Sfrgttk

Is it possible to do 50 km distance without any previous training?

Can I ask the recruiters in my resume to put the reason why I am rejected?

Why is 150k or 200k jobs considered good when there's 300k+ births a month?

Is it unprofessional to ask if a job posting on GlassDoor is real?

How much of data wrangling is a data scientist's job?

What's that red-plus icon near a text?

What are these boxed doors outside store fronts in New York?

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)

Why is consensus so controversial in Britain?

Why does Kotter return in Welcome Back Kotter?

Cross compiling for RPi - error while loading shared libraries

Can I make popcorn with any corn?

dbcc cleantable batch size explanation

Are astronomers waiting to see something in an image from a gravitational lens that they've already seen in an adjacent image?

Intersection point of 2 lines defined by 2 points each

If human space travel is limited by the G force vulnerability, is there a way to counter G forces?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

What defenses are there against being summoned by the Gate spell?

LaTeX: Why are digits allowed in environments, but forbidden in commands?

Can a vampire attack twice with their claws using Multiattack?

Perform and show arithmetic with LuaLaTeX

Is it legal for company to use my work email to pretend I still work there?

tikz convert color string to hex value

Today is the Center

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 - $int_{0}^{2pi}frac{1}{5-3cos x} dx$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

4

$begingroup$

I want to calculate definite integral

$$int_{-2}^{-1} frac{1}{x^2}e^{frac{1}{x}} dx = Omega$$

$$int frac{1}{x^2}e^{frac{1}{x}} dx=-e^{frac{1}{x}}+C$$

so:

$$Omega = [-e^{frac{1}{-2}}]-[-e^{frac{1}{-1}}]=-frac{1}{sqrt{e}} + frac{1}{e}$$

which is a negative value. I believe it should be positive.

What went wrong in the process?

calculus integration definite-integrals

share|cite|improve this question

edited 4 hours ago

Eevee Trainer

9,92931740

asked 4 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
  4 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
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
  $endgroup$
  – weno
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  You flipped the interval's endpoints. $-2<-1$
  $endgroup$
  – mr_e_man
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
  $endgroup$
  – Lubin
  4 hours ago
  

  
  
  
  

 | 
show 2 more comments

4

$begingroup$

I want to calculate definite integral

$$int_{-2}^{-1} frac{1}{x^2}e^{frac{1}{x}} dx = Omega$$

$$int frac{1}{x^2}e^{frac{1}{x}} dx=-e^{frac{1}{x}}+C$$

so:

$$Omega = [-e^{frac{1}{-2}}]-[-e^{frac{1}{-1}}]=-frac{1}{sqrt{e}} + frac{1}{e}$$

which is a negative value. I believe it should be positive.

What went wrong in the process?

calculus integration definite-integrals

share|cite|improve this question

edited 4 hours ago

Eevee Trainer

9,92931740

asked 4 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
  4 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
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I have fixed it now. I meant $int frac{1}{x^2} e^{frac{1}{x}}dx$.
  $endgroup$
  – weno
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  You flipped the interval's endpoints. $-2<-1$
  $endgroup$
  – mr_e_man
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
  $endgroup$
  – Lubin
  4 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I want to calculate definite integral

$$int_{-2}^{-1} frac{1}{x^2}e^{frac{1}{x}} dx = Omega$$

$$int frac{1}{x^2}e^{frac{1}{x}} dx=-e^{frac{1}{x}}+C$$

so:

$$Omega = [-e^{frac{1}{-2}}]-[-e^{frac{1}{-1}}]=-frac{1}{sqrt{e}} + frac{1}{e}$$

which is a negative value. I believe it should be positive.

What went wrong in the process?

calculus integration definite-integrals

share|cite|improve this question

edited 4 hours ago

Eevee Trainer

9,92931740

asked 4 hours ago

wenoweno

39611

$endgroup$

share|cite|improve this question

edited 4 hours ago

Eevee Trainer

9,92931740

asked 4 hours ago

wenoweno

39611

share|cite|improve this question

edited 4 hours ago

Eevee Trainer

9,92931740

asked 4 hours ago

wenoweno

39611

edited 4 hours ago

Eevee Trainer

9,92931740

edited 4 hours ago

Eevee Trainer

9,92931740

asked 4 hours ago

wenoweno

39611

asked 4 hours ago

wenoweno

39611

1 Answer
1

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

Eevee TrainerEevee Trainer

9,92931740

$endgroup$

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
});


}
});

draft saved

draft discarded

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

Name

Email

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

Name

Email

Required, but never shown

4

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

Eevee TrainerEevee Trainer

9,92931740

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

Eevee TrainerEevee Trainer

9,92931740

$endgroup$

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.

share|cite|improve this answer

answered 4 hours ago

Eevee TrainerEevee Trainer

9,92931740

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Eevee TrainerEevee Trainer

9,92931740

answered 4 hours ago

Eevee TrainerEevee Trainer

9,92931740

answered 4 hours ago

Eevee TrainerEevee Trainer

9,92931740