How do we calculate integrals without knowing differentiation?
$begingroup$
To calculate the integrals, we use the fundamental theorem of calculus.
i.e for example to find $int f(x)dx$, what we do is we find a function $g(x)$ such that $g'(x) = f(x)$. But, is there any way to find the integrals without finding the answer first? I mean, can we just use the basic definition of integrals on all integrable functions?
PS: I get that we can use the partition and summation of step functions approach on increasing or decreasing function, but how will we be able to find the integral of any random integrable function by that?
calculus definite-integrals indefinite-integrals step-function partitions-for-integration
asked Dec 21 '18 at 17:59
AstikAstik
41
$endgroup$
add a comment |
$begingroup$
To calculate the integrals, we use the fundamental theorem of calculus.
i.e for example to find $int f(x)dx$, what we do is we find a function $g(x)$ such that $g'(x) = f(x)$. But, is there any way to find the integrals without finding the answer first? I mean, can we just use the basic definition of integrals on all integrable functions?
PS: I get that we can use the partition and summation of step functions approach on increasing or decreasing function, but how will we be able to find the integral of any random integrable function by that?
calculus definite-integrals indefinite-integrals step-function partitions-for-integration
asked Dec 21 '18 at 17:59
AstikAstik
41
$endgroup$
$begingroup$
I'm afraid your PS is your answer (but the function need not be monotone).
$endgroup$
– Ethan Bolker
Dec 21 '18 at 18:06
$begingroup$
is this question about techniques of integration (like substitution or integration by parts or using partial fraction expansion)? or is it about finding the area under the curve as done with Riemann summation?
$endgroup$
– robert bristow-johnson
Dec 21 '18 at 19:24
add a comment |
$begingroup$
To calculate the integrals, we use the fundamental theorem of calculus.
i.e for example to find $int f(x)dx$, what we do is we find a function $g(x)$ such that $g'(x) = f(x)$. But, is there any way to find the integrals without finding the answer first? I mean, can we just use the basic definition of integrals on all integrable functions?
PS: I get that we can use the partition and summation of step functions approach on increasing or decreasing function, but how will we be able to find the integral of any random integrable function by that?
calculus definite-integrals indefinite-integrals step-function partitions-for-integration
asked Dec 21 '18 at 17:59
AstikAstik
41
$endgroup$
To calculate the integrals, we use the fundamental theorem of calculus.
i.e for example to find $int f(x)dx$, what we do is we find a function $g(x)$ such that $g'(x) = f(x)$. But, is there any way to find the integrals without finding the answer first? I mean, can we just use the basic definition of integrals on all integrable functions?
PS: I get that we can use the partition and summation of step functions approach on increasing or decreasing function, but how will we be able to find the integral of any random integrable function by that?
calculus definite-integrals indefinite-integrals step-function partitions-for-integration
calculus definite-integrals indefinite-integrals step-function partitions-for-integration
asked Dec 21 '18 at 17:59
AstikAstik
41
asked Dec 21 '18 at 17:59
AstikAstik
41
asked Dec 21 '18 at 17:59
AstikAstik
41
asked Dec 21 '18 at 17:59
AstikAstik
41
asked Dec 21 '18 at 17:59
AstikAstik
41
41
$begingroup$
I'm afraid your PS is your answer (but the function need not be monotone).
$endgroup$
– Ethan Bolker
Dec 21 '18 at 18:06
$begingroup$
is this question about techniques of integration (like substitution or integration by parts or using partial fraction expansion)? or is it about finding the area under the curve as done with Riemann summation?
$endgroup$
– robert bristow-johnson
Dec 21 '18 at 19:24
add a comment |
$begingroup$
I'm afraid your PS is your answer (but the function need not be monotone).
$endgroup$
– Ethan Bolker
Dec 21 '18 at 18:06
$begingroup$
is this question about techniques of integration (like substitution or integration by parts or using partial fraction expansion)? or is it about finding the area under the curve as done with Riemann summation?
$endgroup$
– robert bristow-johnson
Dec 21 '18 at 19:24
$begingroup$
I'm afraid your PS is your answer (but the function need not be monotone).
$endgroup$
– Ethan Bolker
Dec 21 '18 at 18:06
$begingroup$
I'm afraid your PS is your answer (but the function need not be monotone).
$endgroup$
– Ethan Bolker
Dec 21 '18 at 18:06
$begingroup$
is this question about techniques of integration (like substitution or integration by parts or using partial fraction expansion)? or is it about finding the area under the curve as done with Riemann summation?
$endgroup$
– robert bristow-johnson
Dec 21 '18 at 19:24
$begingroup$
is this question about techniques of integration (like substitution or integration by parts or using partial fraction expansion)? or is it about finding the area under the curve as done with Riemann summation?
$endgroup$
– robert bristow-johnson
Dec 21 '18 at 19:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Historically, integration was first done by using the limits of areas of the rectangles under the step functions as their widths go to 0, and the number of them goes to infinity, as was mentioned above. It was only somewhat later that derivatives were introduced, with the fundamental theorem of calculus after that. This, of course, made it much easier to integrate most "well behaved" functions, even if it cannot always be done in closed form, as was also mentioned in another answer.
In my first-year university advanced calculus course, the teacher took this historical approach by first describing the rigorous minimum requirements for integration, how to do it using step function areas and limits, and then having us calculate various integrals. One of the interesting homework problems was to investigate what he called the "ruler function" (but it's also known by various other names), i.e., a function $f(x)$ defined on the real numbers which is $0$ if $x$ is irrational, and $frac{1}{q}$ where $x$ is rational of the form $frac{p}{q}$ with $p in N$ and $q in Z$ and they are coprime to each other.
In particular, we were to show this function is continuous at all irrational points, discontinuous at all rational points, but Riemann integrable everywhere. This shows that for some functions which don't behave well, you need to go back to the very basic definitions to be able to integrate them. Also, later, we learned this function is not differentiable anywhere. For more information about this, see "Thomae's function" at https://en.wikipedia.org/wiki/Thomae%27s_function .
Overall, note that integration is a "smoothing" operation, as opposed to differentiation which requires a smooth function and, if anything, makes it less "smooth". Thus, there are functions which can be integrated but not differentiated.
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
$endgroup$
add a comment |
$begingroup$
I'm afraid there is no general strategy. In fact, it's a theorem (which can be stated more precisely than I do here) that there are functions that have no 'simple' antiderivative. More concretely, consider the function
$$ x mapsto int_0^x e^{-t^2}, mathrm{d}t.$$
The integrand is perfectly smooth, so this is a well-defined function. However, you can prove that you cannot express it in terms of functions like $sin$, $cos$, $exp$, etc.
Of course, this only talks about the practical side of things. Defining things is not the problem here - we know exactly which functions 'can' be integrated, i.e., whose integral we can define, but that is not the same as calculate.
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
$endgroup$
$begingroup$
So, basically, if we cannot calculate integrals by summation of step functions, then we have to rely on knowing the answer first or converting the integrals into some known form. Is that how it is?
$endgroup$
– Astik
Dec 21 '18 at 18:17
$begingroup$
Also, if a function arrises which has an integral function that we have never seen before, then we can't calculate it?
$endgroup$
– Astik
Dec 21 '18 at 18:18
$begingroup$
It all depends on the function. In general, calculating it by hand is likely to be impossible if not notoriously difficult. The fundamental theorem of calculus seems to be about the only tool we have (at least, the only one that I know of, for as much as that's worth).
$endgroup$
– SvanN
Dec 21 '18 at 18:20
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Historically, integration was first done by using the limits of areas of the rectangles under the step functions as their widths go to 0, and the number of them goes to infinity, as was mentioned above. It was only somewhat later that derivatives were introduced, with the fundamental theorem of calculus after that. This, of course, made it much easier to integrate most "well behaved" functions, even if it cannot always be done in closed form, as was also mentioned in another answer.
In my first-year university advanced calculus course, the teacher took this historical approach by first describing the rigorous minimum requirements for integration, how to do it using step function areas and limits, and then having us calculate various integrals. One of the interesting homework problems was to investigate what he called the "ruler function" (but it's also known by various other names), i.e., a function $f(x)$ defined on the real numbers which is $0$ if $x$ is irrational, and $frac{1}{q}$ where $x$ is rational of the form $frac{p}{q}$ with $p in N$ and $q in Z$ and they are coprime to each other.
In particular, we were to show this function is continuous at all irrational points, discontinuous at all rational points, but Riemann integrable everywhere. This shows that for some functions which don't behave well, you need to go back to the very basic definitions to be able to integrate them. Also, later, we learned this function is not differentiable anywhere. For more information about this, see "Thomae's function" at https://en.wikipedia.org/wiki/Thomae%27s_function .
Overall, note that integration is a "smoothing" operation, as opposed to differentiation which requires a smooth function and, if anything, makes it less "smooth". Thus, there are functions which can be integrated but not differentiated.
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
$endgroup$
add a comment |
$begingroup$
Historically, integration was first done by using the limits of areas of the rectangles under the step functions as their widths go to 0, and the number of them goes to infinity, as was mentioned above. It was only somewhat later that derivatives were introduced, with the fundamental theorem of calculus after that. This, of course, made it much easier to integrate most "well behaved" functions, even if it cannot always be done in closed form, as was also mentioned in another answer.
In my first-year university advanced calculus course, the teacher took this historical approach by first describing the rigorous minimum requirements for integration, how to do it using step function areas and limits, and then having us calculate various integrals. One of the interesting homework problems was to investigate what he called the "ruler function" (but it's also known by various other names), i.e., a function $f(x)$ defined on the real numbers which is $0$ if $x$ is irrational, and $frac{1}{q}$ where $x$ is rational of the form $frac{p}{q}$ with $p in N$ and $q in Z$ and they are coprime to each other.
In particular, we were to show this function is continuous at all irrational points, discontinuous at all rational points, but Riemann integrable everywhere. This shows that for some functions which don't behave well, you need to go back to the very basic definitions to be able to integrate them. Also, later, we learned this function is not differentiable anywhere. For more information about this, see "Thomae's function" at https://en.wikipedia.org/wiki/Thomae%27s_function .
Overall, note that integration is a "smoothing" operation, as opposed to differentiation which requires a smooth function and, if anything, makes it less "smooth". Thus, there are functions which can be integrated but not differentiated.
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
$endgroup$
add a comment |
$begingroup$
Historically, integration was first done by using the limits of areas of the rectangles under the step functions as their widths go to 0, and the number of them goes to infinity, as was mentioned above. It was only somewhat later that derivatives were introduced, with the fundamental theorem of calculus after that. This, of course, made it much easier to integrate most "well behaved" functions, even if it cannot always be done in closed form, as was also mentioned in another answer.
In my first-year university advanced calculus course, the teacher took this historical approach by first describing the rigorous minimum requirements for integration, how to do it using step function areas and limits, and then having us calculate various integrals. One of the interesting homework problems was to investigate what he called the "ruler function" (but it's also known by various other names), i.e., a function $f(x)$ defined on the real numbers which is $0$ if $x$ is irrational, and $frac{1}{q}$ where $x$ is rational of the form $frac{p}{q}$ with $p in N$ and $q in Z$ and they are coprime to each other.
In particular, we were to show this function is continuous at all irrational points, discontinuous at all rational points, but Riemann integrable everywhere. This shows that for some functions which don't behave well, you need to go back to the very basic definitions to be able to integrate them. Also, later, we learned this function is not differentiable anywhere. For more information about this, see "Thomae's function" at https://en.wikipedia.org/wiki/Thomae%27s_function .
Overall, note that integration is a "smoothing" operation, as opposed to differentiation which requires a smooth function and, if anything, makes it less "smooth". Thus, there are functions which can be integrated but not differentiated.
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
$endgroup$
Historically, integration was first done by using the limits of areas of the rectangles under the step functions as their widths go to 0, and the number of them goes to infinity, as was mentioned above. It was only somewhat later that derivatives were introduced, with the fundamental theorem of calculus after that. This, of course, made it much easier to integrate most "well behaved" functions, even if it cannot always be done in closed form, as was also mentioned in another answer.
In my first-year university advanced calculus course, the teacher took this historical approach by first describing the rigorous minimum requirements for integration, how to do it using step function areas and limits, and then having us calculate various integrals. One of the interesting homework problems was to investigate what he called the "ruler function" (but it's also known by various other names), i.e., a function $f(x)$ defined on the real numbers which is $0$ if $x$ is irrational, and $frac{1}{q}$ where $x$ is rational of the form $frac{p}{q}$ with $p in N$ and $q in Z$ and they are coprime to each other.
In particular, we were to show this function is continuous at all irrational points, discontinuous at all rational points, but Riemann integrable everywhere. This shows that for some functions which don't behave well, you need to go back to the very basic definitions to be able to integrate them. Also, later, we learned this function is not differentiable anywhere. For more information about this, see "Thomae's function" at https://en.wikipedia.org/wiki/Thomae%27s_function .
Overall, note that integration is a "smoothing" operation, as opposed to differentiation which requires a smooth function and, if anything, makes it less "smooth". Thus, there are functions which can be integrated but not differentiated.
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
edited Dec 21 '18 at 19:17
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
answered Dec 21 '18 at 19:07
John OmielanJohn Omielan
4,9692218
4,9692218
add a comment |
add a comment |
$begingroup$
I'm afraid there is no general strategy. In fact, it's a theorem (which can be stated more precisely than I do here) that there are functions that have no 'simple' antiderivative. More concretely, consider the function
$$ x mapsto int_0^x e^{-t^2}, mathrm{d}t.$$
The integrand is perfectly smooth, so this is a well-defined function. However, you can prove that you cannot express it in terms of functions like $sin$, $cos$, $exp$, etc.
Of course, this only talks about the practical side of things. Defining things is not the problem here - we know exactly which functions 'can' be integrated, i.e., whose integral we can define, but that is not the same as calculate.
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
$endgroup$
$begingroup$
So, basically, if we cannot calculate integrals by summation of step functions, then we have to rely on knowing the answer first or converting the integrals into some known form. Is that how it is?
$endgroup$
– Astik
Dec 21 '18 at 18:17
$begingroup$
Also, if a function arrises which has an integral function that we have never seen before, then we can't calculate it?
$endgroup$
– Astik
Dec 21 '18 at 18:18
$begingroup$
It all depends on the function. In general, calculating it by hand is likely to be impossible if not notoriously difficult. The fundamental theorem of calculus seems to be about the only tool we have (at least, the only one that I know of, for as much as that's worth).
$endgroup$
– SvanN
Dec 21 '18 at 18:20
add a comment |
$begingroup$
I'm afraid there is no general strategy. In fact, it's a theorem (which can be stated more precisely than I do here) that there are functions that have no 'simple' antiderivative. More concretely, consider the function
$$ x mapsto int_0^x e^{-t^2}, mathrm{d}t.$$
The integrand is perfectly smooth, so this is a well-defined function. However, you can prove that you cannot express it in terms of functions like $sin$, $cos$, $exp$, etc.
Of course, this only talks about the practical side of things. Defining things is not the problem here - we know exactly which functions 'can' be integrated, i.e., whose integral we can define, but that is not the same as calculate.
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
$endgroup$
$begingroup$
So, basically, if we cannot calculate integrals by summation of step functions, then we have to rely on knowing the answer first or converting the integrals into some known form. Is that how it is?
$endgroup$
– Astik
Dec 21 '18 at 18:17
$begingroup$
Also, if a function arrises which has an integral function that we have never seen before, then we can't calculate it?
$endgroup$
– Astik
Dec 21 '18 at 18:18
$begingroup$
It all depends on the function. In general, calculating it by hand is likely to be impossible if not notoriously difficult. The fundamental theorem of calculus seems to be about the only tool we have (at least, the only one that I know of, for as much as that's worth).
$endgroup$
– SvanN
Dec 21 '18 at 18:20
add a comment |
$begingroup$
I'm afraid there is no general strategy. In fact, it's a theorem (which can be stated more precisely than I do here) that there are functions that have no 'simple' antiderivative. More concretely, consider the function
$$ x mapsto int_0^x e^{-t^2}, mathrm{d}t.$$
The integrand is perfectly smooth, so this is a well-defined function. However, you can prove that you cannot express it in terms of functions like $sin$, $cos$, $exp$, etc.
Of course, this only talks about the practical side of things. Defining things is not the problem here - we know exactly which functions 'can' be integrated, i.e., whose integral we can define, but that is not the same as calculate.
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
$endgroup$
I'm afraid there is no general strategy. In fact, it's a theorem (which can be stated more precisely than I do here) that there are functions that have no 'simple' antiderivative. More concretely, consider the function
$$ x mapsto int_0^x e^{-t^2}, mathrm{d}t.$$
The integrand is perfectly smooth, so this is a well-defined function. However, you can prove that you cannot express it in terms of functions like $sin$, $cos$, $exp$, etc.
Of course, this only talks about the practical side of things. Defining things is not the problem here - we know exactly which functions 'can' be integrated, i.e., whose integral we can define, but that is not the same as calculate.
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
answered Dec 21 '18 at 18:03
SvanNSvanN
2,0311422
2,0311422
$begingroup$
So, basically, if we cannot calculate integrals by summation of step functions, then we have to rely on knowing the answer first or converting the integrals into some known form. Is that how it is?
$endgroup$
– Astik
Dec 21 '18 at 18:17
$begingroup$
Also, if a function arrises which has an integral function that we have never seen before, then we can't calculate it?
$endgroup$
– Astik
Dec 21 '18 at 18:18
$begingroup$
It all depends on the function. In general, calculating it by hand is likely to be impossible if not notoriously difficult. The fundamental theorem of calculus seems to be about the only tool we have (at least, the only one that I know of, for as much as that's worth).
$endgroup$
– SvanN
Dec 21 '18 at 18:20
add a comment |
$begingroup$
So, basically, if we cannot calculate integrals by summation of step functions, then we have to rely on knowing the answer first or converting the integrals into some known form. Is that how it is?
$endgroup$
– Astik
Dec 21 '18 at 18:17
$begingroup$
Also, if a function arrises which has an integral function that we have never seen before, then we can't calculate it?
$endgroup$
– Astik
Dec 21 '18 at 18:18
$begingroup$
It all depends on the function. In general, calculating it by hand is likely to be impossible if not notoriously difficult. The fundamental theorem of calculus seems to be about the only tool we have (at least, the only one that I know of, for as much as that's worth).
$endgroup$
– SvanN
Dec 21 '18 at 18:20
$begingroup$
So, basically, if we cannot calculate integrals by summation of step functions, then we have to rely on knowing the answer first or converting the integrals into some known form. Is that how it is?
$endgroup$
– Astik
Dec 21 '18 at 18:17
$begingroup$
So, basically, if we cannot calculate integrals by summation of step functions, then we have to rely on knowing the answer first or converting the integrals into some known form. Is that how it is?
$endgroup$
– Astik
Dec 21 '18 at 18:17
$begingroup$
Also, if a function arrises which has an integral function that we have never seen before, then we can't calculate it?
$endgroup$
– Astik
Dec 21 '18 at 18:18
$begingroup$
Also, if a function arrises which has an integral function that we have never seen before, then we can't calculate it?
$endgroup$
– Astik
Dec 21 '18 at 18:18
$begingroup$
It all depends on the function. In general, calculating it by hand is likely to be impossible if not notoriously difficult. The fundamental theorem of calculus seems to be about the only tool we have (at least, the only one that I know of, for as much as that's worth).
$endgroup$
– SvanN
Dec 21 '18 at 18:20
$begingroup$
It all depends on the function. In general, calculating it by hand is likely to be impossible if not notoriously difficult. The fundamental theorem of calculus seems to be about the only tool we have (at least, the only one that I know of, for as much as that's worth).
$endgroup$
– SvanN
Dec 21 '18 at 18:20
add a comment |
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$begingroup$
I'm afraid your PS is your answer (but the function need not be monotone).
$endgroup$
– Ethan Bolker
Dec 21 '18 at 18:06
$begingroup$
is this question about techniques of integration (like substitution or integration by parts or using partial fraction expansion)? or is it about finding the area under the curve as done with Riemann summation?
$endgroup$
– robert bristow-johnson
Dec 21 '18 at 19:24