How to do antiderivatives

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submitted by Constant-Show2229 to Statisticshelpers_ [link] [comments]


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submitted by Repulsive_Union2244 to Statisticshelpers_ [link] [comments]


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IMPORTANT: When reaching out, please try to include the following information in the initial text message or email so that I can have all the important details necessary to determine the rate for my services:
OCT 2021 UPDATE: I am currently offering discount deals for requests for assistance with completing a student's entire course for the Fall 2021 semester (14 - 20 week courses acceptable), as well as discounts for students seeking help with multiple exams and/or multiple classes for Fall 2021. My availability for the Autumn 2021 / Fall 2021 semester will likely become limited very quickly as I receive more and more academic requests. Therefore it would be very advantageous to reach out to me for academic assistance before my schedule becomes too full.
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HOW TO CONTACT ME:
CONCLUSION:
OCT 2021 UPDATE: I am currently offering discount deals for requests for assistance with completing a student's entire course for the Fall 2021 semester (14 - 20 week courses acceptable), as well as discounts for students seeking help with multiple exams and/or multiple classes for Fall 2021. My availability for the Autumn 2021 / Fall 2021 semester will likely become limited very quickly as I receive more and more academic requests. Therefore it would be very advantageous to reach out to me for academic assistance before my schedule becomes too full.
submitted by douglaskeisenman to Studentcorner [link] [comments]


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submitted by Invite_Sprite to Yogscast [link] [comments]


2024.01.31 11:04 Revolutionary-Sky758 Calculus Assignments Help

Calculus Assignments Help


Calculus Assignments Help

Just like geometry is about shapes and algebra abounds in operations computations that determine equations’ accuracy, then calculus develops insight and accounts change as much comprises within.
Our team specializes in providing custom Calculus Assignments Help. We have always been fascinated by calculus, a mathematical branch that delves into rates of change and accumulation. The fundamental concepts of calculus, the derivative and the integral, play a crucial role in understanding how quantities change and accumulate over time.
Developed in the 17th century by brilliant minds like Isaac Newton and Gottfried Leibniz, calculus has evolved into a vital field with numerous real-world applications. Newton aimed to explain physics and motion, while Leibniz sought to formalize mathematical analysis. Since then, mathematicians have expanded calculus, making it an indispensable tool in various disciplines.
Learning calculus is not just about mastering abstract concepts; it’s about gaining the ability to model and comprehend change across diverse fields. It helps derive essential mathematical relationships crucial in STEM fields, fostering analytical thinking, logic, and problem-solving skills. Proficiency in calculus is often a prerequisite for many college majors and careers, particularly in technical fields.
“Nothing takes place in the world whose meaning is not that of some maximum or minimum.”― Leonhard Euler
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Types of Calculus

We delve into the fascinating realm of calculus, unraveling its diverse sub-fields that tackle distinct concepts and applications. Within the core branches, we explore the intricate study of rates of change, accumulation, and the extension of these ideas to functions with multiple variables.
Embarking on this journey, we encounter three main types of calculus. First is the realm of differential calculus, where we scrutinize slopes, derivatives, and rates of change. Integral calculus then beckons, delving into accumulated change, areas under curves, and various integration techniques. Lastly, multivariable calculus broadens our horizons, extending the calculus framework to functions with multiple inputs involving partial derivatives and multiple integrations.
Each facet of calculus equips us with invaluable tools and approaches to model and analyze both mathematical and real-world problems entwined with the dynamics of change. A comprehensive understanding of these calculus types aids in organizing and contextualizing the vast array of topics encapsulated under the calculus umbrella. In the forthcoming sections, let’s delve into each main branch of calculus, exploring its key concepts and the types of problems they unravel.
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Differential Calculus

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Integral Calculus

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Multivariable Calculus

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Key Calculus Concepts and Theorems

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In our pursuit of excellence, we recognize the pivotal role of fundamental concepts and theorems that form the bedrock of calculus. These include limits, derivatives, integrals, and series, each serving as indispensable analytical tools. Limits, for instance, enable us to define rates of change at a specific point, while derivatives form the basis for analyzing instantaneous change and optimizing functions. Integrals allow the accumulation of quantities over intervals and facilitate area calculations, while series sum sequences of terms and approximate functions. The interconnection of these core ideas is exemplified through crucial theorems, such as the fundamental theorem of calculus.
Our commitment to providing calculus assignments help stems from the understanding that a solid grasp of these foundational calculus concepts equips individuals with a framework for modeling dynamic systems and comprehending continuous change. Proficiency in these key concepts and theorems is imperative for tackling theoretical and applied problems across various domains, including mathematics, science, and engineering. Consider this overview as the starting point for delving deeper into the critical foundations of calculus with WritersABC.

Limits and Continuity

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In our pursuit of excellence, we recognize the pivotal role of fundamental concepts and theorems that form the bedrock of calculus. These include limits, derivatives, integrals, and series, each serving as indispensable analytical tools. Limits, for instance, enable us to define rates of change at a specific point, while derivatives form the basis for analyzing instantaneous change and optimizing functions. Integrals allow the accumulation of quantities over intervals and facilitate area calculations, while series sum sequences of terms and approximate functions. The interconnection of these core ideas is exemplified through crucial theorems, such as the fundamental theorem of calculus.
Our commitment to providing calculus assignments help stems from the understanding that a solid grasp of these foundational calculus concepts equips individuals with a framework for modeling dynamic systems and comprehending continuous change. Proficiency in these key concepts and theorems is imperative for tackling theoretical and applied problems across various domains, including mathematics, science, and engineering. Consider this overview as the starting point for delving deeper into the critical foundations of calculus with WritersABC.

Derivatives and Rules of Differentiation

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Integrals and Integration Techniques

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When tackling integrals, we delve into the profound significance of calculating the area under a curve and deciphering the total change involved. Our experts are well-versed in the various integration formulas, such as power, substitution, and integration by parts, equipping students with a diverse toolkit for solving problems.
At the heart of our approach is the fundamental theorem of calculus, which serves as a pivotal link between differentiation and integration. This connection forms the basis for our comprehensive assistance, ensuring that students grasp the intrinsic relationship between these fundamental mathematical operations.
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Series and Sequences

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Getting Help with Specific Calculus Classes

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Calculus I Help

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Understanding and Solving Different Types of Calculus Problems

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Optimization Problems

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Area and Volume Problems

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Motion Problems

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Finding Integrals and Antiderivatives

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Graphing Functions

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Approximating Areas Under Curves

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Infinite Limits

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Improper Integrals

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Parametric Equations

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2024.01.22 04:29 teledude_22 Trying to understand ordinary differential equation with an initial condition of x subscript 0

Hello, so it has been a while since I have taken Calculus 2, and I am trying to refresh my conceptual understanding of differential equation, but have encountered an initial condition I do not know how to approach.
Lets say we have d/dy x(y) = 8 and our given initial condition is x(0) = x with a subcript of 0.
First off, from my understanding is that "ordinary" differential equations are your basic differential equations you learn about in Calculus 2 that are not the more complex partial differential equations.
And then approaching this problem, I am thinking we need to take the antiderivative of both sides through integration. But I am confused about what d/dy means. This means the derivative, but how is "d/dy" different from "dy/dx"?
So we integrate both sides, get rid of the derivative on the left side and are left with x(y) and then on the right side just find the integral of our constant 8?
And then I am confused about what an initial condition of x(0) = x subscript 0. I am used to easy initial conditions like y(0) = 5, where now we know y = 5 and x = 0. But in my case here, we have an x on both sides, so x = 0, but what do I do with the x subscript 0?
Apologies for the messiness here, I am really just trying to re-grasp my conceptual understanding of solutions to ordinary differential equations with initial conditions, and keep getting overwhelmed by how deep online explanations and math youtube channels are going. Could someone please help guide me through how to conceptually think through solving this problem? Thank you!
submitted by teledude_22 to learnmath [link] [comments]


2024.01.17 14:37 MarioVX Manifold Production Delay & Ramp Up Time Analysis

Manifold Production Delay & Ramp Up Time Analysis

Introduction

When distributing a stream of input items to an array of processing buildings, Ficsit employees typically choose between two major design principles for their distribution belt network: manifolds and balancers. Manifolds are widely appreciated for their compactness, simplicity and extensibility.
It is well known that this comes at the (in most cases acceptable) cost of some delay in production behind the whole manifold, as the initially unbalanced distribution relies on the successive machines' internal buffers becoming filled and causing preceding belts to back up, causing the re-distribution of flow to the machines deeper in the manifold. Thus it takes some time for the production of the array as a whole to ramp up to full capacity.
But as the sparse responses to this post I stumbled across a few days ago show, it remains so far largely uninvestigated and unknown how long this delay really is, depending on the setup - even approximately. The purpose of the following analysis is to change that. u/Cris-Formage , consider this an extensive response to your question, and u/Gorlough, a generalization to your correct answer for the specific example discussed.

Method

Goal
For any given manifold, we would like to calculate two quantities of interest:
  1. ramp-up time - the time how long it takes from a cold start with empty buffers for the manifold to reach its maximum output rate, i.e. all attached processing buildings reaching 100% uptime going forward. This was the subject of the original question.
  2. production delay - how many items in total have been passed on to processing after any given time since the cold start, and how much less this is compared to an instantaneous start at maximum output as a balancer would achieve it. After the ramp-up time, this value becomes unchanging for any given manifold. I am introducing this second quantity because I believe it is more expressive of what we as players actually care about - namely by how much (or little) the manifold really sets us back.
Model
As usually in mathematical modeling, we need to make some difficult trade-offs between precision and universality. I want this analysis to be as universal as possible, so I have decided to ignore belt delays. These depend not only on the MK level of the belt, but also the exact lengths of belt segments and spaces between the buildings. If belt speeds are eventually changed or new MKs are introduced, the analysis would become outdated. Instead, we only consider the following:
  • c - peak input consumption rate of an individual processing building in items/min.
  • f - total, constant in-flow of items into the manifold in items/min.
  • n - the (integer) number of processing buildings attached to the manifold. Since this number is selected such that the entirety of the in-flow of items is consumed, and clock speed adequately adjusted, we can always assert that f = n * c.
  • bs - buffer stack size of the processing buildings. The number of items a processing building can load unprocessed before it is full and the preceding belt backs up.
That means in our model, even though the belts run at infinite speed (or equivalently have zero distance), the speed of the fill-up process as a whole is still limited by the in-flow of items and the buffers having to fill up first, which accounts for the majority of the total time. Especially for higher belt MK levels, the precision of this model increases.
Normalization
It turns out there is quite a bit of redundance in the above specification, which can be eliminated by normalization as a pre-processing step. This translates a wide range of manifolds with different recipe speeds and buffer sizes to a small set of canonical standard cases, and hence the results directly transferable:
We divide c, f & bs by c. This fixes c=1. It follows from f=n*c that f=n, hence f can be omitted as a parameter as well. Finally, instead of bs, we define b := bs/c. Since bs is in items and c in items per time, this quantity is a time - namely the buffer time of the individual processing buildings. That is, how many seconds or minutes of its own input consumption rate it would require to burn through its own filled buffer stack.
Example: We make a manifold for smelters smelting copper ore into copper ingots. The smelters consume 30 copper ore/min, this is c. Copper ore stacks up to 100, this is bs. Suppose our total in-flow into the manifold is 180 copper ore /min. Then we have n = 180/30 = 6, and b = 100/(30/min) = 3.(3) min = 200 sec.
This normalization thus reduces the number of relevant quantitative input parameters from 4 to 2. n and b are sufficient specification... except for one thing, and that's independent of the items, buildings and recipes involved:
Topology
As it turns out, there are two topologically distinct ways to construct a manifold:
  • "top-2": All splitters have 2 attached outputs: one goes into one processing building, the other extends the manifold. Without back-up, each splitter thus divides its received flow in two.
  • "top-3": All splitters except the last one have 3 attached outputs: two go into one processing building each, the third extends the manifold. The out-degree of the very last splitter depends on the parity of n: if n is even, it ends with only two outputs to the remaining two buildings. If n is odd, it ends with three, for the three remaining buildings. As we see later this difference is surprisingly impactful.
top-2 manifold
top-3 manifold. connectivity of the last splitter depends on parity of n, in this example even.
Both topologies qualify are manifolds by the usual understanding as they adhere to translational symmetry, making them easy to build, extensible and relatively compact. The at first glance obvious pros & cons are that top-2 is even more compact as it doesn't connect to the splitter outputs on the opposite side of the processing buildings, meanwhile top-3 uses only half as many splitters to connect the same number of machines which saves some system performance and counts up slower to the engine's object limit (splitters consist of multiple objects so this shouldn't be underestimated). But while all of these may be convincing arguments for one or the other in their own right, in this analysis we are only concerned with their behavior during the ramp-up process.
Algorithmic Computation
With all relevant quantitative and structural input parameters in place, it's time to actually perform the computation which will yield us the ramp-up time and later the production delay.
The following lends itself to automation via a script, which is how I got the results I will present later. But for small n, it is quite simple to do these with pen and paper, which is useful for verification purposes and quite instructive to make sure one understands the computational process.
The core idea is to essentially simulate the whole ramp-up process until the maximum output rate is reached. For this, we need to track the following quantities across time:
  • buffer fill state of each of the n buildings (as per our normalization in time worth of its own consumption rate). Initialized with 0 at t=0 and may never exceed b.
  • in-flow rates for each of the n buildings. When the building's buffer is full, this gets capped at the building's consumption rate (so as per our normalization, at most 1).
  • consumption rate for each of the n buildings. The rate at which the items are processed. At most 1 as per normalization. If the buffer is still empty, it is capped at 1 or the building's in-flow rate, whatever is lower.
  • net fill rate for each of the n buildings. This is a useful but not necessary, auxiliary variable. It is simply in-flow rate minus consumption rate and describes how quickly the buffer of the building is filling up.
  • finally, of course, time itself.
As it turns out, the whole process of filling up a manifold can be decomposed into distinct time segments where everything runs at constant rates, separated by critical transition points where some things change in an instant. These transition points are whenever another building's buffer is hitting its capacity limit. We want to evaluate the buffer states at the transition points, and all the inflow, consumption and fill rates during the segments (as the latter remain constant throughout one segment). From the time and buffer fill level at the previous point and the net fill rate for the next segment for the first building that has not yet capped out its buffer, we can calculate the duration of the segment. Finally with the duration of the segment and the net fill rates and previous buffer states of all subsequent buildings, we can calculate their new buffer fill states at the new transition point, and thus the cycle completes. This continues until the consumption rate of all n buildings reaches 1 for a new segment, indicating that the process is complete. The sum over the durations of all segments is the total time of the process, i.e. the ramp-up time of the whole manifold. One of two goals reached.
For the total processed items, we need the previously calculated durations of all segments individually, and in each segment the sum of the consumption rates over all buildings. The total processed items are then a piecewise defined linear function of time. If a queried time lies in segment k, sum up the product of total consumption rate and duration of all segments up to k-1, then add for the k-th segment the product of total consumption rate with just the time difference between the queried time and the last transition point.
For the production delay, we simply compare this production curve to that of a hypothetical load balancer - the linear function n * t. Beyond the last segment of the ramp-up process, the curves are parallel and thus have constant difference. This difference is the terminal production delay. But especially for comparing different manifolds, all the intermediary delays can be interesting too.

If this sounded a little technical or vague, you're invited to the following example. If it was already clear to you, skip ahead to the next section.
We're picking up the old example of a copper core manifold that translated to b=200sec, n=6. Suppose we connect it in top-3.
b_0 = 0, 0, 0, 0, 0, 0 i_0 = 2, 2, 2/3, 2/3, 1/3, 1/3 c_0 = 1, 1, 2/3, 2/3, 1/3, 1/3 n_0 = 1, 1, 0, 0, 0, 0 t_0 = (200 - 0)/1 = 200 b_1 = 200, 200, 0, 0, 0, 0 i_1 = 1, 1, 4/3, 4/3, 2/3, 2/3 c_1 = 1, 1, 1, 1, 2/3, 2/3 n_1 = 0, 0, 1/3, 1/3, 0, 0 t_1 = (200 - 0)/(1/3) = 600 b_2 = 200, 200, 200, 200, 0, 0 i_2 = 1, 1, 1, 1, 1, 1 ; terminal state T = 200 + 600 = 800 PD(t): 0 =< t =< 200: 4 * t 200 =< t =< 800: 800 + (5 + 1/3) * (t - 200) 800 =< t: 4000 + 6 * (t - 800) = -800 + 6 * t TPD = -800 
So it will take this manifold 800 seconds or 13 minutes and 20 seconds - plus the neglected belt delay times - to reach its maximum output rate from a cold start. By then, it will have accumulated a terminal production delay of 800 seconds worth of base consumption rate in items compared to a balancer that had cold started at the same time. To re-convert this into an actual item count, we can multiply with said consumption rate: 800 seconds * 0.5 items/second = 400 items of Copper Ore that it lags behind. If we instead want to convert this delay into a time rather than item delay for the whole manifold, we instead divide by n: 800 seconds / 6 = 133.33 seconds, or 2 minutes 13.33 seconds that the manifold as a whole is behind in production compared to a balancer (plus neglected belt delays).

Results

So, let's see what we got! There are some findings here that are surprisingly simple and seemed obvious to me in hindsight, nevertheless I didn't anticipate them beforehand, so I didn't want to take them away beforehand either. Then some other findings are just surprising, but not simple. Let's go through all of it:
Contribution of Buffer Time
This is a huge one. As complicated as the ramp-up time works out to be, it turns out that the buffer time is a multiplier that can be cleanly factored out to allow even more normalization!
I.e.: T(n,b,top) = b * T(n,1,top)
This translates to the accumulated production function as a stretching in x-direction. The transition points' times are multiplied by b and so are the production amounts at these points. As such, the TPD is multiplied by b as well.
This means that henceforth, the buffer can be ignored. We understand the following time values as multiples of the buffer time, and production quantities as buffer time worth of individual consumption rate in items.
But why is the total ramp-up time proportional to buffer time? Well, the very first segment's time is proportional to it: T_0 = (b-0)/x = b * 1/x, and the subsequent segments are proportional if the preceding segments time and hence buffer fill states are proportional: T_n+1 = (b - b_n,b)/x = (b - b * b_n,1)/x = b * (1 - b_n,1)/x. It follows by induction that the total time is proportional too.
Terminal Production Delay
It turns out there is an easy shortcut to the TPD of a manifold: Think about where the items are going that have entered the manifold but not exited it through processing. Since our belts have no capacity, they must all be hung up in building buffers. So we only need to imagine the buffer fill states in the terminal segment (which has 100% production) and sum them up.
  • In top-2, all but the last two buildings will have full buffers, and the last two buildings will have empty buffers. TPD = (n-2) * b
  • In top-3 with even n, it's the exact same. TPD = (n-2) * b
  • In top-3 with odd n, all but the last three buildings will have full buffers, and the last three buildings have empty buffers. TPD = (n-3) * b
As I prefaced, kind of obvious in hindsight, perhaps you saw it coming, for some reason I did not so here it is.
This means if you compare topologies based on the criterion of TPD alone, top-2 and top-3 are equal for even n, top-3 is only better for odd n.
Transient Production Delays
Perhaps you're not just interested in the terminal delays, as perhaps you already have use for a smaller quantity of produced items that can be obtained before a complete ramp-up of the manifold. So let's look at the ramp-up process output dynamically. As the TPD hints, it is quite important to distinguish by parity of n. The differences are more apparent for smaller n, so here are the production graphs for n=5 and n=6:
https://preview.redd.it/26ve4ooxjzcc1.png?width=600&format=png&auto=webp&s=5017107fcc60aadd1d1f4a977cc65527597783d1
https://preview.redd.it/hapujooxjzcc1.png?width=600&format=png&auto=webp&s=9ab959af7d59459bf0f5e8d09793838587391ddf
As we can see here, top-3 gets a head start on production. For even n, top-2 catches up to be tied in the terminal state by reaching its max production slightly sooner. Nevertheless, at any point in time, top-3 is ahead of or even with top-2 in terms of accumulated production. For odd n, top-3 is also always ahead or even with top-2, but as we know from the previous result maintains a genuine lead in the end.
Ramp-up time dependence on n
Finally, the last and most difficult piece of the puzzle. How does a growing number of attached buildings (and hence depth of the manifold, and multiplicity of the input stream) influence the ramp-up time of the manifold? Well, without further ado:
linear plot of ramp-up times vs n for both topologies, for small n
semi-log plot of ramp-up times vs n for both topologies, with logarithmic regression curves for top-2 and for either parity n with top-3, for larger n
Pay attention to the logarithmic scaling of the x-axis in the second plot. The behavior for large n attunes to a logarithmic function, not a linear function as the scaled plot may suggest at first glance.
The logarithmic regressions don't fit well for very small n. The values may be read off the first plot, but here is a little lookup table with the values to three decimal places for reference:
n top-2 time top-3 time
2 0 0
3 2 0
4 3 3
5 3.5 1.5
6 3.875 4
7 4.163 2.25
8 4.4 4.6
9 4.591 2.75
10 4.754 5
11 4.897 3.083
12 5.024 5.289
13 5.137 3.339
14 5.241 5.518
15 5.336 3.546
16 5.423 5.708
17 5.503 3.721
18 5.578 5.870
19 5.648 3.872
20 5.713 6.011
Any specific n-value you're interested in for your in-game projects? Write it into the comments, I will compute them and add to the table below:
n top-2 time top-3 time note

Discussion

Evaluation of Results, Practical Advice
It is eye-catching how extremely much faster top-3 is for odd n than both for even n and top-2. Even a lot more machines can be ramped up in shorter time this way. The difference is so vast I initially suspected an error in my code, but manually re-calculating with pen & paper revealed these numbers to be correct and this extreme zig-zagging behavior to be genuine. This has an immediate practical application: When concerned with ramp-up time, overbuild to an odd number (possibly underclock) and connect in top-3.
For even n, top-2 reaches maximum output rate slightly faster than top-3 - however keep in mind the previous result that nevertheless, top-3 is still ahead or even at all times in the number of items it has actually outputted. Intuitively, top-3 distributes the items "more evenly" than top-2. This gets buildings further down the manifold working sooner (and hence output up quicker), but it fills the buffers of earlier buildings slower (and hence reach full buffers later). So here the choice depends on how you value stableness versus earliness of the output (and the other considerations briefly hinted at in the introduction, not the topic of this analysis).
Origin of the roughly logarithmic dependence
Finally, one might be wondering, why the hell the ramp-up time depends roughly logarithmically on n?
My best explanation goes like this: Consider a slightly simplified ramp-up process, where only the in-flow into the buildings at the first non-filled splitter (and before) is considered, and the rest - rather than already slightly filling successive buildings - simply vanishes. Let's assume top-2. Then the first building fills up (normalized buffer) in time 1/(n/2) = 2/n. After it is full, the second splitter receives only n-1 flow (because 1 flow goes and is consumed by the first, filled, building). Only (n-1)/2 goes into the second building, so the time needed to fill it in our simplified model is 1/((n-1)/2) = 2/(n-1). The next one will be 2/(n-2), then 2/(n-3), and so on, all the way down to 2/1. When we add these up, we have T = 2/1 + 2/2 + ... + 2/n = 2 * (1/1 + 1/2 + ... + 1/n). The sum in parentheses has a name, it's called the n-th Harmonic number. Famously the Harmonic numbers can be asymptotically approximated with the natural logarithm and the Euler-Mascheroni constant (about 0.577) as H_n ~ ln(n) + 0.577 for large n. For readers familiar with calculus, it may help to consider that the antiderivative of 1/x is ln(x) to make sense of this. If we plug this in for this simplified ramp-up process, we get T_n ~ 1.154 + 2 ln(n).
A closer comparison of the simplified with the more accurate ramp-up process from our full model reveals that this simplified one must always be slower to ramp-up than the complete one, as we only let flow vanish and not create more. This means the times derived from the formula for the simplified process are a reliable upper bound for the times of the accurate process. This means the accurate process' ramp-up time can grow at most logarithmically with n.

Closing Thoughts

This was a surprisingly vast rabbit hole to delve in, but I'm happy with the clarity of the results. We finally got some quantitative estimates on by how much a manifold actually delays your production until it's ramped up to parity with a balancer that instead might have been more elaborate to plan and build and take away more space. This wasn't done before to this extent in the Satisfactory community as of my knowledge.
Some aspects or doubts you want to discuss? Some part of the derivation you wanted to but couldn't quite follow along and want a more thorough explanation? Some specific values you want the time to be computed for? Other thoughts? Please comment!
If you feel like these results are worth buying me a coffee for my time, you can. Thanks!
Now, happy manifolding and back to work, for Ficsit!
submitted by MarioVX to SatisfactoryGame [link] [comments]


2024.01.14 09:19 Friendly-Employ-5595 How do definite integrals work?

I read through some previous posts briefly but couldn’t find one that matched my question.
So I understand how to compute integrals. I understand what a Riemann sum is and how it works but I don’t get why finding an antiderivative helps us accumulate area. I (kinda) know how to compute a Riemann sum and know it will yield the same answer but I do not understand why. Like how is it connected to the antiderivative?
Thanks for any answers yall.
submitted by Friendly-Employ-5595 to calculus [link] [comments]


2023.12.12 20:42 CheshireKat-_- [calc] How do I know when to take the derivative and when to take the antiderivative in a problem where you have to evaluate the integral by making the given substitution? In the first problem, I took the derivative but in the second it wants me to take the antiderivative in the expliantion

[calc] How do I know when to take the derivative and when to take the antiderivative in a problem where you have to evaluate the integral by making the given substitution? In the first problem, I took the derivative but in the second it wants me to take the antiderivative in the expliantion submitted by CheshireKat-_- to HomeworkHelp [link] [comments]


2023.12.05 21:30 BDady Is there a nice way on conceptually understanding the FTOC part 2?

I'm currently a diff eq student, so I'm familiar with the differential and integral calculus your typical university student would learn. I'm currently writing a document for my physics professor that teaches students without calc 1 experience all the calculus they need to be successful in the course. I've started the chapter on integral calculus, and I've come to realize I honestly don't know why the 2nd part of the FTOC is true aside from some of the proofs you'd normally see. Proofs are great, but I'm looking more for a conceptual understanding. So allow me to better formulate my question.
Forget about definite/indefinite integrals. Lets say we have these three mathematical tools:
  1. Derivatives
  2. Antiderivatives
  3. "Area under a curve" tool
(Yes I know 2 and 3 are indefinite and definite integrals, but my point is to remove any indicator that they're related)
Suppose we have some function, f(x), and we want to know what the area under its curve is between points a and b. Lets say we denote this with the function A(x), which is the area under f(x) from x=a to x. So the area under f(x) from a to b is A(b).
Denoting the antiderivative f(x) as F(x) as usual, the FTOC Part 2 says:
A(b) = F(b) - F(a)
Why? Conceptually, how does an antiderivative have anything to do with the area under a curve? Just to reiterate, I've seen the proofs for it, and while there are some digestible ones I could throw into this document, I would much rather conceptually explain why its true rather than showing some seemingly abstract math.
submitted by BDady to askmath [link] [comments]


2023.11.23 16:34 hpxvzhjfgb Things taught in high school math classes that are false or incompatible with real math

I'm collecting a list of things that are commonly taught in high school math classes that are either objectively false, or use notation, terminology, definitions, etc. in a way that is incompatible with how they are used in actual math (university level math and beyond, i.e. what mathematicians actually do in practise).
Note: I'm NOT looking for instances where your high school math teacher taught the wrong thing by mistake or because they were incompetent, I'm only looking for examples of where the thing that they were actually supposed to teach you was wrong or inconsistent with real math. E.g. if your teacher taught you that log(a+b) = log(a)+log(b) because they are incompetent, that's not a valid example, but if they taught it to you because that's what is actually in the curriculum, then that would be an example of what I'm looking for.
Examples that I know of:
  1. Functions are taught in two separate, incompatible ways. In my high school math classes, functions were first introduced as being equations of the form y = [expression in x], which is wrong because the statement that two numbers are equal is not the same thing as a map between sets. Later (maybe more than a year later?), the f(x)-style functions were introduced as a separate concept. Of course in real math, f(x)-style functions are what people actually use.
  2. I can't count how many times I've seen people post problems of the form "find the domain of f(x)". In real math, the domain and codomain are part of the definition of the function, not something that is deduced from a formula.
  3. In one of my A level maths classes, functions were covered yet again for some reason, except this time we were taught the notation f : A -> B to mean that f is a function from A to B. Except we were taught that A is called the domain, and B is called the range, not the codomain. In real math, B is called the codomain, and the range (or image) is a subset of the codomain.
  4. In calculus classes, it's extremely common for integration and antidifferentiation to be conflated to such a degree that people think they are exactly the same thing. Probably calling antiderivatives "the indefinite integral" doesn't help either. People are taught that integration is the inverse of differentiation, which isn't true. It's not even the left inverse or the right inverse. There are functions that can be integrated but which have no antiderivative, and there are functions that have antiderivatives but which can not be integrated.
  5. Before seeing the formal definition of limits and continuity, it's common for people to be taught that 1/x is discontinuous, when it isn't. All elementary functions are continuous.
  6. Apparently, given an expression of the form a + b, high school math says that the conjugate of a + b is a - b. This is obviously not even a well-defined operation (consider the conjugate of b + a). This might be a US-only thing because this was never taught in my high school math classes.
  7. In calculus classes, people are taught that the general form of an antiderivative (or, sigh, the "indefinite integral") of 1/x is ln(x)+c. This is wrong because R\{0} is not connected which means you can add different constants on the positive and negative axes, e.g. ln(x) + (1 if x>0, 2 otherwise).
  8. In calculus classes, people are told that dy/dx isn't a fraction, which is correct, but they are still taught to do manipulations like u = 2x => du/dx = 2 => du = 2dx when learning about integration by substitution. It is barely any more work to do it properly and show that the chain rule is being used.
There are probably several more that I can't think of right now, but you get the idea. Have you experienced any other examples of this?
submitted by hpxvzhjfgb to math [link] [comments]


2023.11.15 02:33 andrewl_ What is the name for this simple and stateful approach to integration? Why isn't it taught?

Suppose I want to simulate 10 seconds of an initially stationary vehicle moving, with acceleration of 1m/s/s.
If I sample once per second, that's 10 samples, and I have acceleration values:
a = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1} 
And velocity would just accumulate acceleration, creating:
v = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 
And position, starting at 0, would just accumulate velocity, creating:
p = {1, 3, 6, 10, 15, 21, 28, 36, 45, 55} 
I think this simple act of accumulating list values is integration, and it's happening without complicated symbols, just addition.
One function contributes to the rate of change to the next, and the daisy-chained configuration kind of looks like a ladder or staircase when drawn like this. Moving up the stairs is integration; moving down is derivation.
If I use integration rules, I get from a(t)=1 to v(t)=x. Then again from v(t)=x to p(t)=(1/2)*t2. Comparing the values in my above list labelled "p" to the function p(t), they are close in shape.
If I sample twice as much (20 samples now, one every half second) and take care to scale each contribution by one half, it results in:
a = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} v = {0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, 9.5, 10.0} p = {0.25, 0.75, 1.5, 2.5, 3.75, 5.25, 7.0, 9.0, 11.25, 13.75, 16.5, 19.5, 22.75, 26.25, 30.0, 34.0, 38.25, 42.75, 47.5, 52.5} 
And the resulting graph is even closer to the true integral! Increasing to 100 samples and it's nearly dead on.
I know different approaches click for different people, but hooking up a bunch of functions so one adds to the rate of change of the next is something I wish I'd seen a long time ago, way before the integral sign. My calc teacher spent a long time drawing area-covering strips, and making the strips slimmer, and slimmer yet, motivating the "area under the curve" interpretation. I never understood how that was related to the antiderivative; it seemed like just a geometer's tool. But with the above exercise, it's clear what those strips are: just the lower function's values, which accumulate in the rate of change of the function its hooked up to.
EDIT: Many people identified this as Riemman sum (RS) and I see that the values I listed are the areas of the rectangles used by RS to approximate the area under a curve. The part that's new to me is what this area was doing (defining the rate of growth of the next function) and how simply it could be calculated. As another example, since repeated derivation of f(x)=x3 results in 3x2 then 6x then 6, we can build f(x) back by simply accumulating a list of 6's. In python:
f = [6]*10 f = accumulate(f, add) f = accumulate(f, add) f = accumulate(f, add) 
Which looks even simpler in this form:
c, d, e, f = 6, 0, 0, 0 for _ in range(10): d += c e += d f += e 
That this works was evidently obvious to everyone except me. I think I would have benefited had my teacher started with "Let me show you just how simple integration can be." and after the demonstration continued "Now let's formalize it a bit with this technique called RS."
submitted by andrewl_ to askmath [link] [comments]


2023.11.09 18:10 felixnice2 PDE problem

I have this kind of solution to the wave equation: https://imgur.com/a/NSpRO2e
It confuses me at some points, maybe someone could help me out.
I don't understand why, after reducing to a transport equation, they are integrating from 0 to s all those ODEs: t'=1, x'=-1 and z'=f(t-x). How do they know that they should be integrating from 0 to s ? I would have just computed the antiderivatives.
And also, at the end, the solution looks incomplete... it's left as an integral.. I don't really understand why that is the end of the solution
submitted by felixnice2 to MathHelp [link] [comments]


2023.11.09 18:10 felixnice2 PDE problem

I have this kind of solution to the wave equation: https://imgur.com/a/NSpRO2e
It confuses me at some points, maybe someone could help me out.
I don't understand why, after reducing to a transport equation, they are integrating from 0 to s all those ODEs: t'=1, x'=-1 and z'=f(t-x). How do they know that they should be integrating from 0 to s ? I would have just computed the antiderivatives.
And also, at the end, the solution looks incomplete... it's left as an integral.. I don't really understand why that is the end of the solution
submitted by felixnice2 to askmath [link] [comments]


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