Evaluating integral with inner derivative
$begingroup$
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
asked Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
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add a comment |
$begingroup$
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
asked Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
$endgroup$
$begingroup$
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
$endgroup$
– dmtri
Jan 3 at 18:44
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@Mark Viola, another proof that servers are not fast enough :)
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– dmtri
Jan 3 at 19:30
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@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
$endgroup$
– Mark Viola
Jan 3 at 20:24
add a comment |
$begingroup$
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
asked Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
$endgroup$
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 Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
asked Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
asked Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
asked Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
asked Jan 3 at 18:29
rodger_kicksrodger_kicks
8110
8110
$begingroup$
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
$endgroup$
– dmtri
Jan 3 at 18:44
$begingroup$
@Mark Viola, another proof that servers are not fast enough :)
$endgroup$
– dmtri
Jan 3 at 19:30
$begingroup$
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
$endgroup$
– Mark Viola
Jan 3 at 20:24
add a comment |
$begingroup$
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
$endgroup$
– dmtri
Jan 3 at 18:44
$begingroup$
@Mark Viola, another proof that servers are not fast enough :)
$endgroup$
– dmtri
Jan 3 at 19:30
$begingroup$
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
$endgroup$
– Mark Viola
Jan 3 at 20:24
$begingroup$
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
$endgroup$
– dmtri
Jan 3 at 18:44
$begingroup$
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
$endgroup$
– dmtri
Jan 3 at 18:44
$begingroup$
@Mark Viola, another proof that servers are not fast enough :)
$endgroup$
– dmtri
Jan 3 at 19:30
$begingroup$
@Mark Viola, another proof that servers are not fast enough :)
$endgroup$
– dmtri
Jan 3 at 19:30
$begingroup$
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
$endgroup$
– Mark Viola
Jan 3 at 20:24
$begingroup$
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
$endgroup$
– Mark Viola
Jan 3 at 20:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
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add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
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 Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
$endgroup$
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 Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
answered Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
answered Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
answered Jan 3 at 18:32
Mark ViolaMark Viola
134k1278176
134k1278176
add a comment |
add a comment |
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$begingroup$
Sorry, I did not see any answers at all that time. I will delete my comments. +5 from me
$endgroup$
– dmtri
Jan 3 at 18:44
$begingroup$
@Mark Viola, another proof that servers are not fast enough :)
$endgroup$
– dmtri
Jan 3 at 19:30
$begingroup$
@dmtri That has happened to me too, so no worry. Happy Holidays! ;-)
$endgroup$
– Mark Viola
Jan 3 at 20:24