Evaluating integral with inner derivative
Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...
$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$
where $f=f(x)$.
Would expanding the inner derivative operator work here?
integration
add a comment |
Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...
$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$
where $f=f(x)$.
Would expanding the inner derivative operator work here?
integration
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago
@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago
add a comment |
Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...
$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$
where $f=f(x)$.
Would expanding the inner derivative operator work here?
integration
Hi I am failing to spot how to integrate this. By parts I'm guessing? Any hints into the right direction would be helpful...
$$ int xfrac{d}{dx}bigg(ffrac{df}{dx}bigg) dx$$
where $f=f(x)$.
Would expanding the inner derivative operator work here?
integration
integration
asked 2 days ago
kroneckerdel69
257
257
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago
@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago
add a comment |
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago
@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago
@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago
@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago
add a comment |
1 Answer
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Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals
$$begin{align}
int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
&=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
end{align}$$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals
$$begin{align}
int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
&=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
end{align}$$
add a comment |
Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals
$$begin{align}
int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
&=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
end{align}$$
add a comment |
Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals
$$begin{align}
int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
&=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
end{align}$$
Note that $f(x)frac{df(x)}{dx}=frac12 frac{df^2(x)}{dx}$. Then, integrating by parts reveals
$$begin{align}
int xfrac{d}{dx}left(f(x)frac{df(x)}{dx}right),dx&=frac12 xfrac{df^2(x)}{dx}-frac12 int frac{df^2(x)}{dx},dx\\
&=frac12xfrac{df^2(x)}{dx}-frac12 f^2(x)+C
end{align}$$
answered 2 days ago
Mark Viola
130k1274170
130k1274170
add a comment |
add a comment |
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Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
– dmtri
2 days ago
@Mark Viola, another proof that servers are not fast enough :)
– dmtri
2 days ago
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
– Mark Viola
2 days ago