- #1

Peppino

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Derivative of f(x): Rate at which the function is changing at point x

Integral of f(x): ?

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- Thread starter Peppino
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- #1

Peppino

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Derivative of f(x): Rate at which the function is changing at point x

Integral of f(x): ?

- #2

pwsnafu

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Let ##a## and ##b## be real numbers with ##a<b##. Let ##f : [a,b]\rightarrow \mathbb{R}## be a positive real function. Then ##\int_a^b f(x) \, dx## is the area of the region bounded by 1) the x-axis, 2) the vertical line ##x=a##, 3) the vertical line ##x=b##, and 4) ##f(x)## for ##a \leq x \leq b##.

Notice this a real number. To obtain a function, you fix one of the terminals and allow the other vary.

Edit: Just a note: we do not*define* the integral using area. We *interpret* the integral this way. But it is sufficient for someone who isn't mathematically minded.

Notice this a real number. To obtain a function, you fix one of the terminals and allow the other vary.

Edit: Just a note: we do not

Last edited:

- #3

Simon Bridge

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eg. the area under the v-t graph is the displacement (note: NOT the total distance traveled.)

if g(x)=f'(x) then [itex]\int g(x)dx = f(x)+c[/itex]

It is the antiderivative ... i.e. given the equation of the slope of the tangent to a point on a graph, what is the graph?

In actual fact it is a summation - where the interval of the sum is very small.

We just use it to find areas.

- #4

Bacle2

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eg. the area under the v-t graph is the displacement (note: NOT the total distance traveled.)

if g(x)=f'(x) then [itex]\int g(x)dx = f(x)+c[/itex]

It is the antiderivative ... i.e. given the equation of the slope of the tangent to a point on a graph, what is the graph?

In actual fact it is a summation - where the interval of the sum is very small.

We just use it to find areas.

Sorry to nitpick, Simon, but there are exceptions to ∫f'dx =f ,like that of the Cantor

function C(x),for which C'(x)=0 a.e., so that ∫C'(x)dx=∫0dx=0 ≠ C(x).

- #5

Simon Bridge

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I'm sorry - I did not claim that ∫f'dx =f ... please read more carefully. I also did not claim that there were no exceptions or that I was making a definition - complete or otherwise.Sorry to nitpick, Simon, but there are exceptions to ∫f'dx =f

That is not correct - [itex]\int 0dx = c[/itex] ... a constant.function C(x),for which C'(x)=0 a.e., so that ∫C'(x)dx=∫0dx=0 ≠ C(x).

Note: the cantor function has zero derivative

There are functions with zero derivative everywhere ... constant functions.

Cantor functions are usually considered stuff for advanced students - is that the level that the question is being asked?

However, having put in the niggle - your challenge is to improve my description in the context of post #1 and relevant to OPs level of understanding :)

- #6

Bacle2

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To the OP:

You may find this interesting:

Look up the case of the Cantor function:

http://en.wikipedia.org/wiki/Cantor_function

Note that the Cantor function is an exception to:

∫f'= f .

In order to have ∫f'=f we need f to be absolutely continuous.

- #7

Simon Bridge

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eg. if [tex]u(x)=\left\{

\begin{array}{lr}

1 & : x > 0\\

0 & : x \leq 0

\end{array}

\right.[/tex]... you can see that the derivative is 0 everywhere except at the origin.

We need to be able to distinguish this situation from a constant function. This is what the Dirac Delta function is for.

But OP has only just learned to integrate polynomials ... all these special functions and such will be so many octarine words hanging in the air.

I'll stick by the conclusion post #2: the integration is really a sum.

- #8

Peppino

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- #9

Bacle2

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tells you is the functions whose rate of change is ln(x). Like Simon said, it is

difficult to generalize without a context.

- #10

Peppino

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tells you is the functions whose rate of change is ln(x). Like Simon said, it is

difficult to generalize without a context.

Oh that is interesting I haven't thought of it like that. Anyway, in my price example, would integrating the function serve any purpose?

- #11

Bacle2

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time, money supply, etc.?

( you can tell I am not much of an investor )

- #12

Peppino

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time, money supply, etc.?

( you can tell I am not much of an investor )

Just a graph of $ over time

- #13

Bacle2

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I'm sorry, I need to head out right away. If no one has answered, I'll get back to you.

- #14

AlephZero

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I don't think it tells you anything with an obvious meaning. Just because you can calculate something, that doesn't mean calculating it is meaningful.

But if you divide the integral by the time interval (end time - start time), that would tell you the average priice during the time interval.

- #15

Peppino

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I don't think it tells you anything with an obvious meaning. Just because you can calculate something, that doesn't mean calculating it is meaningful.

But if you divide the integral by the time interval (end time - start time), that would tell you the average priice during the time interval.

So let's say the time is in days, and the interval goes on for 10 days. Would [itex]\frac{10ln(10)-10}{10}[/itex] give the average price for the 10 days?

Or do you mean [itex]\int_{0}^{10}\frac{xln(x)-x}{10}[/itex]

- #16

AlephZero

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I mean $$\frac{\int_{t_1}^{t_2} f(t)\,dt} { t_2 - t_1 }$$

- #17

Simon Bridge

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In your example (post #16) dividing through by T would just get the price back.

You are dividing the area by the length of the base.

If the price started at Q and ended at P, linearly, then the integral is (P+Q)T/2. Dividing through by T gets you (P+Q)/2 ... the average price.

That should be general to any p(t) ... the integral of p divided by the time period is the time-average of p.

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