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using Leibniz rule to solve definite integral


Leibniz integral rule implementationDerivative w.r.t. limits of integration of double definite integral (Leibniz' rule)Leibniz rule for improper integralA general case - Leibniz ruleMisconception in applying Leibniz' ruleDerivative of Parametric Integral Using LeibnizApplying Leibniz rule to multiple integralUsing Leibniz Rule to Find an ExpressionLeibniz integral rule in SDEWhy can I not use leibniz integral rule in this case?













4












$begingroup$


The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is



I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.






integration definite-integrals







share|cite|improve this question











$endgroup$











  • $begingroup$
    Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
    $endgroup$
    – cmk
    9 hours ago











  • $begingroup$
    why can't i use differentiation
    $endgroup$
    – Lakshay
    9 hours ago






  • 1




    $begingroup$
    You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
    $endgroup$
    – Vizag
    9 hours ago










  • $begingroup$
    This question was asked in IIT JEE (A) in 2016.
    $endgroup$
    – Dr Zafar Ahmed DSc
    8 hours ago















4












$begingroup$


The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is



I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.






integration definite-integrals







share|cite|improve this question











$endgroup$











  • $begingroup$
    Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
    $endgroup$
    – cmk
    9 hours ago











  • $begingroup$
    why can't i use differentiation
    $endgroup$
    – Lakshay
    9 hours ago






  • 1




    $begingroup$
    You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
    $endgroup$
    – Vizag
    9 hours ago










  • $begingroup$
    This question was asked in IIT JEE (A) in 2016.
    $endgroup$
    – Dr Zafar Ahmed DSc
    8 hours ago













4












4








4


1



$begingroup$


The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is



I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.






integration definite-integrals







share|cite|improve this question











$endgroup$




The total number of distinct $x$ in $[0,1]$ for which $$int_0^xfract^21+t^4dt=2x-1$$ holds is



I tried solving using Leibniz rule but that gives $0$ no. of solutions. However the answer is $1$.






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









cmk

62010




62010










asked 9 hours ago









LakshayLakshay

1306




1306











  • $begingroup$
    Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
    $endgroup$
    – cmk
    9 hours ago











  • $begingroup$
    why can't i use differentiation
    $endgroup$
    – Lakshay
    9 hours ago






  • 1




    $begingroup$
    You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
    $endgroup$
    – Vizag
    9 hours ago










  • $begingroup$
    This question was asked in IIT JEE (A) in 2016.
    $endgroup$
    – Dr Zafar Ahmed DSc
    8 hours ago
















  • $begingroup$
    Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
    $endgroup$
    – cmk
    9 hours ago











  • $begingroup$
    why can't i use differentiation
    $endgroup$
    – Lakshay
    9 hours ago






  • 1




    $begingroup$
    You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
    $endgroup$
    – Vizag
    9 hours ago










  • $begingroup$
    This question was asked in IIT JEE (A) in 2016.
    $endgroup$
    – Dr Zafar Ahmed DSc
    8 hours ago















$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago





$begingroup$
Differentiating doesn't really make sense. Try to use the intermediate value theorem (with some bounding of the integral).
$endgroup$
– cmk
9 hours ago













$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago




$begingroup$
why can't i use differentiation
$endgroup$
– Lakshay
9 hours ago




1




1




$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago




$begingroup$
You can't use differentiation because $f'(x) = g'(x) not implies f(x) = g(x)$ in general.
$endgroup$
– Vizag
9 hours ago












$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago




$begingroup$
This question was asked in IIT JEE (A) in 2016.
$endgroup$
– Dr Zafar Ahmed DSc
8 hours ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      it complicates the solution.
      $endgroup$
      – Dr Zafar Ahmed DSc
      8 hours ago











    Your Answer








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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.






    share|cite|improve this answer











    $endgroup$

















      7












      $begingroup$

      Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.






      share|cite|improve this answer











      $endgroup$















        7












        7








        7





        $begingroup$

        Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.






        share|cite|improve this answer











        $endgroup$



        Consider the function $$f(x)=2x-1-intlimits_0^xfract^21+t^4 dt.$$ Clearly, $f$ is continuous on $[0,1].$ Note that $f(0)=-1<0$, and $$f(1)=1-intlimits_0^1fract^21+t^4 dtgeq 1-intlimits_0^1frac11+0 dt=0.$$ If $f(1)$ is zero, we've found a solution. If not, then $f(1)>0.$ In this case, the intermediate value theorem guarantees that there exists $cin (0,1)$ for which $f(c)=0,$ or $$2c-1=intlimits_0^cfract^21+t^4 dt.$$ This shows that there is, at least, one solution. Next, note that $$f'(x)=2-fracx^21+x^4>0$$ for $xin [0,1]$, so the function is strictly increasing, which tells us that the previously-found zero of $f$ is the only such zero. In the end, we did use the fundamental theorem of calculus here, but only to show that the zero of $f$ was unique.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 8 hours ago

























        answered 9 hours ago









        cmkcmk

        62010




        62010





















            2












            $begingroup$

            Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              it complicates the solution.
              $endgroup$
              – Dr Zafar Ahmed DSc
              8 hours ago















            2












            $begingroup$

            Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              it complicates the solution.
              $endgroup$
              – Dr Zafar Ahmed DSc
              8 hours ago













            2












            2








            2





            $begingroup$

            Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$






            share|cite|improve this answer









            $endgroup$



            Hint: Use that $$1+t^4=1+t^4+2t^2-2t^2=(1+t^2)^2-(sqrt2t)^2=(1+t^2-sqrt2t)(1+t^2+sqrt2t)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 9 hours ago









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            81.3k42867




            81.3k42867











            • $begingroup$
              it complicates the solution.
              $endgroup$
              – Dr Zafar Ahmed DSc
              8 hours ago
















            • $begingroup$
              it complicates the solution.
              $endgroup$
              – Dr Zafar Ahmed DSc
              8 hours ago















            $begingroup$
            it complicates the solution.
            $endgroup$
            – Dr Zafar Ahmed DSc
            8 hours ago




            $begingroup$
            it complicates the solution.
            $endgroup$
            – Dr Zafar Ahmed DSc
            8 hours ago

















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