- #1

- 72

- 0

To my understanding the derivative gives the slope of a curve at given point whereas the the integral finds the area under the curve. How are these two opposites?

- Thread starter krackers
- Start date

- #1

- 72

- 0

To my understanding the derivative gives the slope of a curve at given point whereas the the integral finds the area under the curve. How are these two opposites?

- #2

- 647

- 3

- #3

Bacle2

Science Advisor

- 1,089

- 10

are necessary; for Riemann-integrable you need a.e-continuity and boundedness, and

for Leb.-integrable, you need absolute continuity. Try to see what happens in each

case when the conditions are not satisfied.

- #4

- 395

- 14

You're basically asking for the proof of the Fundamental Theorem of Calculus. A quick youtube search will answer your questions.

Last edited by a moderator:

- #5

- 828

- 2

The area of a circle is [itex]a=\pi r^2[/itex] and the circumference of a circle is [itex]c=2 \pi r[/itex]. Now, do you see that the circumference of the circle is just the derivative of the area of the circle?

- #6

- 395

- 14

That's an example, but I don't really think it helps much in understanding the WHY. He probably knows a million examples.

The area of a circle is [itex]a=\pi r^2[/itex] and the circumference of a circle is [itex]c=2 \pi r[/itex]. Now, do you see that the circumference of the circle is just the derivative of the area of the circle?

The answer to your question is the FTC.

Now is there an intuitive, cutesy way of truly understanding? Not that I know of. It's just one of those beautiful results of math.

- #7

- 828

- 2

The idea is that adding a small amount to the area is like adding the value of the derivative at that area. OK, so, let's say [itex]f[/itex] is a well behaved function and let [itex]F(x)[/itex] be the area under [itex]f[/itex] between [itex]0[/itex] and [itex]x[/itex]. Now, let's try to find the area under [itex]f[/itex] from [itex]x[/itex] to [itex]x+h[/itex]. One way of doing this is to compute [itex]F(x+h)-F(x)[/itex]. Another way of estimating this is by [itex]hf(x)[/itex]. That is,That's an example, but I don't really think it helps much in understanding the WHY. He probably knows a million examples.

[tex]f(x) \simeq \frac{F(x+h)-F(x)}{h}[/tex].

Now, let's take the limit as [itex]h \to 0[/itex]. Then we see that [itex]f(x) = F'(x)[/itex]. So, the derivative of the integral is the function inside the integral. Or, the derivative inside the integral is the antiderivative of the integral.

Well, he is basically asking why FTC is true, so appealing to FTC probably isn't very satisfying. And, if I remember correctly, this was first proven by Isaac Barrow - who was Isaac Newton's teacher (if I remember.) I can almost guarantee that they had some intuitive reason to believe that it was true. That is, they just didn't say "hey this is probably true; I have no reason to think it is true, but I'm gonna try to prove it anyway."The answer to your question is the FTC.

Now is there an intuitive, cutesy way of truly understanding? Not that I know of. It's just one of those beautiful results of math.

Another way to think of it is like this. Let's say that you know how fast someone is traveling from [itex]t=0[/itex] to [itex]t=t_0[/itex] where [itex]t[/itex] is measured in seconds. You want to know how far he is gone. How can you do this? Now, try to apply similar reasoning to what I did above.

- #8

- 795

- 7

Easy visualization.

To my understanding the derivative gives the slope of a curve at given point whereas the the integral finds the area under the curve. How are these two opposites?

You're driving in your car. You go 40mph for one hour then you go 50mph for one hour. How far have you gone? You know that D = RT (distance = rate * time) so one hour at 40mph = 40 miles; plus one hour at 50mph = 90 miles total.

Now instead of calculating your total distance by the hour, you see what your speed is during each minute, and add up all the D = RT calculations.

Now you let the time interval go to zero. Conceptually you go a certain speed for an instant; then another speed for an instant; etc. You do the little D = RT calculation and add up all the instants.

But your instantaneous speed is just the first derivative of your position function. It's the speed shown on your odometer. So the Fundamental Theorem of Calculus is just a matter of adding up all the D = RT calculations as t goes to zero. You could whip up this little story into a formal proof if you wanted to.

Hope this helped.

- #9

- 428

- 1

One says derivative of integral, the other says integral of derivative.

As I see, these have two separate intuitions, the former can be said as the area is changing by the next height. The latter says the total flow throughout can be recorded by the flow at the boundaries, which I have never completely been able to tie together with the notation and operations.

- #10

- 1,796

- 53

I wonder if there is not a more fundamental reason. We can reduce the concept to operations on sets. In this particular case, we operate (differentiate) on an element of a set of functions and obtain an element of another set. We then operate (integrate) on that element with another operator and obtain another element which is equal to the first element except for a constant.Is there any clear explanation as to why exactly derivatives and integrals cancel each other [other than the integral is the anti-derivative]?

What intrinsic property of the operators of differentiation and integration upon differentiable functions allow this inversion to exist? Is this property independent of the particular operators and particular sets used in this example? That is, do other sets and operators share this property of inversion which have nothing to do with differentiation and integration? If so, then perhaps the reason differenitation and integration are inverses of each other really have nothing intrinsically to do with the particular details of the operators but rather only upon the particular properties of sets and the allowable operators acting upon them.

- #11

- 23

- 0

This is basically how I ended up understanding this problem.The idea is that adding a small amount to the area is like adding the value of the derivative at that area. OK, so, let's say [itex]f[/itex] is a well behaved function and let [itex]F(x)[/itex] be the area under [itex]f[/itex] between [itex]0[/itex] and [itex]x[/itex]. Now, let's try to find the area under [itex]f[/itex] from [itex]x[/itex] to [itex]x+h[/itex]. One way of doing this is to compute [itex]F(x+h)-F(x)[/itex]. Another way of estimating this is by [itex]hf(x)[/itex]. That is,

[tex]f(x) \simeq \frac{F(x+h)-F(x)}{h}[/tex].

Now, let's take the limit as [itex]h \to 0[/itex]. Then we see that [itex]f(x) = F'(x)[/itex]. So, the derivative of the integral is the function inside the integral. Or, the derivative inside the integral is the antiderivative of the integral.

Well, he is basically asking why FTC is true, so appealing to FTC probably isn't very satisfying. And, if I remember correctly, this was first proven by Isaac Barrow - who was Isaac Newton's teacher (if I remember.) I can almost guarantee that they had some intuitive reason to believe that it was true. That is, they just didn't say "hey this is probably true; I have no reason to think it is true, but I'm gonna try to prove it anyway."

Another way to think of it is like this. Let's say that you know how fast someone is traveling from [itex]t=0[/itex] to [itex]t=t_0[/itex] where [itex]t[/itex] is measured in seconds. You want to know how far he is gone. How can you do this? Now, try to apply similar reasoning to what I did above.

First you set up the derivative and anti-derivative and then you prove that the anti derivative process is the same as Riemann integration :)

- Replies
- 4

- Views
- 2K

- Replies
- 2

- Views
- 1K

- Replies
- 4

- Views
- 907

- Replies
- 1

- Views
- 2K

- Replies
- 6

- Views
- 7K

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 1K

- Replies
- 17

- Views
- 11K

- Replies
- 31

- Views
- 7K

- Replies
- 17

- Views
- 10K