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Definite integral giving negative value as a result?



The 2019 Stack Overflow Developer Survey Results Are InWhy 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










4












$begingroup$


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?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    Apr 6 at 0:14






  • 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
    Apr 6 at 0:15











  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
    $endgroup$
    – weno
    Apr 6 at 0:19







  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    Apr 6 at 0:25










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    Apr 6 at 0:27















4












$begingroup$


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?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    Apr 6 at 0:14






  • 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
    Apr 6 at 0:15











  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
    $endgroup$
    – weno
    Apr 6 at 0:19







  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    Apr 6 at 0:25










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    Apr 6 at 0:27













4












4








4





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 6 at 0:30









Eevee Trainer

10.4k31742




10.4k31742










asked Apr 6 at 0:09









wenoweno

41811




41811







  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    Apr 6 at 0:14






  • 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
    Apr 6 at 0:15











  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
    $endgroup$
    – weno
    Apr 6 at 0:19







  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    Apr 6 at 0:25










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    Apr 6 at 0:27












  • 2




    $begingroup$
    How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
    $endgroup$
    – Eevee Trainer
    Apr 6 at 0:14






  • 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
    Apr 6 at 0:15











  • $begingroup$
    Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
    $endgroup$
    – weno
    Apr 6 at 0:19







  • 5




    $begingroup$
    You flipped the interval's endpoints. $-2<-1$
    $endgroup$
    – mr_e_man
    Apr 6 at 0:25










  • $begingroup$
    And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
    $endgroup$
    – Lubin
    Apr 6 at 0:27







2




2




$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
Apr 6 at 0:14




$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
Apr 6 at 0:14




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
Apr 6 at 0:15





$begingroup$
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
Apr 6 at 0:15













$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
Apr 6 at 0:19





$begingroup$
Thanks. I have fixed it now. I meant $int frac1x^2 e^frac1xdx$.
$endgroup$
– weno
Apr 6 at 0:19





5




5




$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
Apr 6 at 0:25




$begingroup$
You flipped the interval's endpoints. $-2<-1$
$endgroup$
– mr_e_man
Apr 6 at 0:25












$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
Apr 6 at 0:27




$begingroup$
And just to confirm, taking account of @mr_e_man’s comment above, your work seems correct.
$endgroup$
– Lubin
Apr 6 at 0:27










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









$endgroup$













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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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









    $endgroup$

















      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









      $endgroup$















        4












        4








        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









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 6 at 0:26









        Eevee TrainerEevee Trainer

        10.4k31742




        10.4k31742



























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