Integral calculus/ Area under the curve

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Integral calculus/ Area under the curve
« on: July 18, 2020, 11:46:48 PM »
No need to go in detail about the method of exhaustion in which area under the curve is estimated with rectangles.  It is said that a precise result can be obtained if we place the infinite number of thin rectangles under the curve above the x-axis. This means the width dx of each rectangle is infinitesimally small while the height of each rectangle varies (few may repeat depends on curve) says h1, h2, h3, ...........……hn.

Area of an infinitesimally small rectangle of height h1 and width dx = h1dx where dx tends to zero. The resulting area is very close to zero (almost zero or negligible) due to the very small value of h1. The areas of these infinite number of infinitesimally small rectangles are then added /integrated together for estimating the total area under the curve. Summation of infinite number of zeros is zero. Thus 

Area under curve = [h1dx +h2dx +h3dx …….hndx] = very close to zero or zero

Similarly, dx is common. Therefore = dx(h1+h2+h3………..…..hn) =dx(infinity) As dx tends to zero therefore infinity also became tends to zero and hence =0

Another option is undefined as infinity divided by infinity.

I’m not sure how other people think but IMPOV the integration of the Riemann method may not be the right way to determine the area under curve unless I’m conceiving it wrong – Correct if not how?

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rabinoz

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Re: Integral calculus/ Area under the curve
« Reply #1 on: July 19, 2020, 12:36:25 AM »
<< Go back to school and learn integral calculus there! >>

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #2 on: July 19, 2020, 12:53:58 AM »
Νο, you're a little bit confused. First of all, as you noted the areas are not 0, but infinitesimal. We'll get to what exactly that means. But yeah, imagine we are adding n areas of size 1/n, where n is any integer. Obviously they add up to 1, no matter what n is. You can arbitrarily make n as large as you want, which will make 1/n arbitrarily close to 0 (infinitesimal), and it will still be 1. To really understand this, you have to understand the definition of a limit.

You may find the strict definition a bit confusing so I'll just loosely explain what it boils down to. Saying that the limit of f(x) as x tends to a (which can be infinite) is b, means that by getting x close enough to a you can get your function arbitrarily close to b (it's a bit more complex, in reality f has to be arbitrarily close to b within an entire interval of certain width, but whatever). The definition is similar for sequences (functions of integers) tending to infinity.

Consider the sequence a(n)=1/n+1/n^2. Obviously the limit of this as n tends to infinity is 0 (you get arbitrarily close to 0 by getting n closer to infinity. For example, let's say you want your sequence to be at most 2 away from 0. Obviously a(1)=2, and if you increase n it keeps being smaller than 2. Now say you want it to be at most 1/2 away. Then you can just increase n to 3, since a(3)=1/3+1/9<1/2, and then for every n larger than that it is still <1/2. Intuitively you can tell that 1/n and 1/n^2 get closer and closer to 0 as n becomes larger and larger). Let's add together n areas of size a(n). We get 1+1/n. If we take the limit of 1 as n tends to infinity it is... obviously 1, and the limit of 1/n as n tends to infinity is 0, so the limit of n*a(n) is 1. So adding an infinite number of infinitesimal areas in this case gives us 1, but also at no point in the finite case is it 1, there is always an error of 1/n, that gets smaller and smaller as n increases, think of it as the finite rectangles not matching the curve precisely, and progressively estimating the area better and better as you increase their number.
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Re: Integral calculus/ Area under the curve
« Reply #3 on: July 19, 2020, 02:57:27 AM »
Νο, you're a little bit confused. First of all, as you noted the areas are not 0, but infinitesimal. We'll get to what exactly that means. But yeah, imagine we are adding n areas of size 1/n, where n is any integer. Obviously they add up to 1, no matter what n is. You can arbitrarily make n as large as you want, which will make 1/n arbitrarily close to 0 (infinitesimal), and it will still be 1. To really understand this, you have to understand the definition of a limit.

You may find the strict definition a bit confusing so I'll just loosely explain what it boils down to. Saying that the limit of f(x) as x tends to a (which can be infinite) is b, means that by getting x close enough to a you can get your function arbitrarily close to b (it's a bit more complex, in reality f has to be arbitrarily close to b within an entire interval of certain width, but whatever). The definition is similar for sequences (functions of integers) tending to infinity.

Consider the sequence a(n)=1/n+1/n^2. Obviously the limit of this as n tends to infinity is 0 (you get arbitrarily close to 0 by getting n closer to infinity. For example, let's say you want your sequence to be at most 2 away from 0. Obviously a(1)=2, and if you increase n it keeps being smaller than 2. Now say you want it to be at most 1/2 away. Then you can just increase n to 3, since a(3)=1/3+1/9<1/2, and then for every n larger than that it is still <1/2. Intuitively you can tell that 1/n and 1/n^2 get closer and closer to 0 as n becomes larger and larger). Let's add together n areas of size a(n). We get 1+1/n. If we take the limit of 1 as n tends to infinity it is... obviously 1, and the limit of 1/n as n tends to infinity is 0, so the limit of n*a(n) is 1. So adding an infinite number of infinitesimal areas in this case gives us 1, but also at no point in the finite case is it 1, there is always an error of 1/n, that gets smaller and smaller as n increases, think of it as the finite rectangles not matching the curve precisely, and progressively estimating the area better and better as you increase their number.
I know what you say but I am thinking a lil bit differently. There are two opposite infinities of extreme are involved. One side is trying to drag the multiplication tends to zero while the other side stretch it tends to infinity. Seems oxymoron.

We also know

1.   Anything multiplied by zero is zero
2.   Anything multiplied by infinity is infinity other than zero

Which side is true?
« Last Edit: July 19, 2020, 03:08:12 AM by E E K »

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rabinoz

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Re: Integral calculus/ Area under the curve
« Reply #4 on: July 19, 2020, 04:13:08 AM »
Νο, you're a little bit confused. First of all, as you noted the areas are not 0, but infinitesimal. We'll get to what exactly that means. But yeah, imagine we are adding n areas of size 1/n, where n is any integer. Obviously they add up to 1, no matter what n is. You can arbitrarily make n as large as you want, which will make 1/n arbitrarily close to 0 (infinitesimal), and it will still be 1. To really understand this, you have to understand the definition of a limit.

You may find the strict definition a bit confusing so I'll just loosely explain what it boils down to. Saying that the limit of f(x) as x tends to a (which can be infinite) is b, means that by getting x close enough to a you can get your function arbitrarily close to b (it's a bit more complex, in reality f has to be arbitrarily close to b within an entire interval of certain width, but whatever). The definition is similar for sequences (functions of integers) tending to infinity.

Consider the sequence a(n)=1/n+1/n^2. Obviously the limit of this as n tends to infinity is 0 (you get arbitrarily close to 0 by getting n closer to infinity. For example, let's say you want your sequence to be at most 2 away from 0. Obviously a(1)=2, and if you increase n it keeps being smaller than 2. Now say you want it to be at most 1/2 away. Then you can just increase n to 3, since a(3)=1/3+1/9<1/2, and then for every n larger than that it is still <1/2. Intuitively you can tell that 1/n and 1/n^2 get closer and closer to 0 as n becomes larger and larger). Let's add together n areas of size a(n). We get 1+1/n. If we take the limit of 1 as n tends to infinity it is... obviously 1, and the limit of 1/n as n tends to infinity is 0, so the limit of n*a(n) is 1. So adding an infinite number of infinitesimal areas in this case gives us 1, but also at no point in the finite case is it 1, there is always an error of 1/n, that gets smaller and smaller as n increases, think of it as the finite rectangles not matching the curve precisely, and progressively estimating the area better and better as you increase their number.
I know what you say but I am thinking a lil bit differently. There are two opposite infinities of extreme are involved. One side is trying to drag the multiplication tends to zero while the other side stretch it tends to infinity. Seems oxymoron.

We also know

1.   Anything multiplied by zero is zero
2.   Anything multiplied by infinity is infinity other than zero

Which side is true?
Forget your multiplying by zero or infinity! The whole purpose of calculus is to avoid that.

Just follow the excellent piece Pezevenk wrote!

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #5 on: July 19, 2020, 05:08:51 AM »
I know what you say but I am thinking a lil bit differently. There are two opposite infinities of extreme are involved. One side is trying to drag the multiplication tends to zero while the other side stretch it tends to infinity. Seems oxymoron.
How?

Quote
We also know

1.   Anything multiplied by zero is zero
2.   Anything multiplied by infinity is infinity other than zero

Which side is true?
First of all the second isn't true because infinity is not a number and you can't really define multiplication by infinity. You can only use it when discussing behaviors of limits. What you are thinking is that taking the limit of something that tends to infinity multiplied by something that tends to anything else is either infinity, or if that something else is 0 then it might be 0 or it might be infinity or it might be something else. These limits are called indeterminate forms. For example, you are examining lim(a*b) and you know that lim(a)=infinity and lim(b)=0 then you have an indeterminate form. For example, if a(x)=x, b(x)=1/x and c(x)=1/x^2 then lim(a)=infinity, lim(b)=lim(c)=0, but lim(a*b)=1 and lim(a*c)=lim(1/x)=0. If instead of a(x) you had d(x)=x^2 which also tends to infinity then you'd have lim(d*b)=lim(x)=infinity and lim(d*c)=1. In some cases the limit may not even exist.

Your only issue is that you have a bit of trouble understanding the definitions, which, ok, is pretty common for beginners. You should stop trying to think of it as multiplying 0 by infinity because this is not what we are doing (and we can't do it because it is not defined). We are only using the definition of a limit (which is a very strict and very precise definition).
« Last Edit: July 19, 2020, 05:12:27 AM by Pezevenk »
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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #6 on: July 19, 2020, 05:34:32 AM »
You should look at this series by 3Blue1Brown to gain a better intuition:
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Re: Integral calculus/ Area under the curve
« Reply #7 on: July 19, 2020, 01:50:43 PM »
You should look at this series by 3Blue1Brown to gain a better intuition:

Both in differentiation and integration, the exact relationship between the curve and its slope is still unknown to the Illuminati however they can be traced if one of them is known. Here is the derivation of dy/dx in which dx tends to zero. Please see the attached Fig-A.

https://www.mathsisfun.com/calculus/derivatives-dy-dx.html -
Both dy/dx should be tended to zero not just dx in the link- RIGHT. Fig-A, if we look closely at points 1, 2,3,4 and 5 and so on, dx decrease while dy increases if we go up while deriving the equation of slope of the curve at those different points. Take two points at early stages of curve say point #1, both dx and dy don’t have the same length. The infinitesimal length of dx is greater than the infinitesimal length of dy. Perpendicular dy reaches zero earlier than dx if dx tends to zero. So there is a limit of infinitesimal length of dy for dx beyond which it can’t be reduced further toward zero or otherwise dy will be equal to zero if we cross that limit. Similarly, take two points at or above point #5. The infinitesimal length of dx is less than the infinitesimal length of dy. Since our goal is to bring both dx and dy toward zero therefore again we have trouble with the infinitesimal length of dy which will still be existed even if dx is equal to zero.

Similarly, look at the “Try It On A Function” in the aforementioned link, the final derived equation of dy/dx =2x or dy/dx=2x/1 which means dy=2x while dx =1. How come dx =1 when it says dx goes to zero if we look on the RHS of it.

Equation of dy/dx = 2x/1 which we don't know how does it relate to x^2 but it means "Gradient of the Tangent of Slope" = 2x/1 where dy=2x while dx=1 (always 1). The values of dy change when x changes.

The relationship between x and y doesn’t exist on x-axis and y-axis of the Cartesian coordinate system. It is the advancement of the gradient of the tangent of the slope after plugging the values of x in the equation of dy/dx in between the quadrant (any of 4), which trace the type (degree) of curve or vice versa. The gradient of the tangent of slope changes if the power/exponent of x changes and hence its onward motion w.r.t x or y which defines the degree of curve.

Fig-B: For curve y=x, y changes when x increases or decreases. The gradient of the tangent of slope (tangent theta is 45 degrees) is constant. A curve of first degree is obtained with the progression of the foregoing gradient or vice versa. Let A is the origin. B and D are the final values of x and y. Join B and D at C. A rectangle ABCD is always obtained after graphing all the values of x and y. The total area of this rectangle is xy [length (AB=CD) multiplied by height (AD=CD)].

Area under the curve, which we determine with the help of integration, is actually the ˝,1/3rd ,1/4rth  ……….1/nth area of the aforementioned rectangle ABCD for y=x, y=x2,y=x3 ……..……y=xn respectively.

When y=x, we get x^2/2 after integrating xdx or ydx as y=x. The dx method is not true for integration as explained above. Total area of the rectangle ABCD after graphing x and y is xy or x.x=x^2 as y=x.  The obtained diagonal of this rectangle is a straight line, which divides the said rectangle ABCD into half or two equal triangles ABC and ADC. Area of each triangle = area of rectangle ABCD/2 = xy/2 or x.x/2=(x^2)/2 as y=x. This turned out to be the area under the first-degree curve, which is the area of triangle (bh/2).

Fig-C: For y=x^2, we get x^3/3 after integrating x^2.dx or ydx. The diagonal of the rectangle ABCD is a second-degree curve, which divides the said rectangle ABCD into 2/3 and 1/3 though there is no real proof of it. Thus area under curve x^2 is =x^3/3 or total area of rectangle ABCD/3 = xy/3 (as y=x^2) therefore  =x.x^2/3 =x^3 which is the 1/3rd area of the ABCD rectangle.

Fig-D: For y=x^3, we get x^4/4 after integration. The diagonal of the rectangle ABCD is a curve of third degree, which divides the said rectangle ABCD into 3/4 and 1/4. Thus area under the curve x^3 =x^4/4 = total area of rectangle ABCD/4= xy/4 (as y=x^3) therefore = x.x^3/4=x^4/4 which is the 1/4rth area of the rectangle ABCD.

And so on. Hope this helps how dx works. The infinitesimal length dx decreases if we increase the # of rectangular strips in the original post. Although we know their boundaries say 0 and 1 but infinity exists in between 0 and 1 (dx tends to zero while dy tends to 1). Both never reach their destination. The second problem with the integration is the height of each rectangular strips varies. We ignore variation in heights due to infinite numbers of vertical stripes.   

The following may help how to trace a curve with the help of the gradient of the tangent of the slope. Play with point "P" in the following link.
https://www.geogebra.org/m/uh5pegsz
« Last Edit: July 19, 2020, 02:18:39 PM by E E K »

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #8 on: July 19, 2020, 02:18:29 PM »
Both in differentiation and integration, the exact relationship between the curve and its slope is still unknown to the illuminati however they can be traced if one of them is known.

? Are you joking or am I being trolled this entire time?

Quote
Both dy/dx should be tended to zero not just dx in the link- RIGHT. Fig-A, if we look closely at points 1, 2,3,4 and 5 and so on, dx decrease while dy increases if we go up while deriving the equation of slope of the curve at those different points.

Uhhh... No? I'm not sure what you are trying to say, but whichever way I try to interpret it it's not right. Yes, both dx and dy tend to 0. Yes, it is possible that dy might briefly increase as dx decreases. Yes it is possible that dy can increase if you keep dx stable but moving at another point in the function (in which case you are calculating the derivative somewhere else). Neither of those things should be surprising and neither of those things prevent dy from tending to 0 as dx tends to 0. I don't understand the problem.

Quote
If dx tends to zero then dy might have already reached to zero.

That's not how it works. Both 10/n and 1/n tend to 0 as n tends to infinity. 1/n is smaller than 10/n. Neither will "reach" 0 faster though. And even if dy does become 0 at some point before dx does, either briefly or for an entire interval (for example for a constant function, say f(x)=1, for which dy is always 0), that is not an issue.

Quote
So there is a limit of infinitesimal length of dy for dx beyond which it can’t be reduced further toward zero or otherwise dy will be equal to zero if we cross that limit.
That is not true unless your function becomes a straight horizontal line from that point and onwards.

Quote
Similarly, take two points above at or above point #5. The infinitesimal length of dx is less than the infinitesimal length of dy. Since our goal is to bring both dx and dy toward zero therefore again we have trouble with the infinitesimal length of dy which will still be existed even if dx is equal to zero.

This is again wrong for the same reason.

Quote
the final derived equation of dy/dx =2x or dy/dx=2x/1 which means dy=2x while dx =1.

That is not what it means. It only means dy gets closer and closer to become 2x times dx as dx gets closer and closer to 0. At the limit they both tend to 0, since 2x*(something that tends to 0) is 0. It doesn't mean dy is 2x and dx is 1.
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Re: Integral calculus/ Area under the curve
« Reply #9 on: July 19, 2020, 03:32:08 PM »
As said our goal is to bring both dx and dy tend to zero not just dx or dy.

Make it simpler

At Tan theta = 45 degrees, both dx=dy; both dx and dy tend to zero - ok

When Tan theta <45 degrees, dx >dy; dy reaches zero earlier than dx. So there is a limit of infinitesimal length of dy for dx beyond which dx can’t be reduced further toward zero. It means there is still chance for dx to tend to zero but dy diminish beyond that said point.

When Tan theta >45 degrees, dx <dy; dx reaches zero earlier than dy

« Last Edit: July 19, 2020, 03:35:46 PM by E E K »

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rabinoz

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Re: Integral calculus/ Area under the curve
« Reply #10 on: July 19, 2020, 03:32:47 PM »
You should look at this series by 3Blue1Brown to gain a better intuition:

Both in differentiation and integration, the exact relationship between the curve and its slope is still unknown to the Illuminati however they can be traced if one of them is known. Here is the derivation of dy/dx in which dx tends to zero. Please see the attached Fig-A.
You're an ungrateful little so-and-so aren't you?
Pezevenk spends a great deal of time carefully explaining it to you and you throw it back in his face.

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rabinoz

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Re: Integral calculus/ Area under the curve
« Reply #11 on: July 19, 2020, 03:55:47 PM »
As said our goal is to bring both dx and dy tend to zero not just dx or dy.
No it is not! The goal is to find the limit of dy/dx as dx -> 0 whatever that might be.

For the function y = tan x you will find that for x = π/2 dy/dx as dx -> 0 is infinite - as it should be.

Re: Integral calculus/ Area under the curve
« Reply #12 on: July 19, 2020, 09:55:42 PM »
As said our goal is to bring both dx and dy tend to zero not just dx or dy.
No it is not! The goal is to find the limit of dy/dx as dx -> 0 whatever that might be.

For the function y = tan x you will find that for x = π/2 dy/dx as dx -> 0 is infinite - as it should be.
Your and His Highness!

Our goal is to bring Secant line (touching a curve at 2 points) to a level of Tangent line (touching the curve at one point) while deriving the equation of dy/dx (slope of curve).

This is only possible when dy/dx=1 OR tangent theta =45 degree.

There will always be two points on the curve in the rest of the cases despite limiting dx to zero or = 0. This means secant never changes into a tangent line - Here dx or dy don't obey rules of limits.
« Last Edit: July 19, 2020, 10:09:53 PM by E E K »

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rabinoz

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Re: Integral calculus/ Area under the curve
« Reply #13 on: July 20, 2020, 01:50:42 AM »
As said our goal is to bring both dx and dy tend to zero not just dx or dy.
No it is not! The goal is to find the limit of dy/dx as dx -> 0 whatever that might be.

For the function y = tan x you will find that for x = π/2 dy/dx as dx -> 0 is infinite - as it should be.
Your and His Highness!

Our goal is to bring Secant line (touching a curve at 2 points) to a level of Tangent line (touching the curve at one point) while deriving the equation of dy/dx (slope of curve).

This is only possible when dy/dx=1 OR tangent theta =45 degree.

There will always be two points on the curve in the rest of the cases despite limiting dx to zero or = 0. This means secant never changes into a tangent line - Here dx or dy don't obey rules of limits.
Rubbish but have it your own way! Be a know-it-all who knows nothing If you insist!

Many functions (and tan(x) is one) have singular points where the function is not differentiable.

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #14 on: July 20, 2020, 02:40:14 AM »
As said our goal is to bring both dx and dy tend to zero not just dx or dy.
Our goal is to take a limit, which is a very precisely defined thing, nothing more and nothing less.

Quote

When Tan theta <45 degrees, dx >dy; dy reaches zero earlier than dx.
No, I already explained why you are wrong. If you think dy gets to 0 before dx does, feel free to attempt to calculate the number where it does, for any angle you like. You can't, because it is not possible, because it doesn't get to 0 before dx does. Remember, dy is defined by f(x+dx) - f(x). Say our function is f(x)=x/2. That represents a line passing through the origin with an angle greater than 45 degrees. Dy is x/2+dx/2-x/2=dx/2. Feel free to find a number for which dx/2 is equal to 0 but dx isn't.

Same goes for the case where dy>dx.
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Re: Integral calculus/ Area under the curve
« Reply #15 on: July 20, 2020, 04:08:21 AM »
As said our goal is to bring both dx and dy tend to zero not just dx or dy.
Our goal is to take a limit, which is a very precisely defined thing, nothing more and nothing less.
CORRECT – Take two points A and C on a curve of x^2. As mentioned multiple times there is a point beyond which dx can’t be tended to zero as dy diminish or reaches zero earlier than dx because the length of dy < length of dx when Tan theta < 45 degrees. Although there is dx but no dy means no slope. At this point, limiting dx to zero would be meaningless as we lost dy and slope.

Another try but in detail.

The gradient of the graph of y = x2 at any point is twice the value of x thereat. The process of finding the derivation of a gradient / slope of a function y=f(x) or y = x2 is as follow.

Pick any two points A and B close to each other on the curve of y =x2. The coordinates of A on the curve are (x, y) or (x, x2). Add Δx at A as usual. When x increases by Δx, then y increases by Δy. The x changes from x to (x +Δx) while y changes from y to (y + Δy) or f(x) to (x+Δx)2. Thus the x and y coordinates of B on the curve are (x + Δx, y + Δy) or ([x+Δx, (x+Δx)2]. Now the instantaneous rate of change is given by

Δy/ Δx = [(x + Δx)2 – x2] / [x + Δx - x]
Δy/ Δx = [x2 + Δx2+2xΔx − x2] / Δx
Δy/ Δx = [2x + Δx] / 1
Reduce Δx close to zero by taking limit (Δx to dx and Δy to dy)
dy /dx = 2x + dx
dy /dx = 2x------Eq1 OR
dy = 2x.dx --Eq2

ABC is an infinitesimal triangle made by dx, dy, and hypotenuse or slope of tangent where point A and C are always on the curve. Length of AB = Base = dx, Length of BC = Perpendicular= dy and Length of Hypotenuse = AC. Angle CAB or BAC is the slope of a tangent
According to the aforementioned Eq1 or Eq2-

• dy/dx is directly proportional to x or angle CAB is directly proportional to x.
• dx is indirectly proportional to x OR x is inversely proportional to dx
• dy is directly proportional to x.dx or dx

The length of dx > dy when Angle CAB < 45 degrees The length of dx = dy when Angle CAB = 45 degrees The length of dx < dy when Angle CAB > 45 degrees

The proportionality of both the angle CAB and dy with x are in contradiction with the proportionality of x and dx in the triangle ABC after probing the equation of dy/dx = 2x beyond its derivation on a graph of y = x2. When x increases; dx decreases, dy increases, and angle CAB increases. This means AC also increases and ultimately SECANT when x increases. Our goal is to bring dx, dy and AC to zero (not away from zero either positively or negatively - Point C has to be on the curve or secant to tangent by reducing them close to zero but here dx heads toward zero but dy and AC increases when x increases on axis mathematically.

Although the difference in the length of dx and dy can be noticed clearly on the graph if we examine the triangle ABC at two different points for a gradient (dy/dx), say at when an angle BAC = 0.1 degrees and 89.9 degrees on the curve.

RISE = dy = 2x and RUN = dx = 1 (always constant) in a GRADIENT of 1 in 2x which we obtained from the Eq1 of dy/dx=2x /1 at any point on the curve when there is no difference between secant and tangent – No idea how do we get dy/dx = 2x.dx but above said contradiction may be due to the introduction of another curve of y =(x+dx)2 at a point where we seize x or y=x2 deliberately and introduce delta x OR when function y = f(x) changed to y=f(x+Δx)2. The value of x has reached to its maximum value instead of unlimited when a curve y=x2 doesn’t continue anymore at a point where we introduce delta x or dx as y=x2 and y =(x+dx)2 are two different types of curve (two different functions).

Further, integration is the reverse process of differentiation. Although delta x or dx is ignored during the process of derivation of dy/dx because of their small values but we can’t ignore them in the process of integration which makes a lot of difference in summation. They can’t be disappeared forever and should resurface during the process of integration or summation.

Similarly, dy is the small vertical change in y, therefore, we take the sum of all the small vertical lengths [dy(s)] not the whole slice or y-coordinate(s) from zero to its value on the curve when we integrate both sides of the equation of dy = 2xdx but it turns into a function of x2 or area under the graph – no idea how but the summation of vertical lengths on a graph gives vertical length only not curve?

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #16 on: July 20, 2020, 09:57:53 AM »
As mentioned multiple times there is a point beyond which dx can’t be tended to zero as dy diminish or reaches zero earlier than dx because the length of dy < length of dx when Tan theta < 45 degrees.
As mentioned multiple times, including in the post you are quoting but ignored, this is wrong. You should clear up that misunderstanding first instead of confusing yourself further.
« Last Edit: July 20, 2020, 09:59:25 AM by Pezevenk »
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Re: Integral calculus/ Area under the curve
« Reply #17 on: July 21, 2020, 12:32:26 PM »
At origin, the curve y=x^2 pops out. Take a small distance delta x distance from origin. Its y coordinate will be delta y. Since y=x^2, therefore delta y = delta x^2

Now slope = delta y /delta x =  (delta x)^2/delta x = delta x = delta x/1

It depends how small delta x is

Introducing delta x in the middle of curve changes the function i.e. from y=x^2 to y=(x + delta x)^2. Both are two different curves. Curve y=(x + delta x)^2 starts where curve y=x^2 ends.

Here is point: delta y = (delta x)^2 (always) as y = x^2, not 2x * delta x + (delta x)^2
« Last Edit: July 21, 2020, 01:00:48 PM by E E K »

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #18 on: July 21, 2020, 12:58:52 PM »
At origin, the curve y=x^2 pops out. Take a small distance delta x distance from origin. Its y coordinate will be delta y. Since y=x^2, therefore delta y = delta x^2

Now slope = delta y /delta x =  (delta x)^2/delta x = delta x = delta x/1

It depends how small delta x is

Introducing delta x in the middle of curve changes the function i.e. from y=x^2 to y=(x + delta x)^2. Both are two different curves. Curve y=(x + delta x)^2 starts where curve y=x^2 ends.

Here is point: delta y = delta x^2 (always) as y = x^2, not 2x * delta x + (delta x)^2
They're not different curves. I don't even understand what that is supposed to mean.
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Re: Integral calculus/ Area under the curve
« Reply #19 on: July 22, 2020, 02:04:18 PM »
At origin, the curve y=x^2 pops out. Take a small distance delta x distance from origin. Its y coordinate will be delta y. Since y=x^2, therefore delta y = delta x^2

Now slope = delta y /delta x =  (delta x)^2/delta x = delta x = delta x/1

It depends how small delta x is

Introducing delta x in the middle of curve changes the function i.e. from y=x^2 to y=(x + delta x)^2. Both are two different curves. Curve y=(x + delta x)^2 starts where curve y=x^2 ends.

Here is point: delta y = delta x^2 (always) as y = x^2, not 2x * delta x + (delta x)^2
They're not different curves. I don't even understand what that is supposed to mean.
Last try on differential caculus  - sub topic

Fig -C: Would you be able to derive the equation of slope dy/dx for curve y = x^2 in this case when fixed point A is at origin. Take two points A and B on the curve such A is at the origin as shown in Fig -C. Point B slowly getting closer and closer to fixed point A. As line from A to B slowly gets nearer and nearer AND becoming tangent at A or origin as B gets nearer to A.

when x= Δx

Slope = Δy/ Δx = (Δx)^2/ Δx =  Δx where Δx tends to zero

Would you be able to derive the dy/dx in aforementioned scenario?

My original question was about the integral calculus

As asked earlier, integration is the reverse process of differentiation. Although delta x or dx is ignored during the process of derivation of dy/dx because of their small values but we can’t ignore them in the process of integration which makes a lot of difference in summation. They can’t be disappeared forever and should resurface during the process of integration or summation.

Similarly, dy is the small vertical change in y, therefore, we take the sum of all the small vertical lengths [dy(s)] not the whole slice or y-coordinate(s) from zero to its value on the curve when we integrate both sides of the equation of dy = 2xdx but it turns into a function of x2 or area under the graph – no idea how but the summation of vertical Δy(s) on a graph gives vertical length only not area under curve.
« Last Edit: July 22, 2020, 10:26:16 PM by E E K »

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #20 on: July 23, 2020, 02:09:51 AM »
You keep making the same errors over and over again and ignoring my replies. Are you or are you not interested in paying attention to what others say?

Quote
Slope = Δy/ Δx = (Δx)^2/ Δx =  Δx where Δx tends to zero

Would you be able to derive the dy/dx in aforementioned scenario?


Yes, and in fact I did earlier on in this thread. You take the limit of this expression as dx tends to 0, which is obviously 0.

Quote
Although delta x or dx is ignored during the process of derivation of dy/dx because of their small values but we can’t ignore them in the process of integration which makes a lot of difference in summation.


They're not "ignored", we take a limit. The derivative is not a "rough estimate" obtained by "ignoring" things, it's a precise concept defined by means of limits. But yes, they do in fact "resurface" in a sense. Riemann integrals are all about splitting the interval in finite elements of length dx and height dy, calculating and summing their areas and then taking the limit of that as the intervals of length dx that you've split the interval in goes to infinity, therefore dx tends to 0. That it is the inverse process of differentiation comes from the fundamental theorem of calculus. There are many proofs of the theorem that you can find online.

Quote
no idea how but the summation of vertical Δy(s) on a graph gives vertical length only not area under curve.

Simple, because it gives you the "length" of a function that gives you the area. If you want to find the area beneath a function f(x) which is the derivative of another function F(x), integrating f(x) is equivalent to adding up the dF's, not the df's, and as it turns out, F(x) is the function that measures the area under the curve.
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Re: Integral calculus/ Area under the curve
« Reply #21 on: July 23, 2020, 04:34:22 AM »
Quote
Slope = Δy/ Δx = (Δx)^2/ Δx =  Δx where Δx tends to zero

Would you be able to derive the dy/dx in aforementioned scenario?


Yes, and in fact I did earlier on in this thread. You take the limit of this expression as dx tends to 0, which is obviously 0.

Your equation must be equivelent to standard equation of dy/dx derived by Newton or Leibniz which can be applied for finding tangent of the slope any point on the curve . I can't see the said equation derived by you however you are requested to please go ahead and show it to all viewers even if you mentioned it earlier or otherwise this is my last post here

Quote
Although delta x or dx is ignored during the process of derivation of dy/dx because of their small values but we can’t ignore them in the process of integration which makes a lot of difference in summation.


They're not "ignored", we take a limit. The derivative is not a "rough estimate" obtained by "ignoring" things, it's a precise concept defined by means of limits. But yes, they do in fact "resurface" in a sense. Riemann integrals are all about splitting the interval in finite elements of length dx and height dy, calculating and summing their areas and then taking the limit of that as the intervals of length dx that you've split the interval in goes to infinity, therefore dx tends to 0. That it is the inverse process of differentiation comes from the fundamental theorem of calculus. There are many proofs of the theorem that you can find online.

Quote
no idea how but the summation of vertical Δy(s) on a graph gives vertical length only not area under curve.

Simple, because it gives you the "length" of a function that gives you the area. If you want to find the area beneath a function f(x) which is the derivative of another function F(x), integrating f(x) is equivalent to adding up the dF's, not the df's, and as it turns out, F(x) is the function that measures the area under the curve.
Alraedy answered - [hint: unit of area is sq.m, sq.ft etc while unit of length is foot, inches etc]


« Last Edit: July 23, 2020, 04:53:21 AM by E E K »

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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #22 on: July 23, 2020, 05:35:20 AM »
Quote
Slope = Δy/ Δx = (Δx)^2/ Δx =  Δx where Δx tends to zero

Would you be able to derive the dy/dx in aforementioned scenario?


Yes, and in fact I did earlier on in this thread. You take the limit of this expression as dx tends to 0, which is obviously 0.

Your equation must be equivelent to standard equation of dy/dx derived by Newton or Leibniz which can be applied for finding tangent of the slope any point on the curve . I can't see the said equation derived by you however you are requested to please go ahead and show it to all viewers even if you mentioned it earlier or otherwise this is my last post here
Dude I literally did that just now, I told you the result is 0. Idk what else you want or what it is exactly that you don't understand or THINK it has to look like. All you have to do is take the limit of [f(dx) - f(0)]/dx as dx tends to 0. You get lim(dx^2/dx)=lim(dx)=0. It's simple.

Quote
Alraedy answered - [hint: unit of area is sq.m, sq.ft etc while unit of length is foot, inches etc]
Hint: the different functions f and F correspond to different "units". F is a function of area, therefore of units of square meters etc, unlike f, if that's how you want to think about it. It's kind of a bad way to think about it but whatever.
« Last Edit: July 23, 2020, 05:47:24 AM by Pezevenk »
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Re: Integral calculus/ Area under the curve
« Reply #23 on: July 23, 2020, 02:26:13 PM »
Quote
Dude I literally did that just now, I told you the result is 0. Idk what else you want or what it is exactly that you don't understand or THINK it has to look like. All you have to do is take the limit of [f(dx) - f(0)]/dx as dx tends to 0. You get lim(dx^2/dx)=lim(dx)=0. It's simple.
- As y = x^2 therefore equivalent derivation of dy/dx = 2x was expected. Anyway, following may help.


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Pezevenk

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Re: Integral calculus/ Area under the curve
« Reply #24 on: July 24, 2020, 10:49:17 AM »
Quote
Dude I literally did that just now, I told you the result is 0. Idk what else you want or what it is exactly that you don't understand or THINK it has to look like. All you have to do is take the limit of [f(dx) - f(0)]/dx as dx tends to 0. You get lim(dx^2/dx)=lim(dx)=0. It's simple.
- As y = x^2 therefore equivalent derivation of dy/dx = 2x was expected. Anyway, following may help.


Help with what? You're the one who needs help here. What do you mean "equivalent derivation of dy/dx=2x was expected"? Are you complaining that the derivation for the specific case where x=0 does not LOOK identical to the one for the general case?
Member of the BOTD for Anti Fascism and Racism

It is not a scientific fact, it is a scientific fuck!
-Intikam

Read a bit psicology and stick your imo to where it comes from
-Intikam (again)