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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 | 

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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