Evaluating definite integrals using Fundamental Theorem of Calculus












4












$begingroup$


Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




The following, however, seems to give a counterexample.*

Can someone resolve this for me?:



Let $f(x) = frac{1}{4 sin (x)+5}$.



$f$ is continuous on $[0, 2 pi]$:



enter image description here



Consider two antiderivatives of $f$, $F_1$ and $F_2$:



$$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



$$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



However,



$F_1(2pi)-F_1(0)=2pi/3$



$F_2(2pi)-F_2(0)=0$



Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



$F_1 =$



enter image description here



$F_2=$



enter image description here



*I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






calculus definite-integrals







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    4












    $begingroup$


    Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




    If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




    The following, however, seems to give a counterexample.*

    Can someone resolve this for me?:



    Let $f(x) = frac{1}{4 sin (x)+5}$.



    $f$ is continuous on $[0, 2 pi]$:



    enter image description here



    Consider two antiderivatives of $f$, $F_1$ and $F_2$:



    $$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



    $$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



    Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



    However,



    $F_1(2pi)-F_1(0)=2pi/3$



    $F_2(2pi)-F_2(0)=0$



    Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



    $F_1 =$



    enter image description here



    $F_2=$



    enter image description here



    *I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






    calculus definite-integrals







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    theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      4












      4








      4





      $begingroup$


      Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




      If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




      The following, however, seems to give a counterexample.*

      Can someone resolve this for me?:



      Let $f(x) = frac{1}{4 sin (x)+5}$.



      $f$ is continuous on $[0, 2 pi]$:



      enter image description here



      Consider two antiderivatives of $f$, $F_1$ and $F_2$:



      $$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



      $$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



      Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



      However,



      $F_1(2pi)-F_1(0)=2pi/3$



      $F_2(2pi)-F_2(0)=0$



      Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



      $F_1 =$



      enter image description here



      $F_2=$



      enter image description here



      *I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






      calculus definite-integrals







      share|cite|improve this question









      New contributor




      theorist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):




      If $f$ is continuous on $[a,b]$, then $int_a^b f(x) , dx = F(a)-F(b)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.




      The following, however, seems to give a counterexample.*

      Can someone resolve this for me?:



      Let $f(x) = frac{1}{4 sin (x)+5}$.



      $f$ is continuous on $[0, 2 pi]$:



      enter image description here



      Consider two antiderivatives of $f$, $F_1$ and $F_2$:



      $$F_1(x) = frac{x}{3}+frac{2}{3} tan^{-1}left(frac{cos (x)}{sin (x)+2}right)$$



      $$F_2(x)=frac{1}{3} left(tan ^{-1}left(2-frac{3}{tan left(frac{x}{2}right)+2}right)-tan^{-1}left(2-frac{3}{cot left(frac{x}{2}right)+2}right)right).$$



      Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $int_0^{2pi} f (x) , dx = F_1(2pi)-F_1(0)= F_2(2pi)-F_2(0)$



      However,



      $F_1(2pi)-F_1(0)=2pi/3$



      $F_2(2pi)-F_2(0)=0$



      Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $int_a^b f (x) , dx = F(a)-F(b)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.



      $F_1 =$



      enter image description here



      $F_2=$



      enter image description here



      *I've taken this example function from [a wolfram.com blog.] (https://blog.wolfram.com/2008/01/19/mathematica-and-the-fundamental-theorem-of-calculus/)






      calculus definite-integrals




      calculus definite-integrals






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          A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






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            $begingroup$

            A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






            share|cite|improve this answer









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              11












              $begingroup$

              A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






              share|cite|improve this answer









              $endgroup$
















                11












                11








                11





                $begingroup$

                A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.






                share|cite|improve this answer









                $endgroup$



                A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Martin ArgeramiMartin Argerami

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