Definite integral giving negative value as a result?Why do I get a negative value for this integral?Solving a definite integralReal integral giving a complex resultProgression from indefinite integral to definite integral - $int_0^2pifrac15-3cos x dx$Calculation of definite integralWithout calculating the integral decide if integral is positive or negative / which integral is bigger?Definite integral of absolute value function?Variable substitution in definite integralDefinite integral over singularityInner Product, Definite Integral
Why can't I see bouncing of a switch on an oscilloscope?
Why is Minecraft giving an OpenGL error?
Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?
Did Shadowfax go to Valinor?
I'm flying to France today and my passport expires in less than 2 months
Is it unprofessional to ask if a job posting on GlassDoor is real?
What typically incentivizes a professor to change jobs to a lower ranking university?
Fully-Firstable Anagram Sets
Watching something be written to a file live with tail
Important Resources for Dark Age Civilizations?
tikz convert color string to hex value
What's the point of deactivating Num Lock on login screens?
Approximately how much travel time was saved by the opening of the Suez Canal in 1869?
How to efficiently unroll a matrix by value with numpy?
Today is the Center
How to format long polynomial?
Has there ever been an airliner design involving reducing generator load by installing solar panels?
RSA: Danger of using p to create q
What does the "remote control" for a QF-4 look like?
dbcc cleantable batch size explanation
NMaximize is not converging to a solution
How old can references or sources in a thesis be?
Add text to same line using sed
Is it legal for company to use my work email to pretend I still work there?
Definite integral giving negative value as a result?
Why do I get a negative value for this integral?Solving a definite integralReal integral giving a complex resultProgression from indefinite integral to definite integral - $int_0^2pifrac15-3cos x dx$Calculation of definite integralWithout calculating the integral decide if integral is positive or negative / which integral is bigger?Definite integral of absolute value function?Variable substitution in definite integralDefinite integral over singularityInner Product, Definite Integral
$begingroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
edited 6 hours ago
Eevee Trainer
9,93831740
asked 6 hours ago
wenoweno
39611
$endgroup$
|
show 2 more comments
$begingroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
edited 6 hours ago
Eevee Trainer
9,93831740
asked 6 hours ago
wenoweno
39611
$endgroup$
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
6 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
6 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
6 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
6 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
6 hours ago
|
show 2 more comments
$begingroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
edited 6 hours ago
Eevee Trainer
9,93831740
asked 6 hours ago
wenoweno
39611
$endgroup$
I want to calculate definite integral
$$int_-2^-1 frac1x^2e^frac1x dx = Omega$$
$$int frac1x^2e^frac1x dx=-e^frac1x+C$$
so:
$$Omega = [-e^frac1-2]-[-e^frac1-1]=-frac1sqrte + frac1e$$
which is a negative value. I believe it should be positive.
What went wrong in the process?
calculus integration definite-integrals
calculus integration definite-integrals
edited 6 hours ago
Eevee Trainer
9,93831740
asked 6 hours ago
wenoweno
39611
edited 6 hours ago
Eevee Trainer
9,93831740
asked 6 hours ago
wenoweno
39611
edited 6 hours ago
Eevee Trainer
9,93831740
edited 6 hours ago
Eevee Trainer
9,93831740
edited 6 hours ago
Eevee Trainer
9,93831740
9,93831740
asked 6 hours ago
wenoweno
39611
asked 6 hours ago
wenoweno
39611
asked 6 hours ago
wenoweno
39611
39611
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
6 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
6 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
6 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
6 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
6 hours ago
|
show 2 more comments
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
6 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
6 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
6 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
6 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
6 hours ago
2
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
6 hours ago
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
6 hours ago
2
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
6 hours ago
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
6 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
6 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
6 hours ago
5
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
6 hours ago
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
6 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
6 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
6 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
$endgroup$
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%2f3176540%2fdefinite-integral-giving-negative-value-as-a-result%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
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
$endgroup$
add a comment |
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
$endgroup$
add a comment |
$begingroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
$endgroup$
What you effectively did was swap the order of evaluation for the fundamental theorem of calculus. Recall:
$$int_a^b f(x)dx = F(b) - F(a)$$
when the antiderivative of $f$ is $F$. You instead have $F(a) - F(b)$ ($a=-2,b=-1$) in this case. The end result is merely a sign error - you have precisely the negative of the answer which you should expect.
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
answered 6 hours ago
Eevee TrainerEevee Trainer
9,93831740
9,93831740
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%2f3176540%2fdefinite-integral-giving-negative-value-as-a-result%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
2
$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
6 hours ago
2
$begingroup$
Your antiderivative is completely incorrect: The derivative of $e^1/x^2$ is $e^1/x^2 / (-x^3)$. The red flag that you found is indeed a correct one, and shows that your answer cannot be right. This is a good thing to check.
$endgroup$
– T. Bongers
6 hours ago
$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
6 hours ago
5
$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
6 hours ago
$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
6 hours ago