Mixing
Connection between u-substitution change of limit and Riemann sum
$begingroup$
I worked with integration by $u$ substitution for a bit now, but I still have a hard time rationalizing why we change the limits of the integral when doing $u$ substitution (I know we don't have to change the limits, there is another way, but I'm not interested in that in this question).
I always think of the integral as the sum of infinite many rectangles under a curve (as an intuition.) Let's say we have the following more formal definition of a definite integral:
$$int_{{,a}}^{{,b}}{{fleft( x right),dx}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {x_i^*} right)Delta x}
$$
I tried to bring this together with $u$ substitution, because I want to actually understand why we need to change the limits of the integral. Say we have:
$$int_{{,a}}^{{,b}}{{f(g(x))g'(x),dx}}$$
Let $u = g(x)$, then I think we have:
$$int_{{,g(a)}}^{{,g(b)}}{{f(u),du}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {u_i^*} right)Delta u}$$
But what I don't understand is: what is the proper justification to change the limits of the integral that we change $int_{{a}}^{{b}}$ to $int_{{g(a)}}^{{g(b)}}$, and what effect does this have on the sum, the actual definition of the integral? Is there any connection to the Riemann sum?
calculus integration limits intuition riemann-integration
asked Jan 27 at 8:45
MaxMax
674722
$endgroup$
add a comment |
$begingroup$
I worked with integration by $u$ substitution for a bit now, but I still have a hard time rationalizing why we change the limits of the integral when doing $u$ substitution (I know we don't have to change the limits, there is another way, but I'm not interested in that in this question).
I always think of the integral as the sum of infinite many rectangles under a curve (as an intuition.) Let's say we have the following more formal definition of a definite integral:
$$int_{{,a}}^{{,b}}{{fleft( x right),dx}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {x_i^*} right)Delta x}
$$
I tried to bring this together with $u$ substitution, because I want to actually understand why we need to change the limits of the integral. Say we have:
$$int_{{,a}}^{{,b}}{{f(g(x))g'(x),dx}}$$
Let $u = g(x)$, then I think we have:
$$int_{{,g(a)}}^{{,g(b)}}{{f(u),du}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {u_i^*} right)Delta u}$$
But what I don't understand is: what is the proper justification to change the limits of the integral that we change $int_{{a}}^{{b}}$ to $int_{{g(a)}}^{{g(b)}}$, and what effect does this have on the sum, the actual definition of the integral? Is there any connection to the Riemann sum?
calculus integration limits intuition riemann-integration
asked Jan 27 at 8:45
MaxMax
674722
$endgroup$
$begingroup$
after solving the substituted integral $int f(u)du$ you need to backsubstitute $u$ which is expressed by chaning the limits accordingly
$endgroup$
– Zest
Jan 27 at 8:51
add a comment |
$begingroup$
I worked with integration by $u$ substitution for a bit now, but I still have a hard time rationalizing why we change the limits of the integral when doing $u$ substitution (I know we don't have to change the limits, there is another way, but I'm not interested in that in this question).
I always think of the integral as the sum of infinite many rectangles under a curve (as an intuition.) Let's say we have the following more formal definition of a definite integral:
$$int_{{,a}}^{{,b}}{{fleft( x right),dx}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {x_i^*} right)Delta x}
$$
I tried to bring this together with $u$ substitution, because I want to actually understand why we need to change the limits of the integral. Say we have:
$$int_{{,a}}^{{,b}}{{f(g(x))g'(x),dx}}$$
Let $u = g(x)$, then I think we have:
$$int_{{,g(a)}}^{{,g(b)}}{{f(u),du}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {u_i^*} right)Delta u}$$
But what I don't understand is: what is the proper justification to change the limits of the integral that we change $int_{{a}}^{{b}}$ to $int_{{g(a)}}^{{g(b)}}$, and what effect does this have on the sum, the actual definition of the integral? Is there any connection to the Riemann sum?
calculus integration limits intuition riemann-integration
asked Jan 27 at 8:45
MaxMax
674722
$endgroup$
I worked with integration by $u$ substitution for a bit now, but I still have a hard time rationalizing why we change the limits of the integral when doing $u$ substitution (I know we don't have to change the limits, there is another way, but I'm not interested in that in this question).
I always think of the integral as the sum of infinite many rectangles under a curve (as an intuition.) Let's say we have the following more formal definition of a definite integral:
$$int_{{,a}}^{{,b}}{{fleft( x right),dx}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {x_i^*} right)Delta x}
$$
I tried to bring this together with $u$ substitution, because I want to actually understand why we need to change the limits of the integral. Say we have:
$$int_{{,a}}^{{,b}}{{f(g(x))g'(x),dx}}$$
Let $u = g(x)$, then I think we have:
$$int_{{,g(a)}}^{{,g(b)}}{{f(u),du}} = mathop {lim }limits_{n to infty } sumlimits_{i = 1}^n {fleft( {u_i^*} right)Delta u}$$
But what I don't understand is: what is the proper justification to change the limits of the integral that we change $int_{{a}}^{{b}}$ to $int_{{g(a)}}^{{g(b)}}$, and what effect does this have on the sum, the actual definition of the integral? Is there any connection to the Riemann sum?
calculus integration limits intuition riemann-integration
calculus integration limits intuition riemann-integration
asked Jan 27 at 8:45
MaxMax
674722
asked Jan 27 at 8:45
MaxMax
674722
asked Jan 27 at 8:45
MaxMax
674722
asked Jan 27 at 8:45
MaxMax
674722
asked Jan 27 at 8:45
MaxMax
674722
674722
$begingroup$
after solving the substituted integral $int f(u)du$ you need to backsubstitute $u$ which is expressed by chaning the limits accordingly
$endgroup$
– Zest
Jan 27 at 8:51
add a comment |
$begingroup$
after solving the substituted integral $int f(u)du$ you need to backsubstitute $u$ which is expressed by chaning the limits accordingly
$endgroup$
– Zest
Jan 27 at 8:51
$begingroup$
after solving the substituted integral $int f(u)du$ you need to backsubstitute $u$ which is expressed by chaning the limits accordingly
$endgroup$
– Zest
Jan 27 at 8:51
$begingroup$
after solving the substituted integral $int f(u)du$ you need to backsubstitute $u$ which is expressed by chaning the limits accordingly
$endgroup$
– Zest
Jan 27 at 8:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The change-of-variables theorem can be proved under very weak assumptions about $f$ and $g$. However, a standard proof assumes that $g$ is continuously differentiable and uses the fundamental theorem of calculus. The theorem holds even if $g$ is not monotone (increasing or decreasing). We can even have $g(a) = g(b)$.
A proof along the lines of what you are suggesting requires an assumption that $g$ is monotone. We begin with a partition $a = x_0 < x_1 < ldots < x_n = b$ and form the sum
$$tag{*}S(P,fg')= sum_{j=1}^n f(g(xi_j))g'(xi_j)(x_j - x_{j-1})$$
where we use intermediate points $xi_j in [x_{j-1},x_j]$ and which will converge with refinement to $$int_a^bf(g(x)) g'(x) , dx$$
If $g$ is increasing then a partition $P'$ of $[g(a),g(b)]$ is naturally induced by
$$g(a) = g(x_0) < g(x_1) < ldots < g(x_n) = g(b),$$
and using the intermediate points $g(xi_j)$, we have a Riemann sum for the integral of $f$ over $[g(a),g(b)]$ taking the form
$$S(P',f) = sum_{j=1}^n f(g(xi_j))(,g(x_j) - g(x_{j-1}),)$$
We need the monotonicity of $g$ to ensure that $g(xi_j) in [g(x_{j-1}), g(x_j)]$.
Applying the mean value theorem, there exist points $eta_j in (x_{j-1},x_j))$ such that
$$tag{**}S(P',f) = sum_{j=1}^n f(g(xi_j))g'(eta_j)(x_j - x_{j-1})$$
Notice the similarity between the sums in (*) and (**). Aside from the distinction between $eta_j$ and $xi_j$, they are identical. Using the continuity (and uniform continuity) of $g'$ we can show that as the partition is refined and both $|P|, |P'| to 0$ we have
$$lim_{|P|| to 0}|S(P,fg') - S(P',f)| = 0$$
Therefore, $S(P',f)$ converges to both integrals and we have
$$lim_{|P'| to 0}S(P',f) = int_{g(a)}^{g(b)} f(u) , du = int_a^b f(g(x)) g'(x) , dx$$
Again, there are a number of ways to prove the change-of-variables theorem -- without the assumption that $g$ is monotone -- that avoid this association with Riemann sums. In the most general form only integrability and not continuity of $f$ and $g'$ is assumed.
answered Jan 27 at 9:47
RRLRRL
53k42573
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089289%2fconnection-between-u-substitution-change-of-limit-and-riemann-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The change-of-variables theorem can be proved under very weak assumptions about $f$ and $g$. However, a standard proof assumes that $g$ is continuously differentiable and uses the fundamental theorem of calculus. The theorem holds even if $g$ is not monotone (increasing or decreasing). We can even have $g(a) = g(b)$.
A proof along the lines of what you are suggesting requires an assumption that $g$ is monotone. We begin with a partition $a = x_0 < x_1 < ldots < x_n = b$ and form the sum
$$tag{*}S(P,fg')= sum_{j=1}^n f(g(xi_j))g'(xi_j)(x_j - x_{j-1})$$
where we use intermediate points $xi_j in [x_{j-1},x_j]$ and which will converge with refinement to $$int_a^bf(g(x)) g'(x) , dx$$
If $g$ is increasing then a partition $P'$ of $[g(a),g(b)]$ is naturally induced by
$$g(a) = g(x_0) < g(x_1) < ldots < g(x_n) = g(b),$$
and using the intermediate points $g(xi_j)$, we have a Riemann sum for the integral of $f$ over $[g(a),g(b)]$ taking the form
$$S(P',f) = sum_{j=1}^n f(g(xi_j))(,g(x_j) - g(x_{j-1}),)$$
We need the monotonicity of $g$ to ensure that $g(xi_j) in [g(x_{j-1}), g(x_j)]$.
Applying the mean value theorem, there exist points $eta_j in (x_{j-1},x_j))$ such that
$$tag{**}S(P',f) = sum_{j=1}^n f(g(xi_j))g'(eta_j)(x_j - x_{j-1})$$
Notice the similarity between the sums in (*) and (**). Aside from the distinction between $eta_j$ and $xi_j$, they are identical. Using the continuity (and uniform continuity) of $g'$ we can show that as the partition is refined and both $|P|, |P'| to 0$ we have
$$lim_{|P|| to 0}|S(P,fg') - S(P',f)| = 0$$
Therefore, $S(P',f)$ converges to both integrals and we have
$$lim_{|P'| to 0}S(P',f) = int_{g(a)}^{g(b)} f(u) , du = int_a^b f(g(x)) g'(x) , dx$$
Again, there are a number of ways to prove the change-of-variables theorem -- without the assumption that $g$ is monotone -- that avoid this association with Riemann sums. In the most general form only integrability and not continuity of $f$ and $g'$ is assumed.
answered Jan 27 at 9:47
RRLRRL
53k42573
$endgroup$
add a comment |
$begingroup$
The change-of-variables theorem can be proved under very weak assumptions about $f$ and $g$. However, a standard proof assumes that $g$ is continuously differentiable and uses the fundamental theorem of calculus. The theorem holds even if $g$ is not monotone (increasing or decreasing). We can even have $g(a) = g(b)$.
A proof along the lines of what you are suggesting requires an assumption that $g$ is monotone. We begin with a partition $a = x_0 < x_1 < ldots < x_n = b$ and form the sum
$$tag{*}S(P,fg')= sum_{j=1}^n f(g(xi_j))g'(xi_j)(x_j - x_{j-1})$$
where we use intermediate points $xi_j in [x_{j-1},x_j]$ and which will converge with refinement to $$int_a^bf(g(x)) g'(x) , dx$$
If $g$ is increasing then a partition $P'$ of $[g(a),g(b)]$ is naturally induced by
$$g(a) = g(x_0) < g(x_1) < ldots < g(x_n) = g(b),$$
and using the intermediate points $g(xi_j)$, we have a Riemann sum for the integral of $f$ over $[g(a),g(b)]$ taking the form
$$S(P',f) = sum_{j=1}^n f(g(xi_j))(,g(x_j) - g(x_{j-1}),)$$
We need the monotonicity of $g$ to ensure that $g(xi_j) in [g(x_{j-1}), g(x_j)]$.
Applying the mean value theorem, there exist points $eta_j in (x_{j-1},x_j))$ such that
$$tag{**}S(P',f) = sum_{j=1}^n f(g(xi_j))g'(eta_j)(x_j - x_{j-1})$$
Notice the similarity between the sums in (*) and (**). Aside from the distinction between $eta_j$ and $xi_j$, they are identical. Using the continuity (and uniform continuity) of $g'$ we can show that as the partition is refined and both $|P|, |P'| to 0$ we have
$$lim_{|P|| to 0}|S(P,fg') - S(P',f)| = 0$$
Therefore, $S(P',f)$ converges to both integrals and we have
$$lim_{|P'| to 0}S(P',f) = int_{g(a)}^{g(b)} f(u) , du = int_a^b f(g(x)) g'(x) , dx$$
Again, there are a number of ways to prove the change-of-variables theorem -- without the assumption that $g$ is monotone -- that avoid this association with Riemann sums. In the most general form only integrability and not continuity of $f$ and $g'$ is assumed.
answered Jan 27 at 9:47
RRLRRL
53k42573
$endgroup$
add a comment |
$begingroup$
The change-of-variables theorem can be proved under very weak assumptions about $f$ and $g$. However, a standard proof assumes that $g$ is continuously differentiable and uses the fundamental theorem of calculus. The theorem holds even if $g$ is not monotone (increasing or decreasing). We can even have $g(a) = g(b)$.
A proof along the lines of what you are suggesting requires an assumption that $g$ is monotone. We begin with a partition $a = x_0 < x_1 < ldots < x_n = b$ and form the sum
$$tag{*}S(P,fg')= sum_{j=1}^n f(g(xi_j))g'(xi_j)(x_j - x_{j-1})$$
where we use intermediate points $xi_j in [x_{j-1},x_j]$ and which will converge with refinement to $$int_a^bf(g(x)) g'(x) , dx$$
If $g$ is increasing then a partition $P'$ of $[g(a),g(b)]$ is naturally induced by
$$g(a) = g(x_0) < g(x_1) < ldots < g(x_n) = g(b),$$
and using the intermediate points $g(xi_j)$, we have a Riemann sum for the integral of $f$ over $[g(a),g(b)]$ taking the form
$$S(P',f) = sum_{j=1}^n f(g(xi_j))(,g(x_j) - g(x_{j-1}),)$$
We need the monotonicity of $g$ to ensure that $g(xi_j) in [g(x_{j-1}), g(x_j)]$.
Applying the mean value theorem, there exist points $eta_j in (x_{j-1},x_j))$ such that
$$tag{**}S(P',f) = sum_{j=1}^n f(g(xi_j))g'(eta_j)(x_j - x_{j-1})$$
Notice the similarity between the sums in (*) and (**). Aside from the distinction between $eta_j$ and $xi_j$, they are identical. Using the continuity (and uniform continuity) of $g'$ we can show that as the partition is refined and both $|P|, |P'| to 0$ we have
$$lim_{|P|| to 0}|S(P,fg') - S(P',f)| = 0$$
Therefore, $S(P',f)$ converges to both integrals and we have
$$lim_{|P'| to 0}S(P',f) = int_{g(a)}^{g(b)} f(u) , du = int_a^b f(g(x)) g'(x) , dx$$
Again, there are a number of ways to prove the change-of-variables theorem -- without the assumption that $g$ is monotone -- that avoid this association with Riemann sums. In the most general form only integrability and not continuity of $f$ and $g'$ is assumed.
answered Jan 27 at 9:47
RRLRRL
53k42573
$endgroup$
The change-of-variables theorem can be proved under very weak assumptions about $f$ and $g$. However, a standard proof assumes that $g$ is continuously differentiable and uses the fundamental theorem of calculus. The theorem holds even if $g$ is not monotone (increasing or decreasing). We can even have $g(a) = g(b)$.
A proof along the lines of what you are suggesting requires an assumption that $g$ is monotone. We begin with a partition $a = x_0 < x_1 < ldots < x_n = b$ and form the sum
$$tag{*}S(P,fg')= sum_{j=1}^n f(g(xi_j))g'(xi_j)(x_j - x_{j-1})$$
where we use intermediate points $xi_j in [x_{j-1},x_j]$ and which will converge with refinement to $$int_a^bf(g(x)) g'(x) , dx$$
If $g$ is increasing then a partition $P'$ of $[g(a),g(b)]$ is naturally induced by
$$g(a) = g(x_0) < g(x_1) < ldots < g(x_n) = g(b),$$
and using the intermediate points $g(xi_j)$, we have a Riemann sum for the integral of $f$ over $[g(a),g(b)]$ taking the form
$$S(P',f) = sum_{j=1}^n f(g(xi_j))(,g(x_j) - g(x_{j-1}),)$$
We need the monotonicity of $g$ to ensure that $g(xi_j) in [g(x_{j-1}), g(x_j)]$.
Applying the mean value theorem, there exist points $eta_j in (x_{j-1},x_j))$ such that
$$tag{**}S(P',f) = sum_{j=1}^n f(g(xi_j))g'(eta_j)(x_j - x_{j-1})$$
Notice the similarity between the sums in (*) and (**). Aside from the distinction between $eta_j$ and $xi_j$, they are identical. Using the continuity (and uniform continuity) of $g'$ we can show that as the partition is refined and both $|P|, |P'| to 0$ we have
$$lim_{|P|| to 0}|S(P,fg') - S(P',f)| = 0$$
Therefore, $S(P',f)$ converges to both integrals and we have
$$lim_{|P'| to 0}S(P',f) = int_{g(a)}^{g(b)} f(u) , du = int_a^b f(g(x)) g'(x) , dx$$
Again, there are a number of ways to prove the change-of-variables theorem -- without the assumption that $g$ is monotone -- that avoid this association with Riemann sums. In the most general form only integrability and not continuity of $f$ and $g'$ is assumed.
answered Jan 27 at 9:47
RRLRRL
53k42573
edited Jan 27 at 10:04
answered Jan 27 at 9:47
RRLRRL
53k42573
answered Jan 27 at 9:47
RRLRRL
53k42573
answered Jan 27 at 9:47
RRLRRL
53k42573
53k42573
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089289%2fconnection-between-u-substitution-change-of-limit-and-riemann-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
after solving the substituted integral $int f(u)du$ you need to backsubstitute $u$ which is expressed by chaning the limits accordingly
$endgroup$
– Zest
Jan 27 at 8:51