Mixing
Integral of a modulus function vs absolute value of definite integral
$begingroup$
Please refer to the diagram below.
The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.
Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.
I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.
Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...
calculus integration definite-integrals
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
$endgroup$
add a comment |
$begingroup$
Please refer to the diagram below.
The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.
Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.
I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.
Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...
calculus integration definite-integrals
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
$endgroup$
$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52
add a comment |
$begingroup$
Please refer to the diagram below.
The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.
Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.
I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.
Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...
calculus integration definite-integrals
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
$endgroup$
Please refer to the diagram below.
The diagram shows a curve with equation $y = cos {xover 2} cos x$, for 0 $le x$ $le$ $pi$, along with the $x$ and $y$-intercepts of the graph.
Question: By first finding $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, explain why $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ is smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$. You may refer to the graph provided for assistance.
I have found $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ and have done so as shown:
I then start to calculate $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ as shown below.
Now, I realise I would also end up calculating $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ eventually which implies that $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$ = $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$? But the question has already stated that $vert int_{piover 3}^{2piover 3} cos {xover 2} cos x ,dx vert$ will be smaller than $int_{piover 3}^{2piover 3} vert cos {xover 2} cos x vert,dx$, so I'm not really sure how to proceed from here...
calculus integration definite-integrals
calculus integration definite-integrals
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
asked Jan 2 '17 at 14:44
Charlz97Charlz97
12312
12312
$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52
add a comment |
$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52
$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52
$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
$$
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{pi/2}f(x),dx
-
int_{pi/2}^{2pi/3}f(x),dx
$$
On the other hand
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
$$
does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
$$
int_{pi/3}^{2pi/3}|f(x)|,dx=
(F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
-F(2pi/3)-F(pi/2)+2F(pi/2)
$$
whereas
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
|F(2pi/3)-F(pi/3)|
$$
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
$endgroup$
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%2f2080533%2fintegral-of-a-modulus-function-vs-absolute-value-of-definite-integral%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
$begingroup$
Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
$$
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{pi/2}f(x),dx
-
int_{pi/2}^{2pi/3}f(x),dx
$$
On the other hand
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
$$
does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
$$
int_{pi/3}^{2pi/3}|f(x)|,dx=
(F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
-F(2pi/3)-F(pi/2)+2F(pi/2)
$$
whereas
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
|F(2pi/3)-F(pi/3)|
$$
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
$$
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{pi/2}f(x),dx
-
int_{pi/2}^{2pi/3}f(x),dx
$$
On the other hand
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
$$
does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
$$
int_{pi/3}^{2pi/3}|f(x)|,dx=
(F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
-F(2pi/3)-F(pi/2)+2F(pi/2)
$$
whereas
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
|F(2pi/3)-F(pi/3)|
$$
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
$$
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{pi/2}f(x),dx
-
int_{pi/2}^{2pi/3}f(x),dx
$$
On the other hand
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
$$
does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
$$
int_{pi/3}^{2pi/3}|f(x)|,dx=
(F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
-F(2pi/3)-F(pi/2)+2F(pi/2)
$$
whereas
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
|F(2pi/3)-F(pi/3)|
$$
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
$endgroup$
Let $f(x)=cosfrac{x}{2}cos x$. Then $f$ has a change of sign at $x=pi/2$, so
$$
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{2pi/3}|f(x)|,dx
=
int_{pi/3}^{pi/2}f(x),dx
-
int_{pi/2}^{2pi/3}f(x),dx
$$
On the other hand
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|
$$
does not need to be split at the change of sign. If $F(x)$ is an antiderivative of $f(x)$, you have
$$
int_{pi/3}^{2pi/3}|f(x)|,dx=
(F(pi/2)-F(pi/3))-(F(2pi/3)-F(pi/2))=
-F(2pi/3)-F(pi/2)+2F(pi/2)
$$
whereas
$$
left|int_{pi/3}^{2pi/3}|f(x)|,dxright|=
|F(2pi/3)-F(pi/3)|
$$
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
answered Jan 2 '17 at 14:58
egregegreg
180k1485202
180k1485202
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%2f2080533%2fintegral-of-a-modulus-function-vs-absolute-value-of-definite-integral%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
$begingroup$
Think about the cancellation in the areas. Write the area above the $x$-axis as $A$ and the area below as $B$. Now, write the two integrals as sums and differences of $A$ and $B$.
$endgroup$
– Michael Burr
Jan 2 '17 at 14:52