Fundamental Theorem of Calculus...!?!?

[mooeypoo]

Hey guys,

 

Okay, so I had my summer utterly destroyed by a crashcourse in Calculus II (have to take it now so I can take Calculus III in the fall along with Physics and Engineering 203 which I wouldn't been able to take without it). The course is extremely fast moving, and I'm doing all the work and all that, but my teacher's running through material and I'm having some difficulty understanding a fundamental thing I thought, perhaps, maybe, pretty-please-oh-help-me-im-goin-insane, you bright fellows will help me out.

 

What the heck is it with the Fundamental Theorem of Calculus!?

 

I understand the second part of it (which is funny.. it's supposed to rely on the first part..) where, basically, as far as I got it, the integral of a function is equal to the antiderivative of that function.

 

We first learned that the "point" of integrals is to calculate area (technically), while negative area comes off of the positive one. So it's not precisely an area, but a "net change". Great. I got it. That's part 2 of the Theorem, and I understand that part.

 

I just don't get the first part -- I don't understand the element of replacing the "b" in the series [a,b] of an integral with an X, which is your function, while leaving a.. FUNCTION.. in the integral itself.. it's like a function of a function? wtf..?

 

It may be a silly question, I just have trouble grasping what it's good for, or what it means really, and what practical uses does it have. The book asks for technical "solve this using 1st part of the FTC" and I do it technically, just by memorizing what I need to replace and how, but I hate doing that, and I usually end up forgetting stuff if I don't understand what I'm doing.

 

I really appreciate anything you could help me with to wrap my head around this (without feeling too stupid )

 

Thanks!

 

~moo

[Royston]

We first learned that the "point" of integrals is to calculate area (technically), while negative area comes off of the positive one. So it's not precisely an area, but a "net change". Great. I got it. That's part 2 of the Theorem, and I understand that part.

 

Well you find areas through 'definite' integration, but I wouldn't say that was the 'point' of integration, there are a number of uses e.g take motion, differentation can derive position, which leads to velocity and from velocity you can derive acceleration, integration is the reverse of this, so it's useful due to the data you have available. You can use differential equations to model a whole myriad of physical processes, from modelling radioactive decay to population models.

 

It may be a silly question, I just have trouble grasping what it's good for, or what it means really, and what practical uses does it have.

 

See above

[mooeypoo]

Yes, I understand that - Derivative of position is speed, Derivative of Speed is acceleration. Antiderivative of acceleration is speed, and Anti derivative of speed is position.

 

But you do that with the SECOND part of the FTC.

 

Look here, for example: http://archives.math.utk.edu/visual.calculus/4/ftc.9/

 

What eludes me is not the ultimate goal of the FTC - I get that as I said above - I even understand how it works with net change of "areas".. I can't visualize the change from the f(x) function to the A(x) function (note the "x" is "instead of" the 'b' in the second part.

 

This, btw, is different than what my teacher taugh in the sense that the parts are reversed (pt 1 in the pic is 2 in my book, and vice versa). So in this picture, I just can't figure out how part 1 of the theorem - a function "within a function" operates, and what is it good for. Why is this needed...? What you explained, it seems to me, only has to do with the first part.

 

~moo

[Dave]

Okay, so if I'm understanding you correctly, you're unsure about this step:

 

If [math]A(x) = \int_a^x f(x) \, dx[/math] then [math]\frac{d}{dx} A(x) = f(x)[/math]

 

I think you may be getting a little confused with the notation that's being used. All this is saying is that if we evaluate an indefinite integral, then its derivative is the function being integrated. Let's do a simple example, using [math]f(x) = x^2[/math].

 

[math]A(x) = \int_a^x f(t) \, dt = \int_a^x t^2 \, dt = \frac{x^3}{3} - \frac{a^3}{3}[/math]

 

Notice that the value of a is some constant that we are free to pick (i.e. it defines a constant of integration). Hopefully this term is familiar. Now, when we differentiate A, we get:

 

[math]\frac{d}{dx} A(x) = \frac{d}{dx} \left( \frac{x^3}{3} - \frac{a^3}{3} \right) = \frac{d}{dx} \left( \frac{x^3}{3} \right) = x^2 = f(x)[/math]

 

which is precisely the result that we are guaranteed by the FTC.

 

I hope this helps you a bit!

[w=f[z]]

It may be helpful in the future to know that the fundamental theorem of calculus has a few 3-dimensional analogs:

 

Fundamental theorem for gradients:

[math]\int^{\vec{b}}_{\vec{a}}(\vec{\nabla}F)\cdot d\vec{l}=F(\vec{b})-F(\vec{a})[/math]

where the integral is a line integral.

 

Fundamental theorem for divergences (divergence theorem, Green's theorem, or Gauss' theorem):

[math]\int_{\upsilon}(\vec{\nabla} \cdot \vec{v}) d\tau=\oint_{\partial\upsilon}\vec{v}\cdot d\vec{a}[/math]

where the LHS is over a volume and the RHS is over the surface.

 

Fundamental theorem for curls (Stoke's theorem):

[math]\int_{\sigma}(\vec{\nabla} \times \vec{v}) \cdot d\vec{a}=\oint_{\partial\sigma}\vec{v}\cdot d\vec{l}[/math]

where the LHS is over the area and the RHS is over the perimeter.

 

 

 

I use the [math]\partial[/math] symbol in the sense to mean the boundary.

 

 

Compare those to the FTC:

[math]\int^b_a\frac{df}{dx}dx=f(b)-f(a)[/math]

 

Do you see the similarity?

 

The key feature these have in common with the FTC is that the RHS is over the boundary of that of the LHS - just like the FTC.

 

 

Cheers,

w=f[z]

[mooeypoo]

dear god i didn't start three-dimensional stuff... but.. uhh... i'll save it for the future.

 

Thanks

 

~moo

[Dave]

Don't worry about not getting the 3D stuff, it's not essential for you at this point. You need a bit more solid grounding in vector calculus to really get going with higher dimensional integrals.

[Jim Kata]

Just to be an even a bigger dick, the generalization of the fundamental theorem of calculus is known as "Stokes theorem". It says let [math]

\partial V

[/math] be the closed p-dimensional boundary of a (p+1) dimensional surface [math]

V

[/math]. Let [math]

\sigma

[/math] be a p-form defined throughout [math]

V

[/math]

Then

[math]

\int\limits_V {{\mathbf{d}}\sigma } = \int\limits_{\partial V} \sigma

[/math]

The integral of p-form [math]

\sigma

[/math] over the boundary [math]

\partial V[/math] equals integral of (p+1)-form [math]

{\mathbf{d}}\sigma

[/math] over the interior [math]V[/math]

 

Where [math]

\sigma = \frac{1}

{{p!}}\sigma _{i_1 i_2 \cdots i_p } (x^1 , \cdots ,x^n ){\mathbf{d}}x^{i_1 } \wedge \cdots \wedge {\mathbf{d}}x^{i_p }

[/math] and [math]

{\mathbf{d}}\sigma = \frac{1}

{{p!}}{\mathbf{d}}\sigma _{i_1 i_2 \cdots i_p } (x^1 , \cdots ,x^n ){\mathbf{d}}x^{i_1 } \wedge \cdots \wedge {\mathbf{d}}x^{i_p }

[/math]

with wedge operator [math]

\wedge

[/math] defined as [math]

{\mathbf{A}} \wedge {\mathbf{B}} = {\mathbf{A}} \otimes {\mathbf{B}} - {\mathbf{B}} \otimes {\mathbf{A}}

[/math] and [math]

\otimes

[/math] being the tensorial product of the two spaces. The fundametal theorem of Calculus that you are learning about is a corollary of Stoke's law. Namely: the integral of a gradient [math]

{\mathbf{d}}f

[/math] along a curve, [math]

P(x)

[/math] from [math]

P(a)

[/math] to [math]P(b)[/math] is:

 

[math]

\int {{\mathbf{d}}f = \int\limits_a^b {\left\langle {{\mathbf{d}}f,dP/dx} \right\rangle dx = \int\limits_{P(a)}^{P(b)} {\frac{{df}}

{{dx}}dx = f(P(b)) - f(P(a))} } }

[/math] This law is also called the fundamental law of line integrals

 

Where in your case the line is just the x-axis

[BenTheMan]

Sounds like someone has been fooling around with Eguchi, Gilkey and Hanson. (P.S.---If you haven't, then you should!!!)

[mooeypoo]

You guys are mean. I'm never asking anything again.

 

 

 

 

 

 

 

Thanks btw

 

~moo

[Jim Kata]

Alright, let me legitimately try to answer your question. So you know that the integral [math]\int\limits_a^b {f'(s)ds = } f(b) - f(a)[/math]. Yes? You know that the the derivative of a constant is always zero right? [math]\frac{d}{{dx}}C = 0[/math]. So if you are doing an indefinite integral you can only determine the antiderivative up to constant. That is all [math]

f(x) + C [/math] will yield the same derivative, namely [math]f'(x)[/math]. Now pertaining to your question, another way you can write [math]

\int\limits_a^x {f'(s)ds} = f(x) - f(a)[/math](1), although less rigorous, is [math]f(x)=\int {f'(x)dx}+C[/math]. This is an indefinite integral, there are no definite bounds of integration. Now like I said, equation (1) is the same thing as the integral I just wrote. How? The reason is because [math]\int\limits_a^x {f'(s)ds = } f(x) - f(a)[/math] so [math]f(x) = \int\limits_a^x {f'(s)ds} + f(a)[/math]. The reason mathematicians like to write it like equation (1) is because it already satisfies initial conditions, in that [math]C[/math] is already chosen. As far as your question about replacing [math]b[/math] with [math]x[/math], the answer is quite simple. When you write [math]b[/math] you are talking about a specific number namely [math]b[/math]. When I replace [math]b[/math] with [math]x[/math] I am talking about any possible real number. Think about it the same way as you did for definite integrals, but just now [math]b[/math] is not given to you it can be any real number, [math]x[/math].

 

You said you wanted to see some examples heres a few, sorry I only know Physics:

 

A basic one [math]a = -g[/math] where [math]g = 9.8\frac{m}{{s^2 }}[/math] so [math]\int\limits_{v_0 }^v {dv} = - \int\limits_0^t g{dt}[/math] [math]v - v_0 = - gt - g(0)[/math] which implies [math]

v = - gt + v_0 [/math]. See how in this method I didn't have to plug in initial values in order to find [math]C[/math] it is done by my limits of integration on my velocity integral. You could integrate this again [math]\int\limits_{x_0 }^x v = \int\limits_0^t {( - gt + v_0 )dt}[/math] which gives you [math]x = - \frac{1}{2}gt^2 + v_0 t + x_0[/math]. The equation for the free fall motion of an object.

 

Here's a more elaborate integral:

 

[math]

\vec E(\vec r,t) = \frac{1}

{{4\pi \varepsilon _0 }}\left[ {\frac{{\rho (\vec r',t_r )}}

{{\left| {\vec r - \vec r'} \right|^3 }}(\vec r - \vec r') + \frac{{\dot \rho (\vec r',t_r )}}

{{c\left| {\vec r - \vec r'} \right|^2 }}(\vec r - \vec r') - \frac{{\vec \dot J(\vec r',t_r )}}

{{c^2 \left| {\vec r - \vec r'} \right|}}} \right]dv'

[/math] Solving this gives the general solution to any electric field. There's tons and tons more examples of how integrals are used, butj you should read up on it on your own.

[jcarlson]

Yes, I understand that - Derivative of position is speed, Derivative of Speed is acceleration. Antiderivative of acceleration is speed, and Anti derivative of speed is position.

 

But you do that with the SECOND part of the FTC.

 

Look here, for example: http://archives.math.utk.edu/visual.calculus/4/ftc.9/

 

What eludes me is not the ultimate goal of the FTC - I get that as I said above - I even understand how it works with net change of "areas".. I can't visualize the change from the f(x) function to the A(x) function (note the "x" is "instead of" the 'b' in the second part.

 

This, btw, is different than what my teacher taugh in the sense that the parts are reversed (pt 1 in the pic is 2 in my book, and vice versa). So in this picture, I just can't figure out how part 1 of the theorem - a function "within a function" operates, and what is it good for. Why is this needed...? What you explained, it seems to me, only has to do with the first part.

 

~moo

 

Ok.... so you are confused about the equation

 

[math]A(x) = \int_a^x f(t)dt[/math]

 

and how it implies

 

[math]A'(x) = f(x)[/math]

 

right?

 

Ok, this simply follows from the first equation on the page you gave (which you understand, right?)

 

Let [math]F[/math] be the anti-derivative of [math]f[/math], then:

 

[math]A(x) = \int_a^x f(t)dt = F(x) - F(a)[/math]

 

now, since [math]F(a)[/math] is constant, and its derivative is therefore 0:

 

[math]A'(x) = F'(x) - 0[/math]

 

But since [math]F(x)[/math] is the anti-derivative of [math]f(x)[/math], [math]F'(x) = f(x)[/math], so:

 

[math]A'(x) = f(x)[/math]

 

 

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