Why can I not use leibniz integral rule in this case? The 2019 Stack Overflow Developer Survey Results Are InLeibniz rule for an improper integralIntegrate $ sin x /(1 + A sin x)$ over the range $0$,$2 pi$ for $A=0.2$What is wrong with this integration of $ int_0^2pisin x /(1 + A sin x)$Let the function satisfy $f(x)f'(-x)=f(-x)f'(x)$ and $f(0)=3$ for all $x$Leibniz rule or not? improper integral?Is this transformation for the integral $int_0^infty frac1te^-a^2/t^2-b t textdt$ correct?Analytic solution to definite integral problemWhen deriving Leibniz Integral rule, why not take $f$ as another variable?Applying Leibniz rule to multiple integralLeibniz integral rule for higher order derivatives

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Why can I not use leibniz integral rule in this case?



The 2019 Stack Overflow Developer Survey Results Are InLeibniz rule for an improper integralIntegrate $ sin x /(1 + A sin x)$ over the range $0$,$2 pi$ for $A=0.2$What is wrong with this integration of $ int_0^2pisin x /(1 + A sin x)$Let the function satisfy $f(x)f'(-x)=f(-x)f'(x)$ and $f(0)=3$ for all $x$Leibniz rule or not? improper integral?Is this transformation for the integral $int_0^infty frac1te^-a^2/t^2-b t textdt$ correct?Analytic solution to definite integral problemWhen deriving Leibniz Integral rule, why not take $f$ as another variable?Applying Leibniz rule to multiple integralLeibniz integral rule for higher order derivatives










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$begingroup$



$int_0^x f(t)dt to 5 ~textas~ |x|to 1 $find number of integers in range of $p $ so that the equation $2x + int_0^x f(t)dt=p$ has two roots opposite signs in $(-1,1)$




Using leibniz rule on the given equation differentiating gives $$f(x)=-2$$ Therefore $$int_0^x f(t)dt to 5$$



becomes
$$-2x + c to 5$$



Which can not satisfy the condition uniquely for $|x|to 1$ gives two values of c for $^+ _- 1$










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    It’s ‘int_0^x’
    $endgroup$
    – Randall
    Mar 30 at 16:46















0












$begingroup$



$int_0^x f(t)dt to 5 ~textas~ |x|to 1 $find number of integers in range of $p $ so that the equation $2x + int_0^x f(t)dt=p$ has two roots opposite signs in $(-1,1)$




Using leibniz rule on the given equation differentiating gives $$f(x)=-2$$ Therefore $$int_0^x f(t)dt to 5$$



becomes
$$-2x + c to 5$$



Which can not satisfy the condition uniquely for $|x|to 1$ gives two values of c for $^+ _- 1$










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    It’s ‘int_0^x’
    $endgroup$
    – Randall
    Mar 30 at 16:46













0












0








0





$begingroup$



$int_0^x f(t)dt to 5 ~textas~ |x|to 1 $find number of integers in range of $p $ so that the equation $2x + int_0^x f(t)dt=p$ has two roots opposite signs in $(-1,1)$




Using leibniz rule on the given equation differentiating gives $$f(x)=-2$$ Therefore $$int_0^x f(t)dt to 5$$



becomes
$$-2x + c to 5$$



Which can not satisfy the condition uniquely for $|x|to 1$ gives two values of c for $^+ _- 1$










share|cite|improve this question











$endgroup$





$int_0^x f(t)dt to 5 ~textas~ |x|to 1 $find number of integers in range of $p $ so that the equation $2x + int_0^x f(t)dt=p$ has two roots opposite signs in $(-1,1)$




Using leibniz rule on the given equation differentiating gives $$f(x)=-2$$ Therefore $$int_0^x f(t)dt to 5$$



becomes
$$-2x + c to 5$$



Which can not satisfy the condition uniquely for $|x|to 1$ gives two values of c for $^+ _- 1$







definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 30 at 23:01









Gerry Myerson

148k8152306




148k8152306










asked Mar 30 at 16:42









AvkaAvka

706




706







  • 1




    $begingroup$
    It’s ‘int_0^x’
    $endgroup$
    – Randall
    Mar 30 at 16:46












  • 1




    $begingroup$
    It’s ‘int_0^x’
    $endgroup$
    – Randall
    Mar 30 at 16:46







1




1




$begingroup$
It’s ‘int_0^x’
$endgroup$
– Randall
Mar 30 at 16:46




$begingroup$
It’s ‘int_0^x’
$endgroup$
– Randall
Mar 30 at 16:46










0






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