limit of the values of f(x+h) as h-->0. Now if f is continuous this limit is f(x).Or just draw some pictures. Lok and se what you get when you increase the x value from x to x+h, take the corresponding increase in area, and divide by the base. You always get the average height over the interval [x,x+h]. As the interval gets smaller, i.e. as h-->0, that average approaches the height at the endpoint, if the height function is continuous.

 

[in the rye]

Thanks everyone. It clicked overnight. I appreciate it.