What is the purpose of finding Area between Curves? [on hold]
I've seen many explanations on how to find it; however, I can't seem to find what the purpose is of calculating the area between curves.
calculus soft-question area motivation
New contributor
put on hold as off-topic by Lord Shark the Unknown, Clarinetist, DRF, Servaes, TheSimpliFire yesterday
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I've seen many explanations on how to find it; however, I can't seem to find what the purpose is of calculating the area between curves.
calculus soft-question area motivation
New contributor
put on hold as off-topic by Lord Shark the Unknown, Clarinetist, DRF, Servaes, TheSimpliFire yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – DRF, Servaes, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
36
You might as well ask what the point of computing areas is whatsoever
– Wojowu
2 days ago
3
Maybe it's just a neat example of a kind of geometry problem that can be solved using calculus. If you're wondering why integration is useful (who cares about areas, anyway?) one major reason is that integration can be used to find the total change in a quantity if we know the instantaneous rate of change of that quantity. For example, knowing an object's velocity, you can figure out its position.
– littleO
2 days ago
2
Imho you're asking the wrong kind of question. For most common, elementary math concepts you should not worry about the applications, things will come together once you have a broader understanding of mathematics. It's not reasonable to assume that any isolated math concept will have tons of applications that appear in practice , but that doesn't mean that knowing about it isnt useful. You might as well ask what the purpose of learning about division is or, for an analogy outside of mathematics, why you should learn a specific word.
– user159517
2 days ago
2
How exactly do you define area between curves? Indeed, finding the area between curves by Riemann integration can be considered meaningless because at this level the area is pretty much defined to be the respective integral. What could be considered interesting is to show that the Lebesgue measure (the only measure on $mathbb{R}^2$ such that the measure of every rectangle is equal to its area) of the part of the plane between curves is equal to the integral.
– mechanodroid
2 days ago
2
"Area under the curve" is just an informal illustration of the concept of integral. Actually, an integral is the continuous version of a sum. It is as natural an operation as summing a finite set of numbers.
– Giuseppe Negro
yesterday
|
show 3 more comments
I've seen many explanations on how to find it; however, I can't seem to find what the purpose is of calculating the area between curves.
calculus soft-question area motivation
New contributor
I've seen many explanations on how to find it; however, I can't seem to find what the purpose is of calculating the area between curves.
calculus soft-question area motivation
calculus soft-question area motivation
New contributor
New contributor
edited yesterday
Martin Sleziak
44.7k8115271
44.7k8115271
New contributor
asked 2 days ago
EpicNicksEpicNicks
5815
5815
New contributor
New contributor
put on hold as off-topic by Lord Shark the Unknown, Clarinetist, DRF, Servaes, TheSimpliFire yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – DRF, Servaes, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Lord Shark the Unknown, Clarinetist, DRF, Servaes, TheSimpliFire yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – DRF, Servaes, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
36
You might as well ask what the point of computing areas is whatsoever
– Wojowu
2 days ago
3
Maybe it's just a neat example of a kind of geometry problem that can be solved using calculus. If you're wondering why integration is useful (who cares about areas, anyway?) one major reason is that integration can be used to find the total change in a quantity if we know the instantaneous rate of change of that quantity. For example, knowing an object's velocity, you can figure out its position.
– littleO
2 days ago
2
Imho you're asking the wrong kind of question. For most common, elementary math concepts you should not worry about the applications, things will come together once you have a broader understanding of mathematics. It's not reasonable to assume that any isolated math concept will have tons of applications that appear in practice , but that doesn't mean that knowing about it isnt useful. You might as well ask what the purpose of learning about division is or, for an analogy outside of mathematics, why you should learn a specific word.
– user159517
2 days ago
2
How exactly do you define area between curves? Indeed, finding the area between curves by Riemann integration can be considered meaningless because at this level the area is pretty much defined to be the respective integral. What could be considered interesting is to show that the Lebesgue measure (the only measure on $mathbb{R}^2$ such that the measure of every rectangle is equal to its area) of the part of the plane between curves is equal to the integral.
– mechanodroid
2 days ago
2
"Area under the curve" is just an informal illustration of the concept of integral. Actually, an integral is the continuous version of a sum. It is as natural an operation as summing a finite set of numbers.
– Giuseppe Negro
yesterday
|
show 3 more comments
36
You might as well ask what the point of computing areas is whatsoever
– Wojowu
2 days ago
3
Maybe it's just a neat example of a kind of geometry problem that can be solved using calculus. If you're wondering why integration is useful (who cares about areas, anyway?) one major reason is that integration can be used to find the total change in a quantity if we know the instantaneous rate of change of that quantity. For example, knowing an object's velocity, you can figure out its position.
– littleO
2 days ago
2
Imho you're asking the wrong kind of question. For most common, elementary math concepts you should not worry about the applications, things will come together once you have a broader understanding of mathematics. It's not reasonable to assume that any isolated math concept will have tons of applications that appear in practice , but that doesn't mean that knowing about it isnt useful. You might as well ask what the purpose of learning about division is or, for an analogy outside of mathematics, why you should learn a specific word.
– user159517
2 days ago
2
How exactly do you define area between curves? Indeed, finding the area between curves by Riemann integration can be considered meaningless because at this level the area is pretty much defined to be the respective integral. What could be considered interesting is to show that the Lebesgue measure (the only measure on $mathbb{R}^2$ such that the measure of every rectangle is equal to its area) of the part of the plane between curves is equal to the integral.
– mechanodroid
2 days ago
2
"Area under the curve" is just an informal illustration of the concept of integral. Actually, an integral is the continuous version of a sum. It is as natural an operation as summing a finite set of numbers.
– Giuseppe Negro
yesterday
36
36
You might as well ask what the point of computing areas is whatsoever
– Wojowu
2 days ago
You might as well ask what the point of computing areas is whatsoever
– Wojowu
2 days ago
3
3
Maybe it's just a neat example of a kind of geometry problem that can be solved using calculus. If you're wondering why integration is useful (who cares about areas, anyway?) one major reason is that integration can be used to find the total change in a quantity if we know the instantaneous rate of change of that quantity. For example, knowing an object's velocity, you can figure out its position.
– littleO
2 days ago
Maybe it's just a neat example of a kind of geometry problem that can be solved using calculus. If you're wondering why integration is useful (who cares about areas, anyway?) one major reason is that integration can be used to find the total change in a quantity if we know the instantaneous rate of change of that quantity. For example, knowing an object's velocity, you can figure out its position.
– littleO
2 days ago
2
2
Imho you're asking the wrong kind of question. For most common, elementary math concepts you should not worry about the applications, things will come together once you have a broader understanding of mathematics. It's not reasonable to assume that any isolated math concept will have tons of applications that appear in practice , but that doesn't mean that knowing about it isnt useful. You might as well ask what the purpose of learning about division is or, for an analogy outside of mathematics, why you should learn a specific word.
– user159517
2 days ago
Imho you're asking the wrong kind of question. For most common, elementary math concepts you should not worry about the applications, things will come together once you have a broader understanding of mathematics. It's not reasonable to assume that any isolated math concept will have tons of applications that appear in practice , but that doesn't mean that knowing about it isnt useful. You might as well ask what the purpose of learning about division is or, for an analogy outside of mathematics, why you should learn a specific word.
– user159517
2 days ago
2
2
How exactly do you define area between curves? Indeed, finding the area between curves by Riemann integration can be considered meaningless because at this level the area is pretty much defined to be the respective integral. What could be considered interesting is to show that the Lebesgue measure (the only measure on $mathbb{R}^2$ such that the measure of every rectangle is equal to its area) of the part of the plane between curves is equal to the integral.
– mechanodroid
2 days ago
How exactly do you define area between curves? Indeed, finding the area between curves by Riemann integration can be considered meaningless because at this level the area is pretty much defined to be the respective integral. What could be considered interesting is to show that the Lebesgue measure (the only measure on $mathbb{R}^2$ such that the measure of every rectangle is equal to its area) of the part of the plane between curves is equal to the integral.
– mechanodroid
2 days ago
2
2
"Area under the curve" is just an informal illustration of the concept of integral. Actually, an integral is the continuous version of a sum. It is as natural an operation as summing a finite set of numbers.
– Giuseppe Negro
yesterday
"Area under the curve" is just an informal illustration of the concept of integral. Actually, an integral is the continuous version of a sum. It is as natural an operation as summing a finite set of numbers.
– Giuseppe Negro
yesterday
|
show 3 more comments
8 Answers
8
active
oldest
votes
The best application that comes to mind is computing the center of mass of an object.
Let $f(x)geq g(x)$ on $(a,b),$ then the center of mass of the area bound by the curves $f$ and $g$ is given by the coordinates
$$bar{x}=dfrac{int_a^bf(x)^2-g(x)^2dx}{2int_a^b f(x)-g(x)dx}$$
and
$$bar{y}=dfrac{int_a^bx(f(x)-g(x))dx}{int_a^b f(x)-g(x)dx}.$$
So the center of mass is $(bar{x},bar{y}).$
Now, you might wonder "why is that useful?" That's a good question.
Imagine you have designed a flat, or approximately flat object, which you'd like to move using a single suction cup/suction cup like device. For example, the glass which forms the screen of a modern smart phone. To reduce stress on the glass when picking it up and moving it around, you can pick it up by its center of mass, and distribute the stress uniformly.
add a comment |
There are many applications for the area between curves. For example if the curves represent the speeds,then the area is the difference between the total distance traveled.
Another application is for two investments with different interest rates and the area represents the difference between the total interest gained by accounts.
add a comment |
For instance, computing that area of the disk$${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant r^2}$$ is the same thing as computing the areas between the curves$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&sqrt{r^2-x^2}end{array}$$and$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&-sqrt{r^2-x^2}end{array}.$$So, unless you find the problem of computing the area of a disk beneath you, you can see that it is as interesting as the problem of finding the area between two specific curves.
add a comment |
There are loads of examples in the real world outside of pure mathematics.
In terms of materials, finding the area under the stress-strain curve tells that material's toughness - how much abuse (energy) it can take (absorb) before it begins to fracture. If you take the difference between two materials' toughness, you can see which one is tougher, which is something that's not always apparent at just looking at their graphs.
Example: Say we have materials A and B, each with stress-strain curves given by $sigma(epsilon)$.
begin{cases}
begin{align}
sigma_A(epsilon) &= 5 epsilon, &epsilon in (0,1)\
sigma_B(epsilon) &= tanhepsilon, &epsilon in (0,3)
end{align}
end{cases}
These functions, by the way, are wildly unrealistic in most occurrences, but the concept is still relevant.
$hskip 1 in$
It's not exactly clear which material has the higher toughness. Material A's looks like a lot more, but maybe the longevity of material B makes up for its low yield. We could find the area under each and see which is higher but we could assume material A is tougher and find the area between A and B.
$$Delta U = int_0^3 Big( 5epsilon big(1 - h(epsilon - 1)big) - tanh epsilon Big) , depsilon =frac{5}{2} - logcosh3approx 0.19.$$
So material A is actually hardly more tough than material B.
In terms of systems, you can design an integral-feedback control system that reduces the error in the output from the desired output. In practice, these are PID controllers (Proportional, Integral, Derivative), all 3 of which are focused on changing the actual output to what it should be based on the desired output. The goal is for the error to be zero, i.e. in the integral controller, you're looking for the area between the desired output and actual output to be exactly zero.
For instance, suppose you're driving and you click on the cruise control. You've just told the car you want it to keep going the speed that you are going right this moment. Now the road starts leading you uphill. The car is going to slow down initially, but it is also going to do everything it can to keep you going how fast you told it to go. That is, going uphill, it has encountered some error between the actual speed and desired speed and it wants to bring that error back to zero. The same applies for going downhill as well.
Another example could be controlling the temperature inside an oven, perhaps an expensive one with variable heating elements using a similarly styled controller.
These are just a few examples; there's really no shortage of applications.
add a comment |
Economics.
Example 1. The area between the demand (supply) function and the market-clearing (equilibrium) price shows the consumers (producers) surplus:
$hspace{3cm}$
Source: Wikipedia.
Note that the consumers are willing to pay a price for a good up to their demand function, but they pay the equilibrium price and save the difference (red area).
Example 2. The area between the marginal revenue and the marginal cost functions shows the net profit (loss).
$hspace{3cm}$
Source: Slide $13$ at Slideshare.
I may be wrong, but doesn't profit = revenue - total costs? Why would it be the area under revenue - total costs?
– Daniel Pendergast
yesterday
@DanielPendergast, the orange line is total costs, which is the sum of total fixed and total variable costs. The profit is the total revenue minus the total costs. Hence, the red area represent loss, because $pi=TR-TC<0$ and the blue area represents the profit, because $pi=TR-TC>0$. When the firm produces and sells $1000$ products, it breaks even, because $pi=TR-TC=0$.
– farruhota
yesterday
That makes sense, that the signed area notifies if there is a loss or profit, but what does the actual value of the integral mean? $int_0^t (TR(v) - TC(v))dv$ for any given volume, $t$.
– Daniel Pendergast
yesterday
@DanielPendergast, sorry, no, the integral you wrote does not make sense, because the deleted graph (see edit) shows pointwise (at a specific quantity) profit. For the area between curves (or definite integration), one must consider the marginal functions as shown on the updated graph. Thank you for asking/commenting. You can read more about it here.
– farruhota
yesterday
add a comment |
Another aspect worth pointing out is that definite integrals solve many problems other than finding the area between curves- e.g. finding total net change in some system given the instantaneous rates of change. Finding the area between curves is an application that happens to be particularly easy to visualize.
add a comment |
Given two interest rate functions of time (the instantaneous rates of change of the values of two investments), the integral of the difference between them with respect to time is the difference in the returns of the two investments they represent over the interval of integration.
Chromatographic peak assignment associates mass with the area between a sensor response curve and the baseline curve. A few examples are shown here. (The Wikipedia has this
although the baseline is not explicitly shown. That determining the baseline between, say, 20 and 25 minutes doesn't yield something simple should be easy to see.)
add a comment |
Integration is the art of adding together very many, very small things. There are many things other than areas which can be added together this way, but areas are nice to represent visually, and relatively easy to intuit about (as long as your examples aren't too pathological). So that's the analogy most authors and lecturers go for when they introduce integration.
Note that once the things you add become something different than real numbers (like vectors, or complex numbers), then the area analogy breaks down (it even struggles somewhat for negative numbers). For someone who cannot think of integration as anything other than finding areas, that's going to be a difficult transition. So I definitely would advise you to think about integration in other terms as well.
One standard example would be if you had velocity as a function of time. You could draw the graph of that function and say that the area under the graph represents distance travelled. But this becomes very difficult to generalize once, say, you add direction of travel into the mix (i.e. you're traveling in a plane rather than on a line). You can somewhat still recover the intuition for total distance travelled, but for net distance (i.e. the straight line distance from starting point to end point) it becomes a hassle.
On the other hand, one could forgo the area analogy completely and just say that for a very small time interval, you travelled a very small distance, and then add all those small distances together to find the total distance travelled. This is much easier to intuitively generalize from travel along a line to travel in a plane (especially if you're interested in net distance travelled), and on to higher dimensions.
add a comment |
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
The best application that comes to mind is computing the center of mass of an object.
Let $f(x)geq g(x)$ on $(a,b),$ then the center of mass of the area bound by the curves $f$ and $g$ is given by the coordinates
$$bar{x}=dfrac{int_a^bf(x)^2-g(x)^2dx}{2int_a^b f(x)-g(x)dx}$$
and
$$bar{y}=dfrac{int_a^bx(f(x)-g(x))dx}{int_a^b f(x)-g(x)dx}.$$
So the center of mass is $(bar{x},bar{y}).$
Now, you might wonder "why is that useful?" That's a good question.
Imagine you have designed a flat, or approximately flat object, which you'd like to move using a single suction cup/suction cup like device. For example, the glass which forms the screen of a modern smart phone. To reduce stress on the glass when picking it up and moving it around, you can pick it up by its center of mass, and distribute the stress uniformly.
add a comment |
The best application that comes to mind is computing the center of mass of an object.
Let $f(x)geq g(x)$ on $(a,b),$ then the center of mass of the area bound by the curves $f$ and $g$ is given by the coordinates
$$bar{x}=dfrac{int_a^bf(x)^2-g(x)^2dx}{2int_a^b f(x)-g(x)dx}$$
and
$$bar{y}=dfrac{int_a^bx(f(x)-g(x))dx}{int_a^b f(x)-g(x)dx}.$$
So the center of mass is $(bar{x},bar{y}).$
Now, you might wonder "why is that useful?" That's a good question.
Imagine you have designed a flat, or approximately flat object, which you'd like to move using a single suction cup/suction cup like device. For example, the glass which forms the screen of a modern smart phone. To reduce stress on the glass when picking it up and moving it around, you can pick it up by its center of mass, and distribute the stress uniformly.
add a comment |
The best application that comes to mind is computing the center of mass of an object.
Let $f(x)geq g(x)$ on $(a,b),$ then the center of mass of the area bound by the curves $f$ and $g$ is given by the coordinates
$$bar{x}=dfrac{int_a^bf(x)^2-g(x)^2dx}{2int_a^b f(x)-g(x)dx}$$
and
$$bar{y}=dfrac{int_a^bx(f(x)-g(x))dx}{int_a^b f(x)-g(x)dx}.$$
So the center of mass is $(bar{x},bar{y}).$
Now, you might wonder "why is that useful?" That's a good question.
Imagine you have designed a flat, or approximately flat object, which you'd like to move using a single suction cup/suction cup like device. For example, the glass which forms the screen of a modern smart phone. To reduce stress on the glass when picking it up and moving it around, you can pick it up by its center of mass, and distribute the stress uniformly.
The best application that comes to mind is computing the center of mass of an object.
Let $f(x)geq g(x)$ on $(a,b),$ then the center of mass of the area bound by the curves $f$ and $g$ is given by the coordinates
$$bar{x}=dfrac{int_a^bf(x)^2-g(x)^2dx}{2int_a^b f(x)-g(x)dx}$$
and
$$bar{y}=dfrac{int_a^bx(f(x)-g(x))dx}{int_a^b f(x)-g(x)dx}.$$
So the center of mass is $(bar{x},bar{y}).$
Now, you might wonder "why is that useful?" That's a good question.
Imagine you have designed a flat, or approximately flat object, which you'd like to move using a single suction cup/suction cup like device. For example, the glass which forms the screen of a modern smart phone. To reduce stress on the glass when picking it up and moving it around, you can pick it up by its center of mass, and distribute the stress uniformly.
answered 2 days ago
ChickenmancerChickenmancer
3,564724
3,564724
add a comment |
add a comment |
There are many applications for the area between curves. For example if the curves represent the speeds,then the area is the difference between the total distance traveled.
Another application is for two investments with different interest rates and the area represents the difference between the total interest gained by accounts.
add a comment |
There are many applications for the area between curves. For example if the curves represent the speeds,then the area is the difference between the total distance traveled.
Another application is for two investments with different interest rates and the area represents the difference between the total interest gained by accounts.
add a comment |
There are many applications for the area between curves. For example if the curves represent the speeds,then the area is the difference between the total distance traveled.
Another application is for two investments with different interest rates and the area represents the difference between the total interest gained by accounts.
There are many applications for the area between curves. For example if the curves represent the speeds,then the area is the difference between the total distance traveled.
Another application is for two investments with different interest rates and the area represents the difference between the total interest gained by accounts.
edited 2 days ago
answered 2 days ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
41.4k42061
add a comment |
add a comment |
For instance, computing that area of the disk$${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant r^2}$$ is the same thing as computing the areas between the curves$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&sqrt{r^2-x^2}end{array}$$and$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&-sqrt{r^2-x^2}end{array}.$$So, unless you find the problem of computing the area of a disk beneath you, you can see that it is as interesting as the problem of finding the area between two specific curves.
add a comment |
For instance, computing that area of the disk$${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant r^2}$$ is the same thing as computing the areas between the curves$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&sqrt{r^2-x^2}end{array}$$and$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&-sqrt{r^2-x^2}end{array}.$$So, unless you find the problem of computing the area of a disk beneath you, you can see that it is as interesting as the problem of finding the area between two specific curves.
add a comment |
For instance, computing that area of the disk$${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant r^2}$$ is the same thing as computing the areas between the curves$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&sqrt{r^2-x^2}end{array}$$and$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&-sqrt{r^2-x^2}end{array}.$$So, unless you find the problem of computing the area of a disk beneath you, you can see that it is as interesting as the problem of finding the area between two specific curves.
For instance, computing that area of the disk$${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant r^2}$$ is the same thing as computing the areas between the curves$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&sqrt{r^2-x^2}end{array}$$and$$begin{array}{ccc}[-r,r]&longrightarrow&mathbb{R}^2\x&mapsto&-sqrt{r^2-x^2}end{array}.$$So, unless you find the problem of computing the area of a disk beneath you, you can see that it is as interesting as the problem of finding the area between two specific curves.
answered 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123225
152k22123225
add a comment |
add a comment |
There are loads of examples in the real world outside of pure mathematics.
In terms of materials, finding the area under the stress-strain curve tells that material's toughness - how much abuse (energy) it can take (absorb) before it begins to fracture. If you take the difference between two materials' toughness, you can see which one is tougher, which is something that's not always apparent at just looking at their graphs.
Example: Say we have materials A and B, each with stress-strain curves given by $sigma(epsilon)$.
begin{cases}
begin{align}
sigma_A(epsilon) &= 5 epsilon, &epsilon in (0,1)\
sigma_B(epsilon) &= tanhepsilon, &epsilon in (0,3)
end{align}
end{cases}
These functions, by the way, are wildly unrealistic in most occurrences, but the concept is still relevant.
$hskip 1 in$
It's not exactly clear which material has the higher toughness. Material A's looks like a lot more, but maybe the longevity of material B makes up for its low yield. We could find the area under each and see which is higher but we could assume material A is tougher and find the area between A and B.
$$Delta U = int_0^3 Big( 5epsilon big(1 - h(epsilon - 1)big) - tanh epsilon Big) , depsilon =frac{5}{2} - logcosh3approx 0.19.$$
So material A is actually hardly more tough than material B.
In terms of systems, you can design an integral-feedback control system that reduces the error in the output from the desired output. In practice, these are PID controllers (Proportional, Integral, Derivative), all 3 of which are focused on changing the actual output to what it should be based on the desired output. The goal is for the error to be zero, i.e. in the integral controller, you're looking for the area between the desired output and actual output to be exactly zero.
For instance, suppose you're driving and you click on the cruise control. You've just told the car you want it to keep going the speed that you are going right this moment. Now the road starts leading you uphill. The car is going to slow down initially, but it is also going to do everything it can to keep you going how fast you told it to go. That is, going uphill, it has encountered some error between the actual speed and desired speed and it wants to bring that error back to zero. The same applies for going downhill as well.
Another example could be controlling the temperature inside an oven, perhaps an expensive one with variable heating elements using a similarly styled controller.
These are just a few examples; there's really no shortage of applications.
add a comment |
There are loads of examples in the real world outside of pure mathematics.
In terms of materials, finding the area under the stress-strain curve tells that material's toughness - how much abuse (energy) it can take (absorb) before it begins to fracture. If you take the difference between two materials' toughness, you can see which one is tougher, which is something that's not always apparent at just looking at their graphs.
Example: Say we have materials A and B, each with stress-strain curves given by $sigma(epsilon)$.
begin{cases}
begin{align}
sigma_A(epsilon) &= 5 epsilon, &epsilon in (0,1)\
sigma_B(epsilon) &= tanhepsilon, &epsilon in (0,3)
end{align}
end{cases}
These functions, by the way, are wildly unrealistic in most occurrences, but the concept is still relevant.
$hskip 1 in$
It's not exactly clear which material has the higher toughness. Material A's looks like a lot more, but maybe the longevity of material B makes up for its low yield. We could find the area under each and see which is higher but we could assume material A is tougher and find the area between A and B.
$$Delta U = int_0^3 Big( 5epsilon big(1 - h(epsilon - 1)big) - tanh epsilon Big) , depsilon =frac{5}{2} - logcosh3approx 0.19.$$
So material A is actually hardly more tough than material B.
In terms of systems, you can design an integral-feedback control system that reduces the error in the output from the desired output. In practice, these are PID controllers (Proportional, Integral, Derivative), all 3 of which are focused on changing the actual output to what it should be based on the desired output. The goal is for the error to be zero, i.e. in the integral controller, you're looking for the area between the desired output and actual output to be exactly zero.
For instance, suppose you're driving and you click on the cruise control. You've just told the car you want it to keep going the speed that you are going right this moment. Now the road starts leading you uphill. The car is going to slow down initially, but it is also going to do everything it can to keep you going how fast you told it to go. That is, going uphill, it has encountered some error between the actual speed and desired speed and it wants to bring that error back to zero. The same applies for going downhill as well.
Another example could be controlling the temperature inside an oven, perhaps an expensive one with variable heating elements using a similarly styled controller.
These are just a few examples; there's really no shortage of applications.
add a comment |
There are loads of examples in the real world outside of pure mathematics.
In terms of materials, finding the area under the stress-strain curve tells that material's toughness - how much abuse (energy) it can take (absorb) before it begins to fracture. If you take the difference between two materials' toughness, you can see which one is tougher, which is something that's not always apparent at just looking at their graphs.
Example: Say we have materials A and B, each with stress-strain curves given by $sigma(epsilon)$.
begin{cases}
begin{align}
sigma_A(epsilon) &= 5 epsilon, &epsilon in (0,1)\
sigma_B(epsilon) &= tanhepsilon, &epsilon in (0,3)
end{align}
end{cases}
These functions, by the way, are wildly unrealistic in most occurrences, but the concept is still relevant.
$hskip 1 in$
It's not exactly clear which material has the higher toughness. Material A's looks like a lot more, but maybe the longevity of material B makes up for its low yield. We could find the area under each and see which is higher but we could assume material A is tougher and find the area between A and B.
$$Delta U = int_0^3 Big( 5epsilon big(1 - h(epsilon - 1)big) - tanh epsilon Big) , depsilon =frac{5}{2} - logcosh3approx 0.19.$$
So material A is actually hardly more tough than material B.
In terms of systems, you can design an integral-feedback control system that reduces the error in the output from the desired output. In practice, these are PID controllers (Proportional, Integral, Derivative), all 3 of which are focused on changing the actual output to what it should be based on the desired output. The goal is for the error to be zero, i.e. in the integral controller, you're looking for the area between the desired output and actual output to be exactly zero.
For instance, suppose you're driving and you click on the cruise control. You've just told the car you want it to keep going the speed that you are going right this moment. Now the road starts leading you uphill. The car is going to slow down initially, but it is also going to do everything it can to keep you going how fast you told it to go. That is, going uphill, it has encountered some error between the actual speed and desired speed and it wants to bring that error back to zero. The same applies for going downhill as well.
Another example could be controlling the temperature inside an oven, perhaps an expensive one with variable heating elements using a similarly styled controller.
These are just a few examples; there's really no shortage of applications.
There are loads of examples in the real world outside of pure mathematics.
In terms of materials, finding the area under the stress-strain curve tells that material's toughness - how much abuse (energy) it can take (absorb) before it begins to fracture. If you take the difference between two materials' toughness, you can see which one is tougher, which is something that's not always apparent at just looking at their graphs.
Example: Say we have materials A and B, each with stress-strain curves given by $sigma(epsilon)$.
begin{cases}
begin{align}
sigma_A(epsilon) &= 5 epsilon, &epsilon in (0,1)\
sigma_B(epsilon) &= tanhepsilon, &epsilon in (0,3)
end{align}
end{cases}
These functions, by the way, are wildly unrealistic in most occurrences, but the concept is still relevant.
$hskip 1 in$
It's not exactly clear which material has the higher toughness. Material A's looks like a lot more, but maybe the longevity of material B makes up for its low yield. We could find the area under each and see which is higher but we could assume material A is tougher and find the area between A and B.
$$Delta U = int_0^3 Big( 5epsilon big(1 - h(epsilon - 1)big) - tanh epsilon Big) , depsilon =frac{5}{2} - logcosh3approx 0.19.$$
So material A is actually hardly more tough than material B.
In terms of systems, you can design an integral-feedback control system that reduces the error in the output from the desired output. In practice, these are PID controllers (Proportional, Integral, Derivative), all 3 of which are focused on changing the actual output to what it should be based on the desired output. The goal is for the error to be zero, i.e. in the integral controller, you're looking for the area between the desired output and actual output to be exactly zero.
For instance, suppose you're driving and you click on the cruise control. You've just told the car you want it to keep going the speed that you are going right this moment. Now the road starts leading you uphill. The car is going to slow down initially, but it is also going to do everything it can to keep you going how fast you told it to go. That is, going uphill, it has encountered some error between the actual speed and desired speed and it wants to bring that error back to zero. The same applies for going downhill as well.
Another example could be controlling the temperature inside an oven, perhaps an expensive one with variable heating elements using a similarly styled controller.
These are just a few examples; there's really no shortage of applications.
answered 2 days ago
SkipSkip
1,282214
1,282214
add a comment |
add a comment |
Economics.
Example 1. The area between the demand (supply) function and the market-clearing (equilibrium) price shows the consumers (producers) surplus:
$hspace{3cm}$
Source: Wikipedia.
Note that the consumers are willing to pay a price for a good up to their demand function, but they pay the equilibrium price and save the difference (red area).
Example 2. The area between the marginal revenue and the marginal cost functions shows the net profit (loss).
$hspace{3cm}$
Source: Slide $13$ at Slideshare.
I may be wrong, but doesn't profit = revenue - total costs? Why would it be the area under revenue - total costs?
– Daniel Pendergast
yesterday
@DanielPendergast, the orange line is total costs, which is the sum of total fixed and total variable costs. The profit is the total revenue minus the total costs. Hence, the red area represent loss, because $pi=TR-TC<0$ and the blue area represents the profit, because $pi=TR-TC>0$. When the firm produces and sells $1000$ products, it breaks even, because $pi=TR-TC=0$.
– farruhota
yesterday
That makes sense, that the signed area notifies if there is a loss or profit, but what does the actual value of the integral mean? $int_0^t (TR(v) - TC(v))dv$ for any given volume, $t$.
– Daniel Pendergast
yesterday
@DanielPendergast, sorry, no, the integral you wrote does not make sense, because the deleted graph (see edit) shows pointwise (at a specific quantity) profit. For the area between curves (or definite integration), one must consider the marginal functions as shown on the updated graph. Thank you for asking/commenting. You can read more about it here.
– farruhota
yesterday
add a comment |
Economics.
Example 1. The area between the demand (supply) function and the market-clearing (equilibrium) price shows the consumers (producers) surplus:
$hspace{3cm}$
Source: Wikipedia.
Note that the consumers are willing to pay a price for a good up to their demand function, but they pay the equilibrium price and save the difference (red area).
Example 2. The area between the marginal revenue and the marginal cost functions shows the net profit (loss).
$hspace{3cm}$
Source: Slide $13$ at Slideshare.
I may be wrong, but doesn't profit = revenue - total costs? Why would it be the area under revenue - total costs?
– Daniel Pendergast
yesterday
@DanielPendergast, the orange line is total costs, which is the sum of total fixed and total variable costs. The profit is the total revenue minus the total costs. Hence, the red area represent loss, because $pi=TR-TC<0$ and the blue area represents the profit, because $pi=TR-TC>0$. When the firm produces and sells $1000$ products, it breaks even, because $pi=TR-TC=0$.
– farruhota
yesterday
That makes sense, that the signed area notifies if there is a loss or profit, but what does the actual value of the integral mean? $int_0^t (TR(v) - TC(v))dv$ for any given volume, $t$.
– Daniel Pendergast
yesterday
@DanielPendergast, sorry, no, the integral you wrote does not make sense, because the deleted graph (see edit) shows pointwise (at a specific quantity) profit. For the area between curves (or definite integration), one must consider the marginal functions as shown on the updated graph. Thank you for asking/commenting. You can read more about it here.
– farruhota
yesterday
add a comment |
Economics.
Example 1. The area between the demand (supply) function and the market-clearing (equilibrium) price shows the consumers (producers) surplus:
$hspace{3cm}$
Source: Wikipedia.
Note that the consumers are willing to pay a price for a good up to their demand function, but they pay the equilibrium price and save the difference (red area).
Example 2. The area between the marginal revenue and the marginal cost functions shows the net profit (loss).
$hspace{3cm}$
Source: Slide $13$ at Slideshare.
Economics.
Example 1. The area between the demand (supply) function and the market-clearing (equilibrium) price shows the consumers (producers) surplus:
$hspace{3cm}$
Source: Wikipedia.
Note that the consumers are willing to pay a price for a good up to their demand function, but they pay the equilibrium price and save the difference (red area).
Example 2. The area between the marginal revenue and the marginal cost functions shows the net profit (loss).
$hspace{3cm}$
Source: Slide $13$ at Slideshare.
edited yesterday
answered yesterday
farruhotafarruhota
19.6k2737
19.6k2737
I may be wrong, but doesn't profit = revenue - total costs? Why would it be the area under revenue - total costs?
– Daniel Pendergast
yesterday
@DanielPendergast, the orange line is total costs, which is the sum of total fixed and total variable costs. The profit is the total revenue minus the total costs. Hence, the red area represent loss, because $pi=TR-TC<0$ and the blue area represents the profit, because $pi=TR-TC>0$. When the firm produces and sells $1000$ products, it breaks even, because $pi=TR-TC=0$.
– farruhota
yesterday
That makes sense, that the signed area notifies if there is a loss or profit, but what does the actual value of the integral mean? $int_0^t (TR(v) - TC(v))dv$ for any given volume, $t$.
– Daniel Pendergast
yesterday
@DanielPendergast, sorry, no, the integral you wrote does not make sense, because the deleted graph (see edit) shows pointwise (at a specific quantity) profit. For the area between curves (or definite integration), one must consider the marginal functions as shown on the updated graph. Thank you for asking/commenting. You can read more about it here.
– farruhota
yesterday
add a comment |
I may be wrong, but doesn't profit = revenue - total costs? Why would it be the area under revenue - total costs?
– Daniel Pendergast
yesterday
@DanielPendergast, the orange line is total costs, which is the sum of total fixed and total variable costs. The profit is the total revenue minus the total costs. Hence, the red area represent loss, because $pi=TR-TC<0$ and the blue area represents the profit, because $pi=TR-TC>0$. When the firm produces and sells $1000$ products, it breaks even, because $pi=TR-TC=0$.
– farruhota
yesterday
That makes sense, that the signed area notifies if there is a loss or profit, but what does the actual value of the integral mean? $int_0^t (TR(v) - TC(v))dv$ for any given volume, $t$.
– Daniel Pendergast
yesterday
@DanielPendergast, sorry, no, the integral you wrote does not make sense, because the deleted graph (see edit) shows pointwise (at a specific quantity) profit. For the area between curves (or definite integration), one must consider the marginal functions as shown on the updated graph. Thank you for asking/commenting. You can read more about it here.
– farruhota
yesterday
I may be wrong, but doesn't profit = revenue - total costs? Why would it be the area under revenue - total costs?
– Daniel Pendergast
yesterday
I may be wrong, but doesn't profit = revenue - total costs? Why would it be the area under revenue - total costs?
– Daniel Pendergast
yesterday
@DanielPendergast, the orange line is total costs, which is the sum of total fixed and total variable costs. The profit is the total revenue minus the total costs. Hence, the red area represent loss, because $pi=TR-TC<0$ and the blue area represents the profit, because $pi=TR-TC>0$. When the firm produces and sells $1000$ products, it breaks even, because $pi=TR-TC=0$.
– farruhota
yesterday
@DanielPendergast, the orange line is total costs, which is the sum of total fixed and total variable costs. The profit is the total revenue minus the total costs. Hence, the red area represent loss, because $pi=TR-TC<0$ and the blue area represents the profit, because $pi=TR-TC>0$. When the firm produces and sells $1000$ products, it breaks even, because $pi=TR-TC=0$.
– farruhota
yesterday
That makes sense, that the signed area notifies if there is a loss or profit, but what does the actual value of the integral mean? $int_0^t (TR(v) - TC(v))dv$ for any given volume, $t$.
– Daniel Pendergast
yesterday
That makes sense, that the signed area notifies if there is a loss or profit, but what does the actual value of the integral mean? $int_0^t (TR(v) - TC(v))dv$ for any given volume, $t$.
– Daniel Pendergast
yesterday
@DanielPendergast, sorry, no, the integral you wrote does not make sense, because the deleted graph (see edit) shows pointwise (at a specific quantity) profit. For the area between curves (or definite integration), one must consider the marginal functions as shown on the updated graph. Thank you for asking/commenting. You can read more about it here.
– farruhota
yesterday
@DanielPendergast, sorry, no, the integral you wrote does not make sense, because the deleted graph (see edit) shows pointwise (at a specific quantity) profit. For the area between curves (or definite integration), one must consider the marginal functions as shown on the updated graph. Thank you for asking/commenting. You can read more about it here.
– farruhota
yesterday
add a comment |
Another aspect worth pointing out is that definite integrals solve many problems other than finding the area between curves- e.g. finding total net change in some system given the instantaneous rates of change. Finding the area between curves is an application that happens to be particularly easy to visualize.
add a comment |
Another aspect worth pointing out is that definite integrals solve many problems other than finding the area between curves- e.g. finding total net change in some system given the instantaneous rates of change. Finding the area between curves is an application that happens to be particularly easy to visualize.
add a comment |
Another aspect worth pointing out is that definite integrals solve many problems other than finding the area between curves- e.g. finding total net change in some system given the instantaneous rates of change. Finding the area between curves is an application that happens to be particularly easy to visualize.
Another aspect worth pointing out is that definite integrals solve many problems other than finding the area between curves- e.g. finding total net change in some system given the instantaneous rates of change. Finding the area between curves is an application that happens to be particularly easy to visualize.
answered 2 days ago
Gregory J. PuleoGregory J. Puleo
4,49731520
4,49731520
add a comment |
add a comment |
Given two interest rate functions of time (the instantaneous rates of change of the values of two investments), the integral of the difference between them with respect to time is the difference in the returns of the two investments they represent over the interval of integration.
Chromatographic peak assignment associates mass with the area between a sensor response curve and the baseline curve. A few examples are shown here. (The Wikipedia has this
although the baseline is not explicitly shown. That determining the baseline between, say, 20 and 25 minutes doesn't yield something simple should be easy to see.)
add a comment |
Given two interest rate functions of time (the instantaneous rates of change of the values of two investments), the integral of the difference between them with respect to time is the difference in the returns of the two investments they represent over the interval of integration.
Chromatographic peak assignment associates mass with the area between a sensor response curve and the baseline curve. A few examples are shown here. (The Wikipedia has this
although the baseline is not explicitly shown. That determining the baseline between, say, 20 and 25 minutes doesn't yield something simple should be easy to see.)
add a comment |
Given two interest rate functions of time (the instantaneous rates of change of the values of two investments), the integral of the difference between them with respect to time is the difference in the returns of the two investments they represent over the interval of integration.
Chromatographic peak assignment associates mass with the area between a sensor response curve and the baseline curve. A few examples are shown here. (The Wikipedia has this
although the baseline is not explicitly shown. That determining the baseline between, say, 20 and 25 minutes doesn't yield something simple should be easy to see.)
Given two interest rate functions of time (the instantaneous rates of change of the values of two investments), the integral of the difference between them with respect to time is the difference in the returns of the two investments they represent over the interval of integration.
Chromatographic peak assignment associates mass with the area between a sensor response curve and the baseline curve. A few examples are shown here. (The Wikipedia has this
although the baseline is not explicitly shown. That determining the baseline between, say, 20 and 25 minutes doesn't yield something simple should be easy to see.)
answered yesterday
Eric TowersEric Towers
32.2k22267
32.2k22267
add a comment |
add a comment |
Integration is the art of adding together very many, very small things. There are many things other than areas which can be added together this way, but areas are nice to represent visually, and relatively easy to intuit about (as long as your examples aren't too pathological). So that's the analogy most authors and lecturers go for when they introduce integration.
Note that once the things you add become something different than real numbers (like vectors, or complex numbers), then the area analogy breaks down (it even struggles somewhat for negative numbers). For someone who cannot think of integration as anything other than finding areas, that's going to be a difficult transition. So I definitely would advise you to think about integration in other terms as well.
One standard example would be if you had velocity as a function of time. You could draw the graph of that function and say that the area under the graph represents distance travelled. But this becomes very difficult to generalize once, say, you add direction of travel into the mix (i.e. you're traveling in a plane rather than on a line). You can somewhat still recover the intuition for total distance travelled, but for net distance (i.e. the straight line distance from starting point to end point) it becomes a hassle.
On the other hand, one could forgo the area analogy completely and just say that for a very small time interval, you travelled a very small distance, and then add all those small distances together to find the total distance travelled. This is much easier to intuitively generalize from travel along a line to travel in a plane (especially if you're interested in net distance travelled), and on to higher dimensions.
add a comment |
Integration is the art of adding together very many, very small things. There are many things other than areas which can be added together this way, but areas are nice to represent visually, and relatively easy to intuit about (as long as your examples aren't too pathological). So that's the analogy most authors and lecturers go for when they introduce integration.
Note that once the things you add become something different than real numbers (like vectors, or complex numbers), then the area analogy breaks down (it even struggles somewhat for negative numbers). For someone who cannot think of integration as anything other than finding areas, that's going to be a difficult transition. So I definitely would advise you to think about integration in other terms as well.
One standard example would be if you had velocity as a function of time. You could draw the graph of that function and say that the area under the graph represents distance travelled. But this becomes very difficult to generalize once, say, you add direction of travel into the mix (i.e. you're traveling in a plane rather than on a line). You can somewhat still recover the intuition for total distance travelled, but for net distance (i.e. the straight line distance from starting point to end point) it becomes a hassle.
On the other hand, one could forgo the area analogy completely and just say that for a very small time interval, you travelled a very small distance, and then add all those small distances together to find the total distance travelled. This is much easier to intuitively generalize from travel along a line to travel in a plane (especially if you're interested in net distance travelled), and on to higher dimensions.
add a comment |
Integration is the art of adding together very many, very small things. There are many things other than areas which can be added together this way, but areas are nice to represent visually, and relatively easy to intuit about (as long as your examples aren't too pathological). So that's the analogy most authors and lecturers go for when they introduce integration.
Note that once the things you add become something different than real numbers (like vectors, or complex numbers), then the area analogy breaks down (it even struggles somewhat for negative numbers). For someone who cannot think of integration as anything other than finding areas, that's going to be a difficult transition. So I definitely would advise you to think about integration in other terms as well.
One standard example would be if you had velocity as a function of time. You could draw the graph of that function and say that the area under the graph represents distance travelled. But this becomes very difficult to generalize once, say, you add direction of travel into the mix (i.e. you're traveling in a plane rather than on a line). You can somewhat still recover the intuition for total distance travelled, but for net distance (i.e. the straight line distance from starting point to end point) it becomes a hassle.
On the other hand, one could forgo the area analogy completely and just say that for a very small time interval, you travelled a very small distance, and then add all those small distances together to find the total distance travelled. This is much easier to intuitively generalize from travel along a line to travel in a plane (especially if you're interested in net distance travelled), and on to higher dimensions.
Integration is the art of adding together very many, very small things. There are many things other than areas which can be added together this way, but areas are nice to represent visually, and relatively easy to intuit about (as long as your examples aren't too pathological). So that's the analogy most authors and lecturers go for when they introduce integration.
Note that once the things you add become something different than real numbers (like vectors, or complex numbers), then the area analogy breaks down (it even struggles somewhat for negative numbers). For someone who cannot think of integration as anything other than finding areas, that's going to be a difficult transition. So I definitely would advise you to think about integration in other terms as well.
One standard example would be if you had velocity as a function of time. You could draw the graph of that function and say that the area under the graph represents distance travelled. But this becomes very difficult to generalize once, say, you add direction of travel into the mix (i.e. you're traveling in a plane rather than on a line). You can somewhat still recover the intuition for total distance travelled, but for net distance (i.e. the straight line distance from starting point to end point) it becomes a hassle.
On the other hand, one could forgo the area analogy completely and just say that for a very small time interval, you travelled a very small distance, and then add all those small distances together to find the total distance travelled. This is much easier to intuitively generalize from travel along a line to travel in a plane (especially if you're interested in net distance travelled), and on to higher dimensions.
edited yesterday
answered yesterday
ArthurArthur
111k7107189
111k7107189
add a comment |
add a comment |
36
You might as well ask what the point of computing areas is whatsoever
– Wojowu
2 days ago
3
Maybe it's just a neat example of a kind of geometry problem that can be solved using calculus. If you're wondering why integration is useful (who cares about areas, anyway?) one major reason is that integration can be used to find the total change in a quantity if we know the instantaneous rate of change of that quantity. For example, knowing an object's velocity, you can figure out its position.
– littleO
2 days ago
2
Imho you're asking the wrong kind of question. For most common, elementary math concepts you should not worry about the applications, things will come together once you have a broader understanding of mathematics. It's not reasonable to assume that any isolated math concept will have tons of applications that appear in practice , but that doesn't mean that knowing about it isnt useful. You might as well ask what the purpose of learning about division is or, for an analogy outside of mathematics, why you should learn a specific word.
– user159517
2 days ago
2
How exactly do you define area between curves? Indeed, finding the area between curves by Riemann integration can be considered meaningless because at this level the area is pretty much defined to be the respective integral. What could be considered interesting is to show that the Lebesgue measure (the only measure on $mathbb{R}^2$ such that the measure of every rectangle is equal to its area) of the part of the plane between curves is equal to the integral.
– mechanodroid
2 days ago
2
"Area under the curve" is just an informal illustration of the concept of integral. Actually, an integral is the continuous version of a sum. It is as natural an operation as summing a finite set of numbers.
– Giuseppe Negro
yesterday