For line integral, how to prove $int_L f(vec{x})ds=int_L f(vec{x})dl$ while $ds$ represent arc-length and...












1












$begingroup$


Suppose line $L$ point $vec{x}in L$ and function $f(vec{x})$ is continues on $L$.



Giving a partition of $L$ into small curves of $L_i$, the definition of line integral is $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i$ while $s_i$ is the arc-length of small curve $L_i$ , $lambda=max(Delta s_i)$ and $vec{x}_iin L_i$



My question is if it is sustains that $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i=lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i$ while $Delta l_i$ is the chord length of the small curve $L_i$.



Thanks a lot!




The following are my attempt:



It is known to me that $Delta l=Delta s+o(Delta s)$ while $lim_{Delta lto0}frac{o(Delta s)}{Delta s}=0$ Thus



begin{aligned}
&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)o_i(Delta s_i)\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}\
end{aligned}



Therefore, my target would be $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}=0$ .



However I may have difficulty with this due to the unknow sigh of $f(
vec{x})$ .






real-analysis calculus integration limits definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why does the sign of $f$ matter? All you need is that $|f|$ be bounded. Just bound by absolute values and use $o(Delta s_i)le epsilon$ when $lambda<delta$. Don't you mean $limlimits_{Delta sto 0}frac{o(Delta s)}{Delta s} = 0$? But of course if $Delta sto 0$, then $Delta lle Delta s to 0$.
    $endgroup$
    – Ted Shifrin
    Jan 25 at 2:17












  • $begingroup$
    @TedShifrin It matters because I would like to prove that $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilon$ if the function is one sigh then $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilonleft|sum_{i=1}^n f(vec{x}_i)Delta s_iright|$ since $left|sum_{i=1}^n f(vec{x}_i)Delta s_iright| <infty$ thus sustains.
    $endgroup$
    – Shore
    Jan 25 at 2:24


















1












$begingroup$


Suppose line $L$ point $vec{x}in L$ and function $f(vec{x})$ is continues on $L$.



Giving a partition of $L$ into small curves of $L_i$, the definition of line integral is $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i$ while $s_i$ is the arc-length of small curve $L_i$ , $lambda=max(Delta s_i)$ and $vec{x}_iin L_i$



My question is if it is sustains that $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i=lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i$ while $Delta l_i$ is the chord length of the small curve $L_i$.



Thanks a lot!




The following are my attempt:



It is known to me that $Delta l=Delta s+o(Delta s)$ while $lim_{Delta lto0}frac{o(Delta s)}{Delta s}=0$ Thus



begin{aligned}
&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)o_i(Delta s_i)\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}\
end{aligned}



Therefore, my target would be $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}=0$ .



However I may have difficulty with this due to the unknow sigh of $f(
vec{x})$ .






real-analysis calculus integration limits definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why does the sign of $f$ matter? All you need is that $|f|$ be bounded. Just bound by absolute values and use $o(Delta s_i)le epsilon$ when $lambda<delta$. Don't you mean $limlimits_{Delta sto 0}frac{o(Delta s)}{Delta s} = 0$? But of course if $Delta sto 0$, then $Delta lle Delta s to 0$.
    $endgroup$
    – Ted Shifrin
    Jan 25 at 2:17












  • $begingroup$
    @TedShifrin It matters because I would like to prove that $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilon$ if the function is one sigh then $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilonleft|sum_{i=1}^n f(vec{x}_i)Delta s_iright|$ since $left|sum_{i=1}^n f(vec{x}_i)Delta s_iright| <infty$ thus sustains.
    $endgroup$
    – Shore
    Jan 25 at 2:24
















1












1








1





$begingroup$


Suppose line $L$ point $vec{x}in L$ and function $f(vec{x})$ is continues on $L$.



Giving a partition of $L$ into small curves of $L_i$, the definition of line integral is $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i$ while $s_i$ is the arc-length of small curve $L_i$ , $lambda=max(Delta s_i)$ and $vec{x}_iin L_i$



My question is if it is sustains that $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i=lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i$ while $Delta l_i$ is the chord length of the small curve $L_i$.



Thanks a lot!




The following are my attempt:



It is known to me that $Delta l=Delta s+o(Delta s)$ while $lim_{Delta lto0}frac{o(Delta s)}{Delta s}=0$ Thus



begin{aligned}
&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)o_i(Delta s_i)\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}\
end{aligned}



Therefore, my target would be $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}=0$ .



However I may have difficulty with this due to the unknow sigh of $f(
vec{x})$ .






real-analysis calculus integration limits definite-integrals







share|cite|improve this question











$endgroup$




Suppose line $L$ point $vec{x}in L$ and function $f(vec{x})$ is continues on $L$.



Giving a partition of $L$ into small curves of $L_i$, the definition of line integral is $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i$ while $s_i$ is the arc-length of small curve $L_i$ , $lambda=max(Delta s_i)$ and $vec{x}_iin L_i$



My question is if it is sustains that $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i=lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i$ while $Delta l_i$ is the chord length of the small curve $L_i$.



Thanks a lot!




The following are my attempt:



It is known to me that $Delta l=Delta s+o(Delta s)$ while $lim_{Delta lto0}frac{o(Delta s)}{Delta s}=0$ Thus



begin{aligned}
&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_i\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)o_i(Delta s_i)\
=&lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta l_i-lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}\
end{aligned}



Therefore, my target would be $lim_{lambdato0}sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}=0$ .



However I may have difficulty with this due to the unknow sigh of $f(
vec{x})$ .






real-analysis calculus integration limits definite-integrals




real-analysis calculus integration limits definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 2:23







Shore

















asked Jan 25 at 2:06









ShoreShore

183




183












  • $begingroup$
    Why does the sign of $f$ matter? All you need is that $|f|$ be bounded. Just bound by absolute values and use $o(Delta s_i)le epsilon$ when $lambda<delta$. Don't you mean $limlimits_{Delta sto 0}frac{o(Delta s)}{Delta s} = 0$? But of course if $Delta sto 0$, then $Delta lle Delta s to 0$.
    $endgroup$
    – Ted Shifrin
    Jan 25 at 2:17












  • $begingroup$
    @TedShifrin It matters because I would like to prove that $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilon$ if the function is one sigh then $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilonleft|sum_{i=1}^n f(vec{x}_i)Delta s_iright|$ since $left|sum_{i=1}^n f(vec{x}_i)Delta s_iright| <infty$ thus sustains.
    $endgroup$
    – Shore
    Jan 25 at 2:24




















  • $begingroup$
    Why does the sign of $f$ matter? All you need is that $|f|$ be bounded. Just bound by absolute values and use $o(Delta s_i)le epsilon$ when $lambda<delta$. Don't you mean $limlimits_{Delta sto 0}frac{o(Delta s)}{Delta s} = 0$? But of course if $Delta sto 0$, then $Delta lle Delta s to 0$.
    $endgroup$
    – Ted Shifrin
    Jan 25 at 2:17












  • $begingroup$
    @TedShifrin It matters because I would like to prove that $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilon$ if the function is one sigh then $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilonleft|sum_{i=1}^n f(vec{x}_i)Delta s_iright|$ since $left|sum_{i=1}^n f(vec{x}_i)Delta s_iright| <infty$ thus sustains.
    $endgroup$
    – Shore
    Jan 25 at 2:24


















$begingroup$
Why does the sign of $f$ matter? All you need is that $|f|$ be bounded. Just bound by absolute values and use $o(Delta s_i)le epsilon$ when $lambda<delta$. Don't you mean $limlimits_{Delta sto 0}frac{o(Delta s)}{Delta s} = 0$? But of course if $Delta sto 0$, then $Delta lle Delta s to 0$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:17






$begingroup$
Why does the sign of $f$ matter? All you need is that $|f|$ be bounded. Just bound by absolute values and use $o(Delta s_i)le epsilon$ when $lambda<delta$. Don't you mean $limlimits_{Delta sto 0}frac{o(Delta s)}{Delta s} = 0$? But of course if $Delta sto 0$, then $Delta lle Delta s to 0$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:17














$begingroup$
@TedShifrin It matters because I would like to prove that $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilon$ if the function is one sigh then $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilonleft|sum_{i=1}^n f(vec{x}_i)Delta s_iright|$ since $left|sum_{i=1}^n f(vec{x}_i)Delta s_iright| <infty$ thus sustains.
$endgroup$
– Shore
Jan 25 at 2:24






$begingroup$
@TedShifrin It matters because I would like to prove that $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilon$ if the function is one sigh then $left|sum_{i=1}^n f(vec{x}_i)Delta s_ifrac{o_i(Delta s_i)}{Delta s_i}right|<epsilonleft|sum_{i=1}^n f(vec{x}_i)Delta s_iright|$ since $left|sum_{i=1}^n f(vec{x}_i)Delta s_iright| <infty$ thus sustains.
$endgroup$
– Shore
Jan 25 at 2:24












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