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
$begingroup$
$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 $^+ _- 1$
definite-integrals
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
asked Mar 30 at 16:42
AvkaAvka
706
$endgroup$
add a comment |
$begingroup$
$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 $^+ _- 1$
definite-integrals
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
asked Mar 30 at 16:42
AvkaAvka
706
$endgroup$
1
$begingroup$
It’s ‘int_0^x’
$endgroup$
– Randall
Mar 30 at 16:46
add a comment |
$begingroup$
$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 $^+ _- 1$
definite-integrals
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
asked Mar 30 at 16:42
AvkaAvka
706
$endgroup$
$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
definite-integrals
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
asked Mar 30 at 16:42
AvkaAvka
706
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
asked Mar 30 at 16:42
AvkaAvka
706
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
edited Mar 30 at 23:01
Gerry Myerson
148k8152306
148k8152306
asked Mar 30 at 16:42
AvkaAvka
706
asked Mar 30 at 16:42
AvkaAvka
706
asked Mar 30 at 16:42
AvkaAvka
706
706
1
$begingroup$
It’s ‘int_0^x’
$endgroup$
– Randall
Mar 30 at 16:46
add a comment |
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
add a comment |
0
active
oldest
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1
$begingroup$
It’s ‘int_0^x’
$endgroup$
– Randall
Mar 30 at 16:46