Derivative of indefinite integral vs. definite
So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
edited Dec 16 at 1:20
Eevee Trainer
4,090530
asked Dec 16 at 1:13
Josh White
513
add a comment |
So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
edited Dec 16 at 1:20
Eevee Trainer
4,090530
asked Dec 16 at 1:13
Josh White
513
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
Dec 16 at 1:20
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
Dec 16 at 2:35
add a comment |
So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
edited Dec 16 at 1:20
Eevee Trainer
4,090530
asked Dec 16 at 1:13
Josh White
513
So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
integration definite-integrals indefinite-integrals
edited Dec 16 at 1:20
Eevee Trainer
4,090530
asked Dec 16 at 1:13
Josh White
513
edited Dec 16 at 1:20
Eevee Trainer
4,090530
asked Dec 16 at 1:13
Josh White
513
edited Dec 16 at 1:20
Eevee Trainer
4,090530
edited Dec 16 at 1:20
Eevee Trainer
4,090530
edited Dec 16 at 1:20
Eevee Trainer
4,090530
4,090530
asked Dec 16 at 1:13
Josh White
513
asked Dec 16 at 1:13
Josh White
513
asked Dec 16 at 1:13
Josh White
513
513
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
Dec 16 at 1:20
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
Dec 16 at 2:35
add a comment |
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
Dec 16 at 1:20
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
Dec 16 at 2:35
1
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
Dec 16 at 1:20
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
Dec 16 at 1:20
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
Dec 16 at 2:35
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
Dec 16 at 2:35
add a comment |
2 Answers
2
active
oldest
votes
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
answered Dec 16 at 1:18
Eevee Trainer
4,090530
add a comment |
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
answered Dec 16 at 1:19
Michael Wang
228
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%2f3042130%2fderivative-of-indefinite-integral-vs-definite%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
answered Dec 16 at 1:18
Eevee Trainer
4,090530
add a comment |
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
answered Dec 16 at 1:18
Eevee Trainer
4,090530
add a comment |
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
answered Dec 16 at 1:18
Eevee Trainer
4,090530
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
answered Dec 16 at 1:18
Eevee Trainer
4,090530
answered Dec 16 at 1:18
Eevee Trainer
4,090530
answered Dec 16 at 1:18
Eevee Trainer
4,090530
answered Dec 16 at 1:18
Eevee Trainer
4,090530
4,090530
add a comment |
add a comment |
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
answered Dec 16 at 1:19
Michael Wang
228
add a comment |
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
answered Dec 16 at 1:19
Michael Wang
228
add a comment |
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
answered Dec 16 at 1:19
Michael Wang
228
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
answered Dec 16 at 1:19
Michael Wang
228
answered Dec 16 at 1:19
Michael Wang
228
answered Dec 16 at 1:19
Michael Wang
228
answered Dec 16 at 1:19
Michael Wang
228
228
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3042130%2fderivative-of-indefinite-integral-vs-definite%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
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
Dec 16 at 1:20
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
Dec 16 at 2:35