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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











$endgroup$








  • 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
















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











$endgroup$








  • 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














4












4








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











$endgroup$




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




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 4 hours ago









Eevee Trainer

9,92931740




9,92931740










asked 4 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
    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














  • 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








2




2




$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
4 hours ago




$begingroup$
How exactly did you go about calculating the antiderivative? Wolfram Alpha gives a much different result.
$endgroup$
– Eevee Trainer
4 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
4 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
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






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






5




5




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




$begingroup$
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




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










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

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    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 4 hours ago









        Eevee TrainerEevee Trainer

        9,92931740




        9,92931740






























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