Is a definite integral just a summation?
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I am learning about definite integrals and found the formula for finding an average of a function over a given interval:
$$frac{1}{b-a} int_{a}^{b} fleft(xright) dx$$
If we look at the average function for a set of numbers:
$$frac{1}{n} sum_{i=1}^{n} x_i$$
It would seem as if the integral is essentially a summation of all values from $a$ to $b$. Is it correct to think of it this way?
calculus definite-integrals means
asked 2 hours ago
MCMasteryMCMastery
1598
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add a comment |
$begingroup$
I am learning about definite integrals and found the formula for finding an average of a function over a given interval:
$$frac{1}{b-a} int_{a}^{b} fleft(xright) dx$$
If we look at the average function for a set of numbers:
$$frac{1}{n} sum_{i=1}^{n} x_i$$
It would seem as if the integral is essentially a summation of all values from $a$ to $b$. Is it correct to think of it this way?
calculus definite-integrals means
asked 2 hours ago
MCMasteryMCMastery
1598
$endgroup$
2
$begingroup$
Yeah; that's exactly how one would think of it. This becomes more clear if you look at the formal definition of an integral.
$endgroup$
– Hyperion
2 hours ago
1
$begingroup$
Notice that the functional values have been multiplied by $dx$, so $f(x)dx$ is the area of the elemental strip having thickness $dx$ and height $f(x)$. The integral can be treated as a summation of these differential areas
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– Shubham Johri
2 hours ago
add a comment |
$begingroup$
I am learning about definite integrals and found the formula for finding an average of a function over a given interval:
$$frac{1}{b-a} int_{a}^{b} fleft(xright) dx$$
If we look at the average function for a set of numbers:
$$frac{1}{n} sum_{i=1}^{n} x_i$$
It would seem as if the integral is essentially a summation of all values from $a$ to $b$. Is it correct to think of it this way?
calculus definite-integrals means
asked 2 hours ago
MCMasteryMCMastery
1598
$endgroup$
I am learning about definite integrals and found the formula for finding an average of a function over a given interval:
$$frac{1}{b-a} int_{a}^{b} fleft(xright) dx$$
If we look at the average function for a set of numbers:
$$frac{1}{n} sum_{i=1}^{n} x_i$$
It would seem as if the integral is essentially a summation of all values from $a$ to $b$. Is it correct to think of it this way?
calculus definite-integrals means
calculus definite-integrals means
asked 2 hours ago
MCMasteryMCMastery
1598
asked 2 hours ago
MCMasteryMCMastery
1598
asked 2 hours ago
MCMasteryMCMastery
1598
asked 2 hours ago
MCMasteryMCMastery
1598
asked 2 hours ago
MCMasteryMCMastery
1598
1598
2
$begingroup$
Yeah; that's exactly how one would think of it. This becomes more clear if you look at the formal definition of an integral.
$endgroup$
– Hyperion
2 hours ago
1
$begingroup$
Notice that the functional values have been multiplied by $dx$, so $f(x)dx$ is the area of the elemental strip having thickness $dx$ and height $f(x)$. The integral can be treated as a summation of these differential areas
$endgroup$
– Shubham Johri
2 hours ago
add a comment |
2
$begingroup$
Yeah; that's exactly how one would think of it. This becomes more clear if you look at the formal definition of an integral.
$endgroup$
– Hyperion
2 hours ago
1
$begingroup$
Notice that the functional values have been multiplied by $dx$, so $f(x)dx$ is the area of the elemental strip having thickness $dx$ and height $f(x)$. The integral can be treated as a summation of these differential areas
$endgroup$
– Shubham Johri
2 hours ago
2
2
$begingroup$
Yeah; that's exactly how one would think of it. This becomes more clear if you look at the formal definition of an integral.
$endgroup$
– Hyperion
2 hours ago
$begingroup$
Yeah; that's exactly how one would think of it. This becomes more clear if you look at the formal definition of an integral.
$endgroup$
– Hyperion
2 hours ago
1
1
$begingroup$
Notice that the functional values have been multiplied by $dx$, so $f(x)dx$ is the area of the elemental strip having thickness $dx$ and height $f(x)$. The integral can be treated as a summation of these differential areas
$endgroup$
– Shubham Johri
2 hours ago
$begingroup$
Notice that the functional values have been multiplied by $dx$, so $f(x)dx$ is the area of the elemental strip having thickness $dx$ and height $f(x)$. The integral can be treated as a summation of these differential areas
$endgroup$
– Shubham Johri
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's certainly helpful to think of integrals this way, though not strictly correct. An integral takes all the little slivers of area beneath a curve and sums them up into a bigger area - though there's, of course, a lot of technicality needed to think about it this way.
There is an idea being obscured by your idea of an integral as a sum, however. In truth, a sum is a kind of integral and not vice versa - this is somewhat counterintuitive given that sums are much more familiar objects than integrals, but there's an elegant theory known as Lebesgue integration which basically makes the integral a tool which eats two pieces of information:
We start with some indexing set and some way to "weigh" pieces of that set.
We have some function on that set.
Then, the Lebesgue integral spits out the weighted "sum" of that function.
An integral in the most common sense arises when you say, "I have a function on the real numbers, and I want an interval to have weight equal to its length." A finite sum arises when you say "I have a function taking values $x_1,ldots,x_n$ on the index set ${1,ldots,n}$. Each index gets weight $1$" - and, of course, you can change those weights to get a weighted sum or you can extend the indexing set to every natural number to get an infinite sum. But, the basic thing to note is that there's a more general idea of integral that places summation as part of the theory of integration.
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
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add a comment |
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Yes, and no. In so much as the definite integral is a "sum" it is a limit of a Riemann sum as the "mesh" goes to zero. The "mesh" is the largest part of a partition. That is, your domain is broken up into parts over which you take the average value of your function, and multiply by the width of that part.
It's typically helpful to think about definite integrals as sums over "infinitesimal" displacements, but strictly speaking, that is incorrect.
answered 1 hour ago
Steven HattonSteven Hatton
915418
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add a comment |
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2 Answers
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$begingroup$
It's certainly helpful to think of integrals this way, though not strictly correct. An integral takes all the little slivers of area beneath a curve and sums them up into a bigger area - though there's, of course, a lot of technicality needed to think about it this way.
There is an idea being obscured by your idea of an integral as a sum, however. In truth, a sum is a kind of integral and not vice versa - this is somewhat counterintuitive given that sums are much more familiar objects than integrals, but there's an elegant theory known as Lebesgue integration which basically makes the integral a tool which eats two pieces of information:
We start with some indexing set and some way to "weigh" pieces of that set.
We have some function on that set.
Then, the Lebesgue integral spits out the weighted "sum" of that function.
An integral in the most common sense arises when you say, "I have a function on the real numbers, and I want an interval to have weight equal to its length." A finite sum arises when you say "I have a function taking values $x_1,ldots,x_n$ on the index set ${1,ldots,n}$. Each index gets weight $1$" - and, of course, you can change those weights to get a weighted sum or you can extend the indexing set to every natural number to get an infinite sum. But, the basic thing to note is that there's a more general idea of integral that places summation as part of the theory of integration.
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
$endgroup$
add a comment |
$begingroup$
It's certainly helpful to think of integrals this way, though not strictly correct. An integral takes all the little slivers of area beneath a curve and sums them up into a bigger area - though there's, of course, a lot of technicality needed to think about it this way.
There is an idea being obscured by your idea of an integral as a sum, however. In truth, a sum is a kind of integral and not vice versa - this is somewhat counterintuitive given that sums are much more familiar objects than integrals, but there's an elegant theory known as Lebesgue integration which basically makes the integral a tool which eats two pieces of information:
We start with some indexing set and some way to "weigh" pieces of that set.
We have some function on that set.
Then, the Lebesgue integral spits out the weighted "sum" of that function.
An integral in the most common sense arises when you say, "I have a function on the real numbers, and I want an interval to have weight equal to its length." A finite sum arises when you say "I have a function taking values $x_1,ldots,x_n$ on the index set ${1,ldots,n}$. Each index gets weight $1$" - and, of course, you can change those weights to get a weighted sum or you can extend the indexing set to every natural number to get an infinite sum. But, the basic thing to note is that there's a more general idea of integral that places summation as part of the theory of integration.
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
$endgroup$
add a comment |
$begingroup$
It's certainly helpful to think of integrals this way, though not strictly correct. An integral takes all the little slivers of area beneath a curve and sums them up into a bigger area - though there's, of course, a lot of technicality needed to think about it this way.
There is an idea being obscured by your idea of an integral as a sum, however. In truth, a sum is a kind of integral and not vice versa - this is somewhat counterintuitive given that sums are much more familiar objects than integrals, but there's an elegant theory known as Lebesgue integration which basically makes the integral a tool which eats two pieces of information:
We start with some indexing set and some way to "weigh" pieces of that set.
We have some function on that set.
Then, the Lebesgue integral spits out the weighted "sum" of that function.
An integral in the most common sense arises when you say, "I have a function on the real numbers, and I want an interval to have weight equal to its length." A finite sum arises when you say "I have a function taking values $x_1,ldots,x_n$ on the index set ${1,ldots,n}$. Each index gets weight $1$" - and, of course, you can change those weights to get a weighted sum or you can extend the indexing set to every natural number to get an infinite sum. But, the basic thing to note is that there's a more general idea of integral that places summation as part of the theory of integration.
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
$endgroup$
It's certainly helpful to think of integrals this way, though not strictly correct. An integral takes all the little slivers of area beneath a curve and sums them up into a bigger area - though there's, of course, a lot of technicality needed to think about it this way.
There is an idea being obscured by your idea of an integral as a sum, however. In truth, a sum is a kind of integral and not vice versa - this is somewhat counterintuitive given that sums are much more familiar objects than integrals, but there's an elegant theory known as Lebesgue integration which basically makes the integral a tool which eats two pieces of information:
We start with some indexing set and some way to "weigh" pieces of that set.
We have some function on that set.
Then, the Lebesgue integral spits out the weighted "sum" of that function.
An integral in the most common sense arises when you say, "I have a function on the real numbers, and I want an interval to have weight equal to its length." A finite sum arises when you say "I have a function taking values $x_1,ldots,x_n$ on the index set ${1,ldots,n}$. Each index gets weight $1$" - and, of course, you can change those weights to get a weighted sum or you can extend the indexing set to every natural number to get an infinite sum. But, the basic thing to note is that there's a more general idea of integral that places summation as part of the theory of integration.
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
answered 1 hour ago
Milo BrandtMilo Brandt
39.6k475139
39.6k475139
add a comment |
add a comment |
$begingroup$
Yes, and no. In so much as the definite integral is a "sum" it is a limit of a Riemann sum as the "mesh" goes to zero. The "mesh" is the largest part of a partition. That is, your domain is broken up into parts over which you take the average value of your function, and multiply by the width of that part.
It's typically helpful to think about definite integrals as sums over "infinitesimal" displacements, but strictly speaking, that is incorrect.
answered 1 hour ago
Steven HattonSteven Hatton
915418
$endgroup$
add a comment |
$begingroup$
Yes, and no. In so much as the definite integral is a "sum" it is a limit of a Riemann sum as the "mesh" goes to zero. The "mesh" is the largest part of a partition. That is, your domain is broken up into parts over which you take the average value of your function, and multiply by the width of that part.
It's typically helpful to think about definite integrals as sums over "infinitesimal" displacements, but strictly speaking, that is incorrect.
answered 1 hour ago
Steven HattonSteven Hatton
915418
$endgroup$
add a comment |
$begingroup$
Yes, and no. In so much as the definite integral is a "sum" it is a limit of a Riemann sum as the "mesh" goes to zero. The "mesh" is the largest part of a partition. That is, your domain is broken up into parts over which you take the average value of your function, and multiply by the width of that part.
It's typically helpful to think about definite integrals as sums over "infinitesimal" displacements, but strictly speaking, that is incorrect.
answered 1 hour ago
Steven HattonSteven Hatton
915418
$endgroup$
Yes, and no. In so much as the definite integral is a "sum" it is a limit of a Riemann sum as the "mesh" goes to zero. The "mesh" is the largest part of a partition. That is, your domain is broken up into parts over which you take the average value of your function, and multiply by the width of that part.
It's typically helpful to think about definite integrals as sums over "infinitesimal" displacements, but strictly speaking, that is incorrect.
answered 1 hour ago
Steven HattonSteven Hatton
915418
answered 1 hour ago
Steven HattonSteven Hatton
915418
answered 1 hour ago
Steven HattonSteven Hatton
915418
answered 1 hour ago
Steven HattonSteven Hatton
915418
915418
add a comment |
add a comment |
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$begingroup$
Yeah; that's exactly how one would think of it. This becomes more clear if you look at the formal definition of an integral.
$endgroup$
– Hyperion
2 hours ago
1
$begingroup$
Notice that the functional values have been multiplied by $dx$, so $f(x)dx$ is the area of the elemental strip having thickness $dx$ and height $f(x)$. The integral can be treated as a summation of these differential areas
$endgroup$
– Shubham Johri
2 hours ago