# How does calculus work?

Watch
Announcements

In school we are taught how to differentiate and integrate but not why the techniques work.. who came up with the formula? How does it show the rate of change of a function or the area under a curve?

0

reply

Report

#2

(Original post by

In school we are taught how to differentiate and integrate but not why the techniques work.. who came up with the formula? How does it show the rate of change of a function or the area under a curve?

**tanyapotter**)In school we are taught how to differentiate and integrate but not why the techniques work.. who came up with the formula? How does it show the rate of change of a function or the area under a curve?

0

reply

Report

#3

**tanyapotter**)

In school we are taught how to differentiate and integrate but not why the techniques work.. who came up with the formula? How does it show the rate of change of a function or the area under a curve?

1

reply

(Original post by

Calculus is nothing but dividing something large into small bits and forming a series(for integration) and for differenciation you assume the two points on a curve behaving like a straight line to calculate gradient and then analysing what happenes when the horizontal distance between them tends to ZERO(thats what delta x means). Sorry for my unclear explanation

**Duke Glacia**)Calculus is nothing but dividing something large into small bits and forming a series(for integration) and for differenciation you assume the two points on a curve behaving like a straight line to calculate gradient and then analysing what happenes when the horizontal distance between them tends to ZERO(thats what delta x means). Sorry for my unclear explanation

*how*integration gives an exact value. I also don't know why calculating the rate of change of something is the inverse function of calculating the area under the curve of something. I get that if you differentiate x^2, you get 2x but I don't get

*why*the power is multiplied by the coefficient and then decreased by 1 to give this rate of change

0

reply

Report

#5

(Original post by

I understand how to differentiate and integrate but I don't get how integrating a function and substituting values of x in can suddenly just give the area under the curve. I know how the trapezium rule works but I don't know

**tanyapotter**)I understand how to differentiate and integrate but I don't get how integrating a function and substituting values of x in can suddenly just give the area under the curve. I know how the trapezium rule works but I don't know

*how*integration gives an exact value. I also don't know why calculating the rate of change of something is the inverse function of calculating the area under the curve of something. I get that if you differentiate x^2, you get 2x but I don't get*why*the power is multiplied by the coefficient and then decreased by 1 to give this rate of change
0

reply

Report

#6

**tanyapotter**)

I understand how to differentiate and integrate but I don't get how integrating a function and substituting values of x in can suddenly just give the area under the curve. I know how the trapezium rule works but I don't know

*how*integration gives an exact value. I also don't know why calculating the rate of change of something is the inverse function of calculating the area under the curve of something.

**I get that if you differentiate x^2, you get 2x but I don't get**

*why*the power is multiplied by the coefficient and then decreased by 1 to give this rate of changeAs to how integrating gives the area under a curve... You've said you understand how the trapezium rule works - you probably also know that it's only an approximate answer because the straight edges of a trapezium don't exactly match what is obviously a curved line. The idea of integration is basically using trapeziums/strips that are infinitely thin so that you don't get an underestimation but an exact answer.

I have a feeling I've confused you even more.

0

reply

Report

#7

**tanyapotter**)

I understand how to differentiate and integrate but I don't get how integrating a function and substituting values of x in can suddenly just give the area under the curve. I know how the trapezium rule works but I don't know

*how*integration gives an exact value. I also don't know why calculating the rate of change of something is the inverse function of calculating the area under the curve of something. I get that if you differentiate x^2, you get 2x but I don't get

*why*the power is multiplied by the coefficient and then decreased by 1 to give this rate of change

That's still true for curves, except you choose two points that are really, really, really, really, really, really close together and do the difference in their y-coordinates over their x-coordinates "as the difference tends to 0" - the link I posted has this exact example.

0

reply

(Original post by

Did your teacher show you differentiation from first principles?

**16Characters....**)Did your teacher show you differentiation from first principles?

0

reply

(Original post by

Have you covered differentiation from first principles yet?

As to how integrating gives the area under a curve... You've said you understand how the trapezium rule works - you probably also know that it's only an approximate answer because the straight edges of a trapezium don't exactly match what is obviously a curved line. The idea of integration is basically using trapeziums/strips that are infinitely thin so that you don't get an underestimation but an exact answer.

I have a feeling I've confused you even more.

**Hydeman**)Have you covered differentiation from first principles yet?

As to how integrating gives the area under a curve... You've said you understand how the trapezium rule works - you probably also know that it's only an approximate answer because the straight edges of a trapezium don't exactly match what is obviously a curved line. The idea of integration is basically using trapeziums/strips that are infinitely thin so that you don't get an underestimation but an exact answer.

I have a feeling I've confused you even more.

*how*the integration function calculates and sums up all these infinitely thin strips!!! Like, to find the area under a trapezium you simply do (a+b)h/2 but to find the area under a curve you integrate and substitute the limits in and it gives you a value for the area- how?

0

reply

(Original post by

The reason 'why' is that it's a useful shortcut. You know how, when you're finding the gradient of a line, you take two points, find the difference in the y-coordinates and the difference in the x-coordinates then divide them?

That's still true for curves, except you choose two points that are really, really, really, really, really, really close together and do the difference in their y-coordinates over their x-coordinates "as the difference tends to 0" - the link I posted has this exact example.

**Zacken**)The reason 'why' is that it's a useful shortcut. You know how, when you're finding the gradient of a line, you take two points, find the difference in the y-coordinates and the difference in the x-coordinates then divide them?

That's still true for curves, except you choose two points that are really, really, really, really, really, really close together and do the difference in their y-coordinates over their x-coordinates "as the difference tends to 0" - the link I posted has this exact example.

0

reply

Report

#11

(Original post by

Do you mean like drawing tangents and showing that as h tends to 0, the gradient tends to ___? I learnt that in further maths and it was fine, but what I'm asking is why 5x^3 differentiates to 15x^2 - why does the formula work? Sorry If I sound dumb

**tanyapotter**)Do you mean like drawing tangents and showing that as h tends to 0, the gradient tends to ___? I learnt that in further maths and it was fine, but what I'm asking is why 5x^3 differentiates to 15x^2 - why does the formula work? Sorry If I sound dumb

You have

but as we can ignore the in the numerator to get .

You can try the proof with if you want.

To clarify a bit, the first part stems from taking the tiny difference in the y-coordinates and dividing it by the difference in the x coordinates. For any function f(x), the derivative is given by

.

0

reply

Report

#12

**tanyapotter**)

Do you mean like drawing tangents and showing that as h tends to 0, the gradient tends to ___? I learnt that in further maths and it was fine, but what I'm asking is why 5x^3 differentiates to 15x^2 - why does the formula work? Sorry If I sound dumb

The derivative of a function is defined as . as h --> 0. Let and substitute this into that formula, using the binomial expansion to expand . You will then be able to follow through to the correct result of as the derivative.

0

reply

Report

#13

**tanyapotter**)

Do you mean like drawing tangents and showing that as h tends to 0, the gradient tends to ___? I learnt that in further maths and it was fine, but what I'm asking is why 5x^3 differentiates to 15x^2 -

**why does the formula work?**Sorry If I sound dumb

(Original post by

Yes, I get that but I don't get

**tanyapotter**)Yes, I get that but I don't get

*how*the integration function calculates and sums up all these infinitely thin strips!!! Like, to find the area under a trapezium you simply do (a+b)h/2 but to find the area under a curve you integrate and substitute the limits in and it gives you a value for the area- how?
1

reply

(Original post by

Let's work with your why 'x^2 becomes 2x' example.

You have

but as we can ignore the in the numerator to get .

You can try the proof with if you want.

**Zacken**)Let's work with your why 'x^2 becomes 2x' example.

You have

but as we can ignore the in the numerator to get .

You can try the proof with if you want.

(Original post by

Not dumb at all.The derivative of a function is defined as . as h --> 0. Let and substitute this into that formula, using the binomial expansion to expand . You will then be able to follow through to the correct result of as the derivative.

**16Characters....**)Not dumb at all.The derivative of a function is defined as . as h --> 0. Let and substitute this into that formula, using the binomial expansion to expand . You will then be able to follow through to the correct result of as the derivative.

0

reply

Report

#15

**tanyapotter**)

I understand how to differentiate and integrate but I don't get how integrating a function and substituting values of x in can suddenly just give the area under the curve. I know how the trapezium rule works but I don't know

*how*integration gives an exact value. I also don't know why calculating the rate of change of something is the inverse function of calculating the area under the curve of something. I get that if you differentiate x^2, you get 2x but I don't get

*why*the power is multiplied by the coefficient and then decreased by 1 to give this rate of change

A mix of ingenuity and trial and errors. You can read the

De analysi per aequationes numero terminorum infinita (you can find it here in latin, if you know the language) is an interesting essay on "practical knowledge" at the end of the day.

Newton invented calculus in order to associate different laws of motion and math, without a rigorous "demonstration" in the mathematical sense (like Liebnitz will more or less do a bit later), and in fact the work was unpublished (and unpublishable, as it is more a diary to self rather than an actual scientific explanation).

To give you an example, it starts like this:

*Methodum generalem quam de curvarum quantitate per infinitam terminorum seriem mensuranda olim excogitaveram, in sequentibus brevitèr explicatam potiùs quàm accuratè demonstratam habes.*

That (on the spot) means:

I have devised (

*yes, devised!! -my note*) a general method in order to measure the quantity of curving using infinite series of terms, in the following I can explain it shortly what I have accurately demonstrated (

*and that is not exactly true. -my note*).

Then he does some clever drawings like these two:and calculates with simple linear algebra and triangular approximation his derivation and integrations of the polynomial powers that were more or less all you needed for the physics at the time (and even now in a certain sense ).

Liebniz followed a different, much more rigorous path and derived the law of calculus in a relatively more similar way to how we study it today (even though the currently used demonstration that make use of the differential quotient is due to Cauchy a couple of centuries later).

The two discoveries were absolutely independent of each other, since Newton never published his ideas on calculus and Liebniz followed a different path to demonstrate it.

In the end, Newton was there before (when he was 21 and Liebniz 16) to say that "the curvature of x^2 is 2x", but no more and no less. In spite of this, when Liebnitz came out publicly with his demonstration, Newton produced fake evidence about what he did in the past and used all of his huge political and academic power to crush Liebniz and cover the truth.

Nonetheless, the discovery of Newton was brilliant and, even if not publicly released, gave him an incommensurable edge over his colleagues. The discovery of calculus by Newton teaches us how a brilliant discovery doesn't have to be necessarily explained in bloody detail to be effective, and not only a granitic demonstration but also a well rooted intuition can give you the thrust you need to understand the marvels of nature.

1

reply

Report

#16

(Original post by

This is what I was looking for!! Thanks so much

**tanyapotter**)This is what I was looking for!! Thanks so much

**Protip**: next time, post in the maths forum for maths help, we have some seriously talented helpers over there!

0

reply

(Original post by

Haha, no problem at all!

**Zacken**)Haha, no problem at all!

**Protip**: next time, post in the maths forum for maths help, we have some seriously talented helpers over there!I'll have had like 3 proofs off of you in the space of 30 minutes, apologies

0

reply

*really*doing maths, even if it is at A-Level! I could pretty easily teach my 10 year old sister how to differentiate and integrate but I couldn't prove the formulae very easily at all and I think that's how I know I wasn't born to do maths

0

reply

Report

#20

(Original post by

Will do! Would you mind giving a proof for integration too?

I'll have had like 3 proofs off of you in the space of 30 minutes, apologies

**tanyapotter**)Will do! Would you mind giving a proof for integration too?

I'll have had like 3 proofs off of you in the space of 30 minutes, apologies

Okay, so integration is basically the area under the graph, let's integrate so we're finding the area under x^n between 0 and b.

We have a real number and we consider a subdividion of the interval given by .

Now, we take the upper sum starting from the right, let's call it - so we have:

by the sum of a geometric series.

Now, we let to get so

0

reply

X

### Quick Reply

Back

to top

to top