integrating and differentiating

[between3and26characterslon]

Could someone explain, in as simple and non technical terms as possible, what integrating and differentiating are. Is there a simple everyday example where these two things are used. I've had a look in various places and they all tell you how to do it but not what it is you are actually doing i.e. the penny hasn't dropped for me yet.

 

Your help would be much appreciated.

[Bignose]

So, you have a curve. Differentiation is an operation/calculation to find the rate of change of that curve. The curve's rise over run, is a good way to think about it at the beginning.

 

Integration is just the inverse operation of differentiation. That is, given a rate of change of a curve -- given a rise over run -- what is the curve?

 

The best example is probably position, velocity, and acceleration.

 

Let a point move on the x-axis as a function of time. Let that point's position be called X(t), and let's further assume that the form of this function is known (i.e. X(t) = sin(t) or X(t) = t + 4.5, the exact form is not needed now).

 

The velocity of that point at any period of time is the derivative of that function with respect to time. That is, the rate of change of position with respect to time. If you think about it, that is exactly what velocity is -- Z m/s means that one's position is changing Z meters every second. Let the velocity be called V(t), and denoting that it is equal to the derivative with respect to time of position is written [math]V(t) = \frac{dX(t)}{dt}[/math].

 

Acceleration and velocity have the same relationship velocity and position do. That is, acceleration is the change in velocity with respect to time. Thusly, acceleration, let is be called A(t) follows this equation: [math]A(t) = \frac{dV(t)}{dt}[/math]. But, since we already know that velocity is a derivative with respect to position, we can go ahead of show how acceleration is related to position directly:

 

[math]A(t) = \frac{d}{dt}\frac{dX(t)}{dt}[/math]. That is, the derivative of the derivative of position as a function of time. This is often written more succinctly as [math]A(t) = \frac{d^2X(t)}{dt^2}[/math], and is also called the second derivative.

 

Now, as above, integration is just the inverse operation of differentiation. So, it would be used to find velocities if the acceleration is known, and to find position if the velocity is known. That is, if you are only given the velocity V(t), you can integrate that function to find X(t). There is one caveat to this however, and that is that integration only finds the answer to an additive constant. That is, if X(t) = 5t is a valid position given the velocity, then X(t) = 5t - 60 and X(t) = 5t + 100 are also both valid. This is generally written X(t) = 5t + C, where C is any constant. In order to lock down the exact value of C, you need to be given an additional piece of information, such as the position at time 0 was 4 --> X(0) = 4. With that, you can find the exact value of C. Why this is so is probably above your current level of understanding; without going into more details, I'm going to ask that you just accept that this is the way these operations work.

[Cap'n Refsmmat]

You can also find what I hope is an intuitive explanation of the meaning of differentiation here:

 

http://www.scienceforums.net/topic/29473-introduction-to-calculus-differentiation/

[DrRocket]

Could someone explain, in as simple and non technical terms as possible, what integrating and differentiating are. Is there a simple everyday example where these two things are used. I've had a look in various places and they all tell you how to do it but not what it is you are actually doing i.e. the penny hasn't dropped for me yet.

 

Your help would be much appreciated.

 

At any given point the derivative of a function is the slope of a line tangent to the curve at that point.

 

 

If a function is non-negative the integral of that function over the interval is the area between the graph of the function and the x-axis. Chemists have been known to integrate a curve by cutting it out of a piece of graph paper and weighing it on a sensitive laboratory balance.

 

If the function takes on positive and negative values, areas above the x-axis are positive and areas below the x-axis are negative.

 

Derivatives and integals are connected by the Fundamental Theorem of calculus, which relates integrals directly to "ant-derivatives" (if f is the derivative of g then g is an anti-derivative of f).

 

 

There are much more sophisticated and more general ways to approach differential and integral calculus, but this is the basic idea.

[mississippichem]

Derivatives and integals are connected by the Fundamental Theorem of calculus, which relates integrals directly to "ant-derivatives" (if f is the derivative of g then g is an anti-derivative of f).

 

I think many students get confused by the calculus notation. So I decided to post the fundamental theorem of calculus symbolically and explain the symbols:

 

[math] \int \frac{d f(x)}{dx} dx = f(x) + C [/math]

 

Basically this says, that if you take the antiderivative (indefinite integral) with respect to x, [math] \int (...) dx[/math], of the derivative of a function of x, [math] \frac {df}{dx} [/math], then you get the original function [math] f(x) [/math] whose derivative you took the anti-derivative of (The big [math] C [/math] stands for any constant term that may have disappeared from taking a derivative) Here is a simple example:

 

We have a function, [math] x^2 + 2x + 3 [/math]. Lets take the antiderivative of that function (don't worry about how I'm computing this, just try and follow the concept):

 

[math] \int \left ( x^{2} + 2x + 3 \right ) dx= \frac{1}{3}x^{3} + x^{2} + 3x + C [/math]

 

Alright, so now lets take the derivative of the right side of that last equation, (we can't derive the C really, just hang with me here).

 

[math] \frac{d}{dx} \left ( \frac{1}{3}x^{3} + x^{2} + 3x \right ) = x^{2} + 2x + 3 [/math], so know we have the original function again!

 

I hope I'm not confusing you with the notation. Here you can see though, how the derivative and antiderivative are really opposite operations. Take a derivative of a function, take the integral of that derivative, and you are back to where you started. Had we differentiated the original function first, then integrated, we wouldn't have recovered our constant term though. That's a different lesson though, (definite integrals).

 

I'm not a mathematician like some of these guys above, but this is how I think about these things as I use this math quite often.

 

Chemists have been known to integrate a curve by cutting it out of a piece of graph paper and weighing it on a sensitive laboratory balance.

 

That gives new meaning to the phrase "numerical integration". Now days in the 21st century we've advanced to being able to hit the "integrate between __ and __" button in spec. programs at least. What's really appreciated is the automated Fourier transform!

Edited June 12, 2011 by mississippichem

[between3and26characterslon]

Thanks everyone for your replies, unfortunately you have all underestimated my ignorance in this field. Is it possible to give a very simplistic physical example where integrating or differentiating gives you real results.

 

You will need to lower your standards a bit....

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

...nope, a bit lower....

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

...lower...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

...bit more...

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

...try here

[DrRocket]

Thanks everyone for your replies, unfortunately you have all underestimated my ignorance in this field. Is it possible to give a very simplistic physical example where integrating or differentiating gives you real results.

 

You will need to lower your standards a bit.......try here

 

The time derivative of position is velocity. The time derivative of velocity is acceleration.

 

The time integral of velocity is change in position. The time integral of acceleration is change in velocity.

[baxtrom]

...try here

 

 

Imagine you're a plumber. Let's say you put up a pipe on a wall, and you want some slope on it so that excess water will collect at some point were you can drain it. Perhaps the pipe is 5 ft from the floor at one end of the wall, and 6 ft at the other end. Say the pipe is 10 ft long. Then, the derivative of the vertical position of the pipe is 0.1 ft per ft.

 

The integral is just as simple. Lets say you know you want to start at 5 ft above the floor, and that the slope (derivative) should be 0.1 ft per ft. If you put the pipe up properly, you'll end up at 6 ft vertical position at the other end of the wall. That's integrating.

 

[DrRocket]

Imagine you're a plumber. Let's say you put up a pipe on a wall, and you want some slope on it so that excess water will collect at some point were you can drain it. Perhaps the pipe is 5 ft from the floor at one end of the wall, and 6 ft at the other end. Say the pipe is 10 ft long. Then, the derivative of the vertical position of the pipe is 0.1 ft per ft.

 

The integral is just as simple. Lets say you know you want to start at 5 ft above the floor, and that the slope (derivative) should be 0.1 ft per ft. If you put the pipe up properly, you'll end up at 6 ft vertical position at the other end of the wall. That's integrating.

 

 

That is differentiating functions whose graph is a straight line. But that is an important class of functions.

 

The whole point of differential calculus s that a great deal can be said about a very large class of functions by approximating them locally by straight lines.

[baric]

 

...try here

 

 

ok, here you go.

 

Think of a curve on a graph... maybe a curve representing the square of a number

 

 

For example, at x=1, then y =1. at x=2, y=4. at x=3, y=9.

 

Got that?

 

Notice how the value for y goes up faster than the value of x? That means that the SLOPE of the curve is increasing as x increases.

 

What differentiating does is give us a new equation that tells us the particular slope of the line at any x value.

 

In the real world, this slope is often called the rate of change.

 

 

 

Integration goes the other way. Imagine that same curved line as before, but also visualize the area underneath it. What integration tells us is how much area is under the curve between two x points.

 

In the real world, this can give you specific values at points in time for something that is changing. For example, the integration of a velocity equation will tell us how far something has traveled in a specified amount of time.

 

 

These two concepts are the fundamental basis of calculus and have a tremendous amount of use in the natural world. What calculus boils down to is understanding how to differentiate or integrate the various COMPLICATED equations that are used to describe things in the real world.

[between3and26characterslon]

Imagine you're a plumber....

 

ok, here you go...

 

Thanks, this is more my level. Thanks to everyone else as well.

[Zone Ranger]

I think many students get confused by the calculus notation. So I decided to post the fundamental theorem of calculus symbolically and explain the symbols:

 

[math] \int \frac{d f(x)}{dx} dx = f(x) + C [/math]

 

Basically this says, that if you take the antiderivative (indefinite integral) with respect to x, [math] \int (...) dx[/math], of the derivative of a function of x, [math] \frac {df}{dx} [/math], then you get the original function [math] f(x) [/math] whose derivative you took the anti-derivative of (The big [math] C [/math] stands for any constant term that may have disappeared from taking a derivative) Here is a simple example:

 

We have a function, [math] x^2 + 2x + 3 [/math]. Lets take the antiderivative of that function (don't worry about how I'm computing this, just try and follow the concept):

 

[math] \int \left ( x^{2} + 2x + 3 \right ) dx= \frac{1}{3}x^{3} + x^{2} + 3x + C [/math]

 

Alright, so now lets take the derivative of the right side of that last equation, (we can't derive the C really, just hang with me here).

 

[math] \frac{d}{dx} \left ( \frac{1}{3}x^{3} + x^{2} + 3x \right ) = x^{2} + 2x + 3 [/math], so know we have the original function again!

 

I hope I'm not confusing you with the notation. Here you can see though, how the derivative and antiderivative are really opposite operations. Take a derivative of a function, take the integral of that derivative, and you are back to where you started. Had we differentiated the original function first, then integrated, we wouldn't have recovered our constant term though. That's a different lesson though, (definite integrals).

 

I'm not a mathematician like some of these guys above, but this is how I think about these things as I use this math quite often.

 

 

 

That gives new meaning to the phrase "numerical integration". Now days in the 21st century we've advanced to being able to hit the "integrate between __ and __" button in spec. programs at least. What's really appreciated is the automated Fourier transform!

 

 

This is not (either part) the fundamental theorem of calculus

[mississippichem]

This is not (either part) the fundamental theorem of calculus

 

Not rigorously no. What is the fundamental theorem of calculus then?

[Zone Ranger]

Not rigorously no. What is the fundamental theorem of calculus then?

 

 

 

Fundamental_theorem_of_calculus

[mississippichem]

Fundamental_theorem_of_calculus

 

From your link:

 

There are two parts to the Fundamental Theorem of Calculus. Loosely put, the first part deals with the derivative of an antiderivative, while the second part deals with the relationship between antiderivatives and definite integrals.

 

also from your link:

 

The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an indefinite integration[1] can be reversed by a differentiation. The first part is also important because it guarantees the existence of antiderivatives for continuous functions.[2]

[Zone Ranger]

From your link:

 

 

 

also from your link:

 

this

[math]\int\frac{df(x)}{dx}dx=f(x)+C[/math] is NOT the FTC

 

 

 

The actual theorem is provided in the subsection of the link I provided.

 

Here is another link FTC

 

Could you please point out how what you typed is the FTC.

[DrRocket]

this

[math]\int\frac{df(x)}{dx}dx=f(x)+C[/math] is NOT the FTC

 

 

 

The actual theorem is provided in the subsection of the link I provided.

 

Here is another link FTC

 

Could you please point out how what you typed is the FTC.

 

 

The fundamental theorem is usually stated in terms of the definite, rather than the indefinite integral. However, the expression as presented by mississippichem does an adequate job of communicating the underlying concept for the pedagogic task at hand. You are splitting hairs needlessly.

 

I would be more precise in a formal class, but for forum purposes, and given the level of the original question, there is no point in full rigor.

Edited June 14, 2011 by DrRocket

[Zone Ranger]

The fundamental theorem is usually stated in terms of the definite, rather than the indefinite integral. However, the expression as presented by mississippichem does an adequate job of communicating the underlying concept for the pedagogic task at hand. You are splitting hairs needlessly.

 

 

What he has in not the FTC so he shouldn't call it that.

[mississippichem]

this

[math]\int\frac{df(x)}{dx}dx=f(x)+C[/math] is NOT the FTC

 

 

 

The actual theorem is provided in the subsection of the link I provided.

 

Here is another link FTC

 

Could you please point out how what you typed is the FTC.

 

Alright,

 

I realize that the more formal statement of the FTC is:

 

[math] \int_{a}^{b} f(x) dx=F(b) - F(a) [/math], for a continuous [math] f(x) [/math] on [a,b]. Everyone knows that. But again, from your new link:

 

This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.

 

I was trying to explain the relationship between differentiation and anti-differentiation that is a result (Corollary?) of the fundamental theorem of calculus simply to someone who didn't have any knowledge of the matter. If my post was somehow incorrect please explain.

 

Your link also states that:

 

if [math]F(x) = \int_{a}^{x}f(t) dt[/math]

 

then [math] F'(x) = f(x) [/math]

 

Which is basically what I stated. I'm not a mathematician so if you are more knowledgeable please clear up my apparent confusion.

Edited June 14, 2011 by mississippichem

[DrRocket]

Your link also states that:

 

if [math]F(x) = \int_{a}^{x}f(t) dt[/math]

 

then [math] F'(x) = f(x) [/math]

 

Which is basically what I stated. I'm not a mathematician so if you are more knowledgeable please clear up my apparent confusion.

 

If the link actually states that without further qualification, then the link is wrong. It is true for continuous functions f. It is not true if f has a "jump" discontinuity. Derivatives need not be continuous, but all derivatives satisfy the intermediate value properety.

 

What he has in not the FTC so he shouldn't call it that.

 

 

You are being overly pedantic and quite silly.

Edited June 14, 2011 by DrRocket

[Zone Ranger]

You are being overly pedantic and quite silly.

 

Last Comment:

 

This in not a small detail...which I'm sure you are aware of. The two statements are quite different.

[mississippichem]

Last Comment:

 

This in not a small detail...which I'm sure you are aware of. The two statements are quite different.

 

You would not be splitting hairs in the context of a formal math classroom. I was aware of the correction Dr. Rocket posted, but given that this thread is for the most elementary level of knowledge transfer my explanation was sufficient and I will stand by it. The original poster doesn't even know what continuity and jump discontinuities are probably, no insult to him.

 

If you wish to criticize me on my lack of rigor then you can, but for what?

[DrRocket]

OK, for those who demand the best theorem, rigorously stated, at this level, here are the real fundamental theorems of calculus (there are 2 of them)

 

First fundamental theorem of calculus. Let [math][a,b][/math] be an arbitrary closed interval and let [math]f:[a,b] \rightarrow \mathbb R[/math] be such that[math]f[/math] is bounded and the set of points of discontinuity of [math]f[/math] are of Lebesgue measure 0. Suppose that there exists a function [math]F:(c,d) \rightarrow \mathbb R[/math] where [math]c<a<b<d[/math] such that [math]F'(x)=f(x) \forall x \in [a,b] [/math] Then [math] \int_a^b f(x) \ dx = F(b)-F(a)[/math]

 

 

 

Second fundamental theorem of calculus. Let [math][a,b][/math] be an arbitrary closed interval and let [math]f:[a,b] \rightarrow \mathbb R[/math] be such that[math]f[/math] is bounded and the set of points of discontinuity of [math]f[/math] are of Lebesgue measure 0. For [math] x \in [a,b] [/math] set [math]F(x) = \int_a^x f(t) dt [/math] then

 

1) [math]F[/math] is continuous on [math][a,b][/math]

 

and

 

2) if [math]f[/math] is continuous at [math]x_0 \in (a,b)[/math] then [math]F[/math] is differentiable at [math]x_0[/math] and [math]F'(x_0)=f(x_0)[/math]

 

 

This is absolutely correct, and absolutely opaque to the OP who is asking what derviatives and integrals are at the intuitive level.

 

There are two parts to mathematics. 1) The intuitive idea that helps one conceptualize an idea. 2) The rigorous definitions that make the intuitive ideas precise and useable. You need both of them, but often the first element is the most difficult to really understand. (And you don't really have 1vuntil you understand 2, but one must avoid putting the cart before the horse.)

[baric]

OK, for those who demand the best theorem, rigorously stated

 

This is absolutely out of place in this particular thread.

[DrRocket]

This is absolutely out of place in this particular thread.

 

 

That is precisely the point.