tanyapotter
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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?
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Zacken
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(Original post by 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?
I hope you don't mine me giving you a link, but I found that the nrich website has a very good section on explaining differentiation from an intuitive viewpoint. You might want to have a read through it here.
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Duke Glacia
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(Original post by 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?
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
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tanyapotter
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(Original post by 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
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
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16Characters....
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(Original post by 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
Did your teacher show you differentiation from first principles?
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troubadour.
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(Original post by 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
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. :getmecoat:
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(Original post by 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
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.
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tanyapotter
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(Original post by 16Characters....)
Did your teacher show you differentiation from first principles?
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
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tanyapotter
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(Original post by 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. :getmecoat:
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?
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tanyapotter
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(Original post by 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.
Yeah, we learnt the numerical methods for calculus in further maths I just don't get why the formula for differentiating or integrating is what it is if that makes sense
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Zacken
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(Original post by 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
Let's work with your why 'x^2 becomes 2x' example.

You have \displaystyle \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2hx + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2hx - h^2}{h}

but as h \to 0 \implies h^2 \to 0 we can ignore the h^2 in the numerator to get \displaystyle \frac{2hx}{h} = 2x.

You can try the proof with x^n 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

\displaystyle \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
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16Characters....
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(Original post by 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
Not dumb at all.

The derivative of a function is defined as f'(x) = \frac{f(x+h) - f(x)}{h}. as h --> 0. Let  f(x) = ax^n and substitute this into that formula, using the binomial expansion to expand (x + h)^n. You will then be able to follow through to the correct result of  anx^{n-1} as the derivative.
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troubadour.
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(Original post by 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
Didn't you do the whole thing with the graph and the explanation under it when doing it from first principles? That's how we did it although I struggle to remember the basics of it now...

(Original post by 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?
You're probably better off looking at that link somebody mentioned earlier - I'm a very poor teacher if I'm honest. D:
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tanyapotter
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(Original post by Zacken)
Let's work with your why 'x^2 becomes 2x' example.

You have \displaystyle \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2hx + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2hx - h^2}{h}

but as h \to 0 \implies h^2 \to 0 we can ignore the h^2 in the numerator to get \displaystyle \frac{2hx}{h} = 2x.

You can try the proof with x^n if you want.
(Original post by 16Characters....)
Not dumb at all.The derivative of a function is defined as f'(x) = \frac{f(x+h) - f(x)}{h}. as h --> 0. Let  f(x) = ax^n and substitute this into that formula, using the binomial expansion to expand (x + h)^n. You will then be able to follow through to the correct result of  anx^{n-1} as the derivative.
This is what I was looking for!! Thanks so much
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Duke Glacia
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(Original post by 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
If you have the edexcel book its clearly explained.Sir Isaac newtons inventionof calculus was trial and error. I found this on Quora. Hope it helps.

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:Imageand 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.
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Zacken
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(Original post by tanyapotter)
This is what I was looking for!! Thanks so much
Haha, no problem at all!

Protip: next time, post in the maths forum for maths help, we have some seriously talented helpers over there!
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tanyapotter
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(Original post by 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!
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
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Andy98
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I'm not sure many people actually understand why it works.
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tanyapotter
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(Original post by Andy98)
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I'm not sure many people actually understand why it works.
You're definitely not taught it at my school at AS or A-Level. This makes me seriously doubt my mathematical ability - I am simply following instructions and not 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
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Zacken
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(Original post by 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
Haha, I don't mind at all - it's good practice!

Okay, so integration is basically the area under the graph, let's integrate \int_0^b x^n \, \mathrm{d}x so we're finding the area under x^n between 0 and b.

We have a real number 0< r < 1 and we consider a subdividion of the interval [0,b] given by S = \{0, \ldots, r^2b, rb, b\}.

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



U(s) = \displaystyle b^n \left({b - r b}\right) + \left({rb}\right)^n \left({r b - r^2 b}\right) + \left({r^2b}\right)^n \left({r^2 b - r^3 b}\right) + \cdots
=\displaystyle b^{n+1} \left({1 - r}\right) + b^{n+1} r^{n+1} \left({1 - r}\right) +  b^{n+1} r^{2n+2}\left({1 - r}\right) + \cdots

=\displaystyle b^{n+1} \left({1 - r}\right) \left({1 + r^{n+1} + r^{\left({n+1}\right)^2} + \cdots}\right)
=\displaystyle \frac {b^{n+1} }{1 + r + r^2 + \cdots + r^n}

by the sum of a geometric series.

Now, we let r \to 1 to get \frac{b^{n+1}}{n+1} so

\int_0^b x^n \, \mathrm{d}x = \frac{b^{n+1}}{n+1}
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