Mixing
How to find $int_a^bx^mdx$ using the limit sum definition of definite integral?
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
We have to evaluate
$$int_a^bx^mdx$$
Converting it to a sum as limit tends to infinity
$$=lim_hto0hsum_r=1^n f(a+rh)=lim_hto0hsum_r=1^n (a+rh)^m$$
with $h=(b-a)/n$ with $ntoinfty$.
Now, I have having trouble going on with this. What to do next?
calculus integration definite-integrals riemann-sum
edited Jul 28 at 14:37
Henning Makholm
225k16290516
asked Jul 28 at 14:21
tatan
5,01442053
 |Â
show 5 more comments
up vote
5
down vote
favorite
We have to evaluate
$$int_a^bx^mdx$$
Converting it to a sum as limit tends to infinity
$$=lim_hto0hsum_r=1^n f(a+rh)=lim_hto0hsum_r=1^n (a+rh)^m$$
with $h=(b-a)/n$ with $ntoinfty$.
Now, I have having trouble going on with this. What to do next?
calculus integration definite-integrals riemann-sum
edited Jul 28 at 14:37
Henning Makholm
225k16290516
asked Jul 28 at 14:21
tatan
5,01442053
1
The sum doesn't appear correct, at least not without restrictions. Is $n$ intended to be $(b-a)/h$?
– B. Mehta
Jul 28 at 14:32
@B.Mehta Yep.... Edited that
– tatan
Jul 28 at 14:33
In that case, is the limit as $n to infty$ also?
– B. Mehta
Jul 28 at 14:34
@B.Mehta $hto 0$ directly implies $ntoinfty$... doesn't it?
– tatan
Jul 28 at 14:35
More likely $h=(b-a)/n$ and the limit should be taken as $ntoinfty,$ I would think. It's not clear what the sum means if $n$ isn't an integer.
– saulspatz
Jul 28 at 14:35
 |Â
show 5 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
We have to evaluate
$$int_a^bx^mdx$$
Converting it to a sum as limit tends to infinity
$$=lim_hto0hsum_r=1^n f(a+rh)=lim_hto0hsum_r=1^n (a+rh)^m$$
with $h=(b-a)/n$ with $ntoinfty$.
Now, I have having trouble going on with this. What to do next?
calculus integration definite-integrals riemann-sum
edited Jul 28 at 14:37
Henning Makholm
225k16290516
asked Jul 28 at 14:21
tatan
5,01442053
We have to evaluate
$$int_a^bx^mdx$$
Converting it to a sum as limit tends to infinity
$$=lim_hto0hsum_r=1^n f(a+rh)=lim_hto0hsum_r=1^n (a+rh)^m$$
with $h=(b-a)/n$ with $ntoinfty$.
Now, I have having trouble going on with this. What to do next?
calculus integration definite-integrals riemann-sum
edited Jul 28 at 14:37
Henning Makholm
225k16290516
asked Jul 28 at 14:21
tatan
5,01442053
edited Jul 28 at 14:37
Henning Makholm
225k16290516
edited Jul 28 at 14:37
Henning Makholm
225k16290516
edited Jul 28 at 14:37
Henning Makholm
225k16290516
225k16290516
asked Jul 28 at 14:21
tatan
5,01442053
asked Jul 28 at 14:21
tatan
5,01442053
asked Jul 28 at 14:21
tatan
5,01442053
5,01442053
1
The sum doesn't appear correct, at least not without restrictions. Is $n$ intended to be $(b-a)/h$?
– B. Mehta
Jul 28 at 14:32
@B.Mehta Yep.... Edited that
– tatan
Jul 28 at 14:33
In that case, is the limit as $n to infty$ also?
– B. Mehta
Jul 28 at 14:34
@B.Mehta $hto 0$ directly implies $ntoinfty$... doesn't it?
– tatan
Jul 28 at 14:35
More likely $h=(b-a)/n$ and the limit should be taken as $ntoinfty,$ I would think. It's not clear what the sum means if $n$ isn't an integer.
– saulspatz
Jul 28 at 14:35
 |Â
show 5 more comments
1
The sum doesn't appear correct, at least not without restrictions. Is $n$ intended to be $(b-a)/h$?
– B. Mehta
Jul 28 at 14:32
@B.Mehta Yep.... Edited that
– tatan
Jul 28 at 14:33
In that case, is the limit as $n to infty$ also?
– B. Mehta
Jul 28 at 14:34
@B.Mehta $hto 0$ directly implies $ntoinfty$... doesn't it?
– tatan
Jul 28 at 14:35
More likely $h=(b-a)/n$ and the limit should be taken as $ntoinfty,$ I would think. It's not clear what the sum means if $n$ isn't an integer.
– saulspatz
Jul 28 at 14:35
1
1
The sum doesn't appear correct, at least not without restrictions. Is $n$ intended to be $(b-a)/h$?
– B. Mehta
Jul 28 at 14:32
The sum doesn't appear correct, at least not without restrictions. Is $n$ intended to be $(b-a)/h$?
– B. Mehta
Jul 28 at 14:32
@B.Mehta Yep.... Edited that
– tatan
Jul 28 at 14:33
@B.Mehta Yep.... Edited that
– tatan
Jul 28 at 14:33
In that case, is the limit as $n to infty$ also?
– B. Mehta
Jul 28 at 14:34
In that case, is the limit as $n to infty$ also?
– B. Mehta
Jul 28 at 14:34
@B.Mehta $hto 0$ directly implies $ntoinfty$... doesn't it?
– tatan
Jul 28 at 14:35
@B.Mehta $hto 0$ directly implies $ntoinfty$... doesn't it?
– tatan
Jul 28 at 14:35
More likely $h=(b-a)/n$ and the limit should be taken as $ntoinfty,$ I would think. It's not clear what the sum means if $n$ isn't an integer.
– saulspatz
Jul 28 at 14:35
More likely $h=(b-a)/n$ and the limit should be taken as $ntoinfty,$ I would think. It's not clear what the sum means if $n$ isn't an integer.
– saulspatz
Jul 28 at 14:35
 |Â
show 5 more comments
3 Answers
3
active
oldest
votes
up vote
4
down vote
Assuming $a>0$, I consider the special case $m=-1$.
Here $f(x)=frac1x$, and we divide the interval $ale xle b$ so that the points of division are in a geometric progression as $$a,ar,ar^2,cdots,ar^n=b,;quadtext so r=left(fracbaright)^1/n$$
The length of the $k$th ($1le kle n$) sub-interval is $$delta_k=ar^k-ar^k-1=ar^k-1(r-1)$$
Let $delta$ be the maximum among $delta_1,delta_2,cdots,delta_n$. Then as $ntoinfty, rto1$ and so $deltato0$. We take $$delta_k=ar^k-1quad,,1le kle n$$
So by definition,
beginalign
int_a^b frac1x,dx&=lim_deltato0sum_k=1^n f(xi_k)delta_k
\&=lim_ntoinftysum_k=1^nfracar^k-1(r-1)ar^k-1
\&=lim_ntoinfty n(r-1)
\&=lim_1/nto0fracleft(fracbaright)^1/n-11/n
\&=ln fracbaqquadqquadleft[because, lim_xto0fraca^x-1x=ln aright]
endalign
For the general problem, we can use the general definition of definite integral of a continuous (or integrable) function. I roughly state the definition so that the notation is clear.
Let $f(x)$ be a bounded function defined on a closed interval $[a,b]$. We divide the interval $[a,b]$ into $n$ subintervals $[x_0,x_1],,[x_1,x_2],cdots,[x_n-1,x_n]$ by introducing the points $x_0,x_2,cdots,x_n$ satisfying $a=x_0<x_1<cdots<x_n=b$. Let $$delta_r=x_r-x_r-1quad,r=1,2,cdots,n$$
Choose any point $xi_r$ of the $r$th subinterval, i.e. $$x_r-1le xi_rle x_r$$
Let $delta=max(delta_1,cdots,delta_n)$, the norm of the subdivision of the partition.
We then form the sum $sum_r=1^nf(xi_r)delta_r$, which depends on the choice of the points $x_r$'s and $xi_r$'s.
We now let $delta$ tend to zero, so that $n$, the number of subdivisions, goes to infinity. If now the sum $sum_r=1^nf(xi_r)delta_r$ tends to a definite limit which is independent of the choice of the points $x_r$'s and $xi_r$'s in $[x_r-1,x_r]$, then this limit is defined to be the definite (Riemann) integral of $f(x)$ over $[a,b]$.
That is, $$int_a^b f(x),dx=lim_deltato0sum_r=1^n f(xi_r)delta_r$$
Many expressions for $int_a^b f(x),dx$ can be found by taking $x_r$'s and $xi_r$'s in suitable manner. We need one such particular choice here.
We choose $x_r$'s in such a way that they form a G.P. with common ratio $rho$ (say), like $$a=x_0,x_1=arho,x_2=arho^2,cdots,x_n=arho^n=b$$
So, $rho=left(fracbaright)^1/n$ and as $ntoinfty$, $rhoto 1$.
For $xi_r$'s, we choose the left hand end point of the $r$th subinterval $[x_r-1,x_r]$. That is, $$xi_r=x_r-1=arho^r-1quad,,r=1,2,cdots,n$$
Then for each $r$, $$delta_r=x_r-x_r-1=arho^r-1(rho-1)to0qquad[because rhoto1]$$
Hence we arrive at a new expression for $int_a^b f(x),dx$, given by $$int_a^b f(x),dx=lim_ntoinfty\ (rhoto1)a(rho -1)sum_r=1^n rho^r-1f(arho^r-1)$$
This form is particularly helpful for evaluating integrals of the form $int_a^b x^m,dx$, for $0<a<b$ and any rational number $m$.
For $mne -1$, we have by taking $f(x)=x^m$,
beginalign
int_a^b x^m,dx&=lim_ntoinfty\( rhoto1)a(rho -1)sum_r=1^n rho^r-1(arho^r-1)^m
\&=a^m+1lim_ntoinftyleft[fracrho -1rho^m+1sum_r=1^n rho^r(1+m)right]
\&=a^m+1lim_ntoinftyleft[frac(rho -1)rho^m+1rho^m+1fracrho^n(m+1)-1rho^m+1-1right]qquadleft[because, mne -1,, rho^m+1-1ne 0right]
\&=a^m+1lim_ntoinftyfrac(rho-1)rho^m+1-1left[left(fracbaright)^m+1-1right]qquadquadleft[because rho^n=b/aright]
\&=(b^m+1-a^m+1)lim_rhoto1frac1left(fracrho^m+1-1rho-1right)
\&=fracb^m+1-a^m+1m+1
endalign
In fact, the case $m=-1$ in the beginning has been done in an exactly similar manner. So the case distinction need not be made as mentioned in a comment below.
I had learnt this approach as Walli's method from my high school textbook, although I could not find a reference online.
answered Jul 28 at 16:31
StubbornAtom
3,69611134
Your technique applies even when $mneq - 1$. +1 for simple approach.
– Paramanand Singh
Jul 28 at 17:20
add a comment |Â
up vote
1
down vote
From the Euler-Maclaurin Summation Formula, $ colorbluesum_k=1^n k^m-ell=fracn^m-ell+1m-ell+1+Oleft(n^m-ellright)$. Using this result along with the binomial expansion $ colorred(x+y)^m=sum_ell=0^m x^ell y^m-ell$ and the relationship $ binommellfrac1m-ell+1=frac1m+1binomm+1ell$, we have for $mge 1$
$$beginalign
fracb-ansum_k=1^n colorredleft(a+fracb-ankright)^m&=fracb-ansum_k=1^n colorredsum_ell=0^mbinommella^ell left(fracb-ankright)^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorbluesum_k=1^n k^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorblueleft(fracn^m-ell+1m-ell+1 +O(n^m-
ell)right)\\
&=frac(b-a)^m+1m+1sum_ell=0^mbinomm+1ell left(fracab-aright)^ell+Oleft(frac1nright)\\
&=frac(b-a)^m+1m+1left(left(1+fracab-aright)^m+1-fraca^m+1(b-a)^m+1right)+Oleft(frac1nright)\\
&=fracb^m+1-a^m+1m+1+Oleft(frac1nright)
endalign$$
answered Jul 28 at 17:25
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 28 at 19:20
add a comment |Â
up vote
0
down vote
I will assume $m>0$. Using MVT, for a partition $P:a=x_1<x_2<...<x_n+1=b$, we have $exists xi_i in (x_i,x_i+1)$ s.t. $xi_i^m(x_i+1-x_i)=fracx_i+1^m+1m+1-fracx_i^m+1m+1$. As a result
$$S^*(P)=sumlimits_i=1^n xi_i^m(x_i+1-x_i)=
sumlimits_i=1^n left(fracx_i+1^m+1m+1-fracx_i^m+1m+1right)=\
fracx_n+1^m+1m+1-fracx_1^m+1m+1=fracb^m+1-a^m+1m+1$$
Because $inflimits_xin[x_i,x_i+1]leftx^mright leq xi_i^m leq suplimits_xin[x_i,x_i+1]x^m$, we also have
$$L(x^m,P)leq S^*(P)leq U(x^m,P)$$
But $x^m$ is continuous on $[a,b]$, thus uniform continuous, so $L(x^m,P)$ and $U(x^m,P)$ will squeeze to $S^*(P)$, because
$$0< U(x^m, P) - L(x^m, P)=sumlimits_i=1^n left(suplimits_xin[x_i,x_i+1]x^m-inflimits_xin[x_i,x_i+1]x^mright)(x_i+1-x_i)leq \
fracvarepsilonb-a (b-a) = varepsilon$$
assuming we choose the partition such that $|y^m-x^m|<fracvarepsilona-b, forall x,y in[x_i,x_i+1], |x_i+1-x_i|<delta$ (see the definition of uniform continuity). This makes $x^m$ integrable (see Riemann's criterion, theorem 3 here, for more details).
answered Jul 28 at 15:42
rtybase
8,77221333
I think you would have to explain the notations you have used a bit more explicitly as I am not familiar with them
– tatan
Jul 28 at 15:43
I have provided many links explaining those notations. $S^*(P)$ is just a particular sum I constructed based on MVT, i.e. just an arbitrary notation. However $L$ (lower) and $U$ (upper) are very well known (see the link coming with "... we also have").
– rtybase
Jul 28 at 16:20
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Assuming $a>0$, I consider the special case $m=-1$.
Here $f(x)=frac1x$, and we divide the interval $ale xle b$ so that the points of division are in a geometric progression as $$a,ar,ar^2,cdots,ar^n=b,;quadtext so r=left(fracbaright)^1/n$$
The length of the $k$th ($1le kle n$) sub-interval is $$delta_k=ar^k-ar^k-1=ar^k-1(r-1)$$
Let $delta$ be the maximum among $delta_1,delta_2,cdots,delta_n$. Then as $ntoinfty, rto1$ and so $deltato0$. We take $$delta_k=ar^k-1quad,,1le kle n$$
So by definition,
beginalign
int_a^b frac1x,dx&=lim_deltato0sum_k=1^n f(xi_k)delta_k
\&=lim_ntoinftysum_k=1^nfracar^k-1(r-1)ar^k-1
\&=lim_ntoinfty n(r-1)
\&=lim_1/nto0fracleft(fracbaright)^1/n-11/n
\&=ln fracbaqquadqquadleft[because, lim_xto0fraca^x-1x=ln aright]
endalign
For the general problem, we can use the general definition of definite integral of a continuous (or integrable) function. I roughly state the definition so that the notation is clear.
Let $f(x)$ be a bounded function defined on a closed interval $[a,b]$. We divide the interval $[a,b]$ into $n$ subintervals $[x_0,x_1],,[x_1,x_2],cdots,[x_n-1,x_n]$ by introducing the points $x_0,x_2,cdots,x_n$ satisfying $a=x_0<x_1<cdots<x_n=b$. Let $$delta_r=x_r-x_r-1quad,r=1,2,cdots,n$$
Choose any point $xi_r$ of the $r$th subinterval, i.e. $$x_r-1le xi_rle x_r$$
Let $delta=max(delta_1,cdots,delta_n)$, the norm of the subdivision of the partition.
We then form the sum $sum_r=1^nf(xi_r)delta_r$, which depends on the choice of the points $x_r$'s and $xi_r$'s.
We now let $delta$ tend to zero, so that $n$, the number of subdivisions, goes to infinity. If now the sum $sum_r=1^nf(xi_r)delta_r$ tends to a definite limit which is independent of the choice of the points $x_r$'s and $xi_r$'s in $[x_r-1,x_r]$, then this limit is defined to be the definite (Riemann) integral of $f(x)$ over $[a,b]$.
That is, $$int_a^b f(x),dx=lim_deltato0sum_r=1^n f(xi_r)delta_r$$
Many expressions for $int_a^b f(x),dx$ can be found by taking $x_r$'s and $xi_r$'s in suitable manner. We need one such particular choice here.
We choose $x_r$'s in such a way that they form a G.P. with common ratio $rho$ (say), like $$a=x_0,x_1=arho,x_2=arho^2,cdots,x_n=arho^n=b$$
So, $rho=left(fracbaright)^1/n$ and as $ntoinfty$, $rhoto 1$.
For $xi_r$'s, we choose the left hand end point of the $r$th subinterval $[x_r-1,x_r]$. That is, $$xi_r=x_r-1=arho^r-1quad,,r=1,2,cdots,n$$
Then for each $r$, $$delta_r=x_r-x_r-1=arho^r-1(rho-1)to0qquad[because rhoto1]$$
Hence we arrive at a new expression for $int_a^b f(x),dx$, given by $$int_a^b f(x),dx=lim_ntoinfty\ (rhoto1)a(rho -1)sum_r=1^n rho^r-1f(arho^r-1)$$
This form is particularly helpful for evaluating integrals of the form $int_a^b x^m,dx$, for $0<a<b$ and any rational number $m$.
For $mne -1$, we have by taking $f(x)=x^m$,
beginalign
int_a^b x^m,dx&=lim_ntoinfty\( rhoto1)a(rho -1)sum_r=1^n rho^r-1(arho^r-1)^m
\&=a^m+1lim_ntoinftyleft[fracrho -1rho^m+1sum_r=1^n rho^r(1+m)right]
\&=a^m+1lim_ntoinftyleft[frac(rho -1)rho^m+1rho^m+1fracrho^n(m+1)-1rho^m+1-1right]qquadleft[because, mne -1,, rho^m+1-1ne 0right]
\&=a^m+1lim_ntoinftyfrac(rho-1)rho^m+1-1left[left(fracbaright)^m+1-1right]qquadquadleft[because rho^n=b/aright]
\&=(b^m+1-a^m+1)lim_rhoto1frac1left(fracrho^m+1-1rho-1right)
\&=fracb^m+1-a^m+1m+1
endalign
In fact, the case $m=-1$ in the beginning has been done in an exactly similar manner. So the case distinction need not be made as mentioned in a comment below.
I had learnt this approach as Walli's method from my high school textbook, although I could not find a reference online.
answered Jul 28 at 16:31
StubbornAtom
3,69611134
Your technique applies even when $mneq - 1$. +1 for simple approach.
– Paramanand Singh
Jul 28 at 17:20
add a comment |Â
up vote
4
down vote
Assuming $a>0$, I consider the special case $m=-1$.
Here $f(x)=frac1x$, and we divide the interval $ale xle b$ so that the points of division are in a geometric progression as $$a,ar,ar^2,cdots,ar^n=b,;quadtext so r=left(fracbaright)^1/n$$
The length of the $k$th ($1le kle n$) sub-interval is $$delta_k=ar^k-ar^k-1=ar^k-1(r-1)$$
Let $delta$ be the maximum among $delta_1,delta_2,cdots,delta_n$. Then as $ntoinfty, rto1$ and so $deltato0$. We take $$delta_k=ar^k-1quad,,1le kle n$$
So by definition,
beginalign
int_a^b frac1x,dx&=lim_deltato0sum_k=1^n f(xi_k)delta_k
\&=lim_ntoinftysum_k=1^nfracar^k-1(r-1)ar^k-1
\&=lim_ntoinfty n(r-1)
\&=lim_1/nto0fracleft(fracbaright)^1/n-11/n
\&=ln fracbaqquadqquadleft[because, lim_xto0fraca^x-1x=ln aright]
endalign
For the general problem, we can use the general definition of definite integral of a continuous (or integrable) function. I roughly state the definition so that the notation is clear.
Let $f(x)$ be a bounded function defined on a closed interval $[a,b]$. We divide the interval $[a,b]$ into $n$ subintervals $[x_0,x_1],,[x_1,x_2],cdots,[x_n-1,x_n]$ by introducing the points $x_0,x_2,cdots,x_n$ satisfying $a=x_0<x_1<cdots<x_n=b$. Let $$delta_r=x_r-x_r-1quad,r=1,2,cdots,n$$
Choose any point $xi_r$ of the $r$th subinterval, i.e. $$x_r-1le xi_rle x_r$$
Let $delta=max(delta_1,cdots,delta_n)$, the norm of the subdivision of the partition.
We then form the sum $sum_r=1^nf(xi_r)delta_r$, which depends on the choice of the points $x_r$'s and $xi_r$'s.
We now let $delta$ tend to zero, so that $n$, the number of subdivisions, goes to infinity. If now the sum $sum_r=1^nf(xi_r)delta_r$ tends to a definite limit which is independent of the choice of the points $x_r$'s and $xi_r$'s in $[x_r-1,x_r]$, then this limit is defined to be the definite (Riemann) integral of $f(x)$ over $[a,b]$.
That is, $$int_a^b f(x),dx=lim_deltato0sum_r=1^n f(xi_r)delta_r$$
Many expressions for $int_a^b f(x),dx$ can be found by taking $x_r$'s and $xi_r$'s in suitable manner. We need one such particular choice here.
We choose $x_r$'s in such a way that they form a G.P. with common ratio $rho$ (say), like $$a=x_0,x_1=arho,x_2=arho^2,cdots,x_n=arho^n=b$$
So, $rho=left(fracbaright)^1/n$ and as $ntoinfty$, $rhoto 1$.
For $xi_r$'s, we choose the left hand end point of the $r$th subinterval $[x_r-1,x_r]$. That is, $$xi_r=x_r-1=arho^r-1quad,,r=1,2,cdots,n$$
Then for each $r$, $$delta_r=x_r-x_r-1=arho^r-1(rho-1)to0qquad[because rhoto1]$$
Hence we arrive at a new expression for $int_a^b f(x),dx$, given by $$int_a^b f(x),dx=lim_ntoinfty\ (rhoto1)a(rho -1)sum_r=1^n rho^r-1f(arho^r-1)$$
This form is particularly helpful for evaluating integrals of the form $int_a^b x^m,dx$, for $0<a<b$ and any rational number $m$.
For $mne -1$, we have by taking $f(x)=x^m$,
beginalign
int_a^b x^m,dx&=lim_ntoinfty\( rhoto1)a(rho -1)sum_r=1^n rho^r-1(arho^r-1)^m
\&=a^m+1lim_ntoinftyleft[fracrho -1rho^m+1sum_r=1^n rho^r(1+m)right]
\&=a^m+1lim_ntoinftyleft[frac(rho -1)rho^m+1rho^m+1fracrho^n(m+1)-1rho^m+1-1right]qquadleft[because, mne -1,, rho^m+1-1ne 0right]
\&=a^m+1lim_ntoinftyfrac(rho-1)rho^m+1-1left[left(fracbaright)^m+1-1right]qquadquadleft[because rho^n=b/aright]
\&=(b^m+1-a^m+1)lim_rhoto1frac1left(fracrho^m+1-1rho-1right)
\&=fracb^m+1-a^m+1m+1
endalign
In fact, the case $m=-1$ in the beginning has been done in an exactly similar manner. So the case distinction need not be made as mentioned in a comment below.
I had learnt this approach as Walli's method from my high school textbook, although I could not find a reference online.
answered Jul 28 at 16:31
StubbornAtom
3,69611134
Your technique applies even when $mneq - 1$. +1 for simple approach.
– Paramanand Singh
Jul 28 at 17:20
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Assuming $a>0$, I consider the special case $m=-1$.
Here $f(x)=frac1x$, and we divide the interval $ale xle b$ so that the points of division are in a geometric progression as $$a,ar,ar^2,cdots,ar^n=b,;quadtext so r=left(fracbaright)^1/n$$
The length of the $k$th ($1le kle n$) sub-interval is $$delta_k=ar^k-ar^k-1=ar^k-1(r-1)$$
Let $delta$ be the maximum among $delta_1,delta_2,cdots,delta_n$. Then as $ntoinfty, rto1$ and so $deltato0$. We take $$delta_k=ar^k-1quad,,1le kle n$$
So by definition,
beginalign
int_a^b frac1x,dx&=lim_deltato0sum_k=1^n f(xi_k)delta_k
\&=lim_ntoinftysum_k=1^nfracar^k-1(r-1)ar^k-1
\&=lim_ntoinfty n(r-1)
\&=lim_1/nto0fracleft(fracbaright)^1/n-11/n
\&=ln fracbaqquadqquadleft[because, lim_xto0fraca^x-1x=ln aright]
endalign
For the general problem, we can use the general definition of definite integral of a continuous (or integrable) function. I roughly state the definition so that the notation is clear.
Let $f(x)$ be a bounded function defined on a closed interval $[a,b]$. We divide the interval $[a,b]$ into $n$ subintervals $[x_0,x_1],,[x_1,x_2],cdots,[x_n-1,x_n]$ by introducing the points $x_0,x_2,cdots,x_n$ satisfying $a=x_0<x_1<cdots<x_n=b$. Let $$delta_r=x_r-x_r-1quad,r=1,2,cdots,n$$
Choose any point $xi_r$ of the $r$th subinterval, i.e. $$x_r-1le xi_rle x_r$$
Let $delta=max(delta_1,cdots,delta_n)$, the norm of the subdivision of the partition.
We then form the sum $sum_r=1^nf(xi_r)delta_r$, which depends on the choice of the points $x_r$'s and $xi_r$'s.
We now let $delta$ tend to zero, so that $n$, the number of subdivisions, goes to infinity. If now the sum $sum_r=1^nf(xi_r)delta_r$ tends to a definite limit which is independent of the choice of the points $x_r$'s and $xi_r$'s in $[x_r-1,x_r]$, then this limit is defined to be the definite (Riemann) integral of $f(x)$ over $[a,b]$.
That is, $$int_a^b f(x),dx=lim_deltato0sum_r=1^n f(xi_r)delta_r$$
Many expressions for $int_a^b f(x),dx$ can be found by taking $x_r$'s and $xi_r$'s in suitable manner. We need one such particular choice here.
We choose $x_r$'s in such a way that they form a G.P. with common ratio $rho$ (say), like $$a=x_0,x_1=arho,x_2=arho^2,cdots,x_n=arho^n=b$$
So, $rho=left(fracbaright)^1/n$ and as $ntoinfty$, $rhoto 1$.
For $xi_r$'s, we choose the left hand end point of the $r$th subinterval $[x_r-1,x_r]$. That is, $$xi_r=x_r-1=arho^r-1quad,,r=1,2,cdots,n$$
Then for each $r$, $$delta_r=x_r-x_r-1=arho^r-1(rho-1)to0qquad[because rhoto1]$$
Hence we arrive at a new expression for $int_a^b f(x),dx$, given by $$int_a^b f(x),dx=lim_ntoinfty\ (rhoto1)a(rho -1)sum_r=1^n rho^r-1f(arho^r-1)$$
This form is particularly helpful for evaluating integrals of the form $int_a^b x^m,dx$, for $0<a<b$ and any rational number $m$.
For $mne -1$, we have by taking $f(x)=x^m$,
beginalign
int_a^b x^m,dx&=lim_ntoinfty\( rhoto1)a(rho -1)sum_r=1^n rho^r-1(arho^r-1)^m
\&=a^m+1lim_ntoinftyleft[fracrho -1rho^m+1sum_r=1^n rho^r(1+m)right]
\&=a^m+1lim_ntoinftyleft[frac(rho -1)rho^m+1rho^m+1fracrho^n(m+1)-1rho^m+1-1right]qquadleft[because, mne -1,, rho^m+1-1ne 0right]
\&=a^m+1lim_ntoinftyfrac(rho-1)rho^m+1-1left[left(fracbaright)^m+1-1right]qquadquadleft[because rho^n=b/aright]
\&=(b^m+1-a^m+1)lim_rhoto1frac1left(fracrho^m+1-1rho-1right)
\&=fracb^m+1-a^m+1m+1
endalign
In fact, the case $m=-1$ in the beginning has been done in an exactly similar manner. So the case distinction need not be made as mentioned in a comment below.
I had learnt this approach as Walli's method from my high school textbook, although I could not find a reference online.
answered Jul 28 at 16:31
StubbornAtom
3,69611134
Assuming $a>0$, I consider the special case $m=-1$.
Here $f(x)=frac1x$, and we divide the interval $ale xle b$ so that the points of division are in a geometric progression as $$a,ar,ar^2,cdots,ar^n=b,;quadtext so r=left(fracbaright)^1/n$$
The length of the $k$th ($1le kle n$) sub-interval is $$delta_k=ar^k-ar^k-1=ar^k-1(r-1)$$
Let $delta$ be the maximum among $delta_1,delta_2,cdots,delta_n$. Then as $ntoinfty, rto1$ and so $deltato0$. We take $$delta_k=ar^k-1quad,,1le kle n$$
So by definition,
beginalign
int_a^b frac1x,dx&=lim_deltato0sum_k=1^n f(xi_k)delta_k
\&=lim_ntoinftysum_k=1^nfracar^k-1(r-1)ar^k-1
\&=lim_ntoinfty n(r-1)
\&=lim_1/nto0fracleft(fracbaright)^1/n-11/n
\&=ln fracbaqquadqquadleft[because, lim_xto0fraca^x-1x=ln aright]
endalign
For the general problem, we can use the general definition of definite integral of a continuous (or integrable) function. I roughly state the definition so that the notation is clear.
Let $f(x)$ be a bounded function defined on a closed interval $[a,b]$. We divide the interval $[a,b]$ into $n$ subintervals $[x_0,x_1],,[x_1,x_2],cdots,[x_n-1,x_n]$ by introducing the points $x_0,x_2,cdots,x_n$ satisfying $a=x_0<x_1<cdots<x_n=b$. Let $$delta_r=x_r-x_r-1quad,r=1,2,cdots,n$$
Choose any point $xi_r$ of the $r$th subinterval, i.e. $$x_r-1le xi_rle x_r$$
Let $delta=max(delta_1,cdots,delta_n)$, the norm of the subdivision of the partition.
We then form the sum $sum_r=1^nf(xi_r)delta_r$, which depends on the choice of the points $x_r$'s and $xi_r$'s.
We now let $delta$ tend to zero, so that $n$, the number of subdivisions, goes to infinity. If now the sum $sum_r=1^nf(xi_r)delta_r$ tends to a definite limit which is independent of the choice of the points $x_r$'s and $xi_r$'s in $[x_r-1,x_r]$, then this limit is defined to be the definite (Riemann) integral of $f(x)$ over $[a,b]$.
That is, $$int_a^b f(x),dx=lim_deltato0sum_r=1^n f(xi_r)delta_r$$
Many expressions for $int_a^b f(x),dx$ can be found by taking $x_r$'s and $xi_r$'s in suitable manner. We need one such particular choice here.
We choose $x_r$'s in such a way that they form a G.P. with common ratio $rho$ (say), like $$a=x_0,x_1=arho,x_2=arho^2,cdots,x_n=arho^n=b$$
So, $rho=left(fracbaright)^1/n$ and as $ntoinfty$, $rhoto 1$.
For $xi_r$'s, we choose the left hand end point of the $r$th subinterval $[x_r-1,x_r]$. That is, $$xi_r=x_r-1=arho^r-1quad,,r=1,2,cdots,n$$
Then for each $r$, $$delta_r=x_r-x_r-1=arho^r-1(rho-1)to0qquad[because rhoto1]$$
Hence we arrive at a new expression for $int_a^b f(x),dx$, given by $$int_a^b f(x),dx=lim_ntoinfty\ (rhoto1)a(rho -1)sum_r=1^n rho^r-1f(arho^r-1)$$
This form is particularly helpful for evaluating integrals of the form $int_a^b x^m,dx$, for $0<a<b$ and any rational number $m$.
For $mne -1$, we have by taking $f(x)=x^m$,
beginalign
int_a^b x^m,dx&=lim_ntoinfty\( rhoto1)a(rho -1)sum_r=1^n rho^r-1(arho^r-1)^m
\&=a^m+1lim_ntoinftyleft[fracrho -1rho^m+1sum_r=1^n rho^r(1+m)right]
\&=a^m+1lim_ntoinftyleft[frac(rho -1)rho^m+1rho^m+1fracrho^n(m+1)-1rho^m+1-1right]qquadleft[because, mne -1,, rho^m+1-1ne 0right]
\&=a^m+1lim_ntoinftyfrac(rho-1)rho^m+1-1left[left(fracbaright)^m+1-1right]qquadquadleft[because rho^n=b/aright]
\&=(b^m+1-a^m+1)lim_rhoto1frac1left(fracrho^m+1-1rho-1right)
\&=fracb^m+1-a^m+1m+1
endalign
In fact, the case $m=-1$ in the beginning has been done in an exactly similar manner. So the case distinction need not be made as mentioned in a comment below.
I had learnt this approach as Walli's method from my high school textbook, although I could not find a reference online.
answered Jul 28 at 16:31
StubbornAtom
3,69611134
edited Jul 29 at 5:50
answered Jul 28 at 16:31
StubbornAtom
3,69611134
answered Jul 28 at 16:31
StubbornAtom
3,69611134
answered Jul 28 at 16:31
StubbornAtom
3,69611134
3,69611134
Your technique applies even when $mneq - 1$. +1 for simple approach.
– Paramanand Singh
Jul 28 at 17:20
add a comment |Â
Your technique applies even when $mneq - 1$. +1 for simple approach.
– Paramanand Singh
Jul 28 at 17:20
Your technique applies even when $mneq - 1$. +1 for simple approach.
– Paramanand Singh
Jul 28 at 17:20
Your technique applies even when $mneq - 1$. +1 for simple approach.
– Paramanand Singh
Jul 28 at 17:20
add a comment |Â
up vote
1
down vote
From the Euler-Maclaurin Summation Formula, $ colorbluesum_k=1^n k^m-ell=fracn^m-ell+1m-ell+1+Oleft(n^m-ellright)$. Using this result along with the binomial expansion $ colorred(x+y)^m=sum_ell=0^m x^ell y^m-ell$ and the relationship $ binommellfrac1m-ell+1=frac1m+1binomm+1ell$, we have for $mge 1$
$$beginalign
fracb-ansum_k=1^n colorredleft(a+fracb-ankright)^m&=fracb-ansum_k=1^n colorredsum_ell=0^mbinommella^ell left(fracb-ankright)^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorbluesum_k=1^n k^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorblueleft(fracn^m-ell+1m-ell+1 +O(n^m-
ell)right)\\
&=frac(b-a)^m+1m+1sum_ell=0^mbinomm+1ell left(fracab-aright)^ell+Oleft(frac1nright)\\
&=frac(b-a)^m+1m+1left(left(1+fracab-aright)^m+1-fraca^m+1(b-a)^m+1right)+Oleft(frac1nright)\\
&=fracb^m+1-a^m+1m+1+Oleft(frac1nright)
endalign$$
answered Jul 28 at 17:25
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 28 at 19:20
add a comment |Â
up vote
1
down vote
From the Euler-Maclaurin Summation Formula, $ colorbluesum_k=1^n k^m-ell=fracn^m-ell+1m-ell+1+Oleft(n^m-ellright)$. Using this result along with the binomial expansion $ colorred(x+y)^m=sum_ell=0^m x^ell y^m-ell$ and the relationship $ binommellfrac1m-ell+1=frac1m+1binomm+1ell$, we have for $mge 1$
$$beginalign
fracb-ansum_k=1^n colorredleft(a+fracb-ankright)^m&=fracb-ansum_k=1^n colorredsum_ell=0^mbinommella^ell left(fracb-ankright)^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorbluesum_k=1^n k^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorblueleft(fracn^m-ell+1m-ell+1 +O(n^m-
ell)right)\\
&=frac(b-a)^m+1m+1sum_ell=0^mbinomm+1ell left(fracab-aright)^ell+Oleft(frac1nright)\\
&=frac(b-a)^m+1m+1left(left(1+fracab-aright)^m+1-fraca^m+1(b-a)^m+1right)+Oleft(frac1nright)\\
&=fracb^m+1-a^m+1m+1+Oleft(frac1nright)
endalign$$
answered Jul 28 at 17:25
Mark Viola
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 28 at 19:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
From the Euler-Maclaurin Summation Formula, $ colorbluesum_k=1^n k^m-ell=fracn^m-ell+1m-ell+1+Oleft(n^m-ellright)$. Using this result along with the binomial expansion $ colorred(x+y)^m=sum_ell=0^m x^ell y^m-ell$ and the relationship $ binommellfrac1m-ell+1=frac1m+1binomm+1ell$, we have for $mge 1$
$$beginalign
fracb-ansum_k=1^n colorredleft(a+fracb-ankright)^m&=fracb-ansum_k=1^n colorredsum_ell=0^mbinommella^ell left(fracb-ankright)^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorbluesum_k=1^n k^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorblueleft(fracn^m-ell+1m-ell+1 +O(n^m-
ell)right)\\
&=frac(b-a)^m+1m+1sum_ell=0^mbinomm+1ell left(fracab-aright)^ell+Oleft(frac1nright)\\
&=frac(b-a)^m+1m+1left(left(1+fracab-aright)^m+1-fraca^m+1(b-a)^m+1right)+Oleft(frac1nright)\\
&=fracb^m+1-a^m+1m+1+Oleft(frac1nright)
endalign$$
answered Jul 28 at 17:25
Mark Viola
126k1172167
From the Euler-Maclaurin Summation Formula, $ colorbluesum_k=1^n k^m-ell=fracn^m-ell+1m-ell+1+Oleft(n^m-ellright)$. Using this result along with the binomial expansion $ colorred(x+y)^m=sum_ell=0^m x^ell y^m-ell$ and the relationship $ binommellfrac1m-ell+1=frac1m+1binomm+1ell$, we have for $mge 1$
$$beginalign
fracb-ansum_k=1^n colorredleft(a+fracb-ankright)^m&=fracb-ansum_k=1^n colorredsum_ell=0^mbinommella^ell left(fracb-ankright)^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorbluesum_k=1^n k^m-ell\\
&=fracb-ansum_ell=0^mbinommella^ell left(fracb-anright)^m-ellcolorblueleft(fracn^m-ell+1m-ell+1 +O(n^m-
ell)right)\\
&=frac(b-a)^m+1m+1sum_ell=0^mbinomm+1ell left(fracab-aright)^ell+Oleft(frac1nright)\\
&=frac(b-a)^m+1m+1left(left(1+fracab-aright)^m+1-fraca^m+1(b-a)^m+1right)+Oleft(frac1nright)\\
&=fracb^m+1-a^m+1m+1+Oleft(frac1nright)
endalign$$
answered Jul 28 at 17:25
Mark Viola
126k1172167
edited Jul 28 at 17:36
answered Jul 28 at 17:25
Mark Viola
126k1172167
answered Jul 28 at 17:25
Mark Viola
126k1172167
answered Jul 28 at 17:25
Mark Viola
126k1172167
126k1172167
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 28 at 19:20
add a comment |Â
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 28 at 19:20
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 28 at 19:20
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Jul 28 at 19:20
add a comment |Â
up vote
0
down vote
I will assume $m>0$. Using MVT, for a partition $P:a=x_1<x_2<...<x_n+1=b$, we have $exists xi_i in (x_i,x_i+1)$ s.t. $xi_i^m(x_i+1-x_i)=fracx_i+1^m+1m+1-fracx_i^m+1m+1$. As a result
$$S^*(P)=sumlimits_i=1^n xi_i^m(x_i+1-x_i)=
sumlimits_i=1^n left(fracx_i+1^m+1m+1-fracx_i^m+1m+1right)=\
fracx_n+1^m+1m+1-fracx_1^m+1m+1=fracb^m+1-a^m+1m+1$$
Because $inflimits_xin[x_i,x_i+1]leftx^mright leq xi_i^m leq suplimits_xin[x_i,x_i+1]x^m$, we also have
$$L(x^m,P)leq S^*(P)leq U(x^m,P)$$
But $x^m$ is continuous on $[a,b]$, thus uniform continuous, so $L(x^m,P)$ and $U(x^m,P)$ will squeeze to $S^*(P)$, because
$$0< U(x^m, P) - L(x^m, P)=sumlimits_i=1^n left(suplimits_xin[x_i,x_i+1]x^m-inflimits_xin[x_i,x_i+1]x^mright)(x_i+1-x_i)leq \
fracvarepsilonb-a (b-a) = varepsilon$$
assuming we choose the partition such that $|y^m-x^m|<fracvarepsilona-b, forall x,y in[x_i,x_i+1], |x_i+1-x_i|<delta$ (see the definition of uniform continuity). This makes $x^m$ integrable (see Riemann's criterion, theorem 3 here, for more details).
answered Jul 28 at 15:42
rtybase
8,77221333
I think you would have to explain the notations you have used a bit more explicitly as I am not familiar with them
– tatan
Jul 28 at 15:43
I have provided many links explaining those notations. $S^*(P)$ is just a particular sum I constructed based on MVT, i.e. just an arbitrary notation. However $L$ (lower) and $U$ (upper) are very well known (see the link coming with "... we also have").
– rtybase
Jul 28 at 16:20
add a comment |Â
up vote
0
down vote
I will assume $m>0$. Using MVT, for a partition $P:a=x_1<x_2<...<x_n+1=b$, we have $exists xi_i in (x_i,x_i+1)$ s.t. $xi_i^m(x_i+1-x_i)=fracx_i+1^m+1m+1-fracx_i^m+1m+1$. As a result
$$S^*(P)=sumlimits_i=1^n xi_i^m(x_i+1-x_i)=
sumlimits_i=1^n left(fracx_i+1^m+1m+1-fracx_i^m+1m+1right)=\
fracx_n+1^m+1m+1-fracx_1^m+1m+1=fracb^m+1-a^m+1m+1$$
Because $inflimits_xin[x_i,x_i+1]leftx^mright leq xi_i^m leq suplimits_xin[x_i,x_i+1]x^m$, we also have
$$L(x^m,P)leq S^*(P)leq U(x^m,P)$$
But $x^m$ is continuous on $[a,b]$, thus uniform continuous, so $L(x^m,P)$ and $U(x^m,P)$ will squeeze to $S^*(P)$, because
$$0< U(x^m, P) - L(x^m, P)=sumlimits_i=1^n left(suplimits_xin[x_i,x_i+1]x^m-inflimits_xin[x_i,x_i+1]x^mright)(x_i+1-x_i)leq \
fracvarepsilonb-a (b-a) = varepsilon$$
assuming we choose the partition such that $|y^m-x^m|<fracvarepsilona-b, forall x,y in[x_i,x_i+1], |x_i+1-x_i|<delta$ (see the definition of uniform continuity). This makes $x^m$ integrable (see Riemann's criterion, theorem 3 here, for more details).
answered Jul 28 at 15:42
rtybase
8,77221333
I think you would have to explain the notations you have used a bit more explicitly as I am not familiar with them
– tatan
Jul 28 at 15:43
I have provided many links explaining those notations. $S^*(P)$ is just a particular sum I constructed based on MVT, i.e. just an arbitrary notation. However $L$ (lower) and $U$ (upper) are very well known (see the link coming with "... we also have").
– rtybase
Jul 28 at 16:20
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I will assume $m>0$. Using MVT, for a partition $P:a=x_1<x_2<...<x_n+1=b$, we have $exists xi_i in (x_i,x_i+1)$ s.t. $xi_i^m(x_i+1-x_i)=fracx_i+1^m+1m+1-fracx_i^m+1m+1$. As a result
$$S^*(P)=sumlimits_i=1^n xi_i^m(x_i+1-x_i)=
sumlimits_i=1^n left(fracx_i+1^m+1m+1-fracx_i^m+1m+1right)=\
fracx_n+1^m+1m+1-fracx_1^m+1m+1=fracb^m+1-a^m+1m+1$$
Because $inflimits_xin[x_i,x_i+1]leftx^mright leq xi_i^m leq suplimits_xin[x_i,x_i+1]x^m$, we also have
$$L(x^m,P)leq S^*(P)leq U(x^m,P)$$
But $x^m$ is continuous on $[a,b]$, thus uniform continuous, so $L(x^m,P)$ and $U(x^m,P)$ will squeeze to $S^*(P)$, because
$$0< U(x^m, P) - L(x^m, P)=sumlimits_i=1^n left(suplimits_xin[x_i,x_i+1]x^m-inflimits_xin[x_i,x_i+1]x^mright)(x_i+1-x_i)leq \
fracvarepsilonb-a (b-a) = varepsilon$$
assuming we choose the partition such that $|y^m-x^m|<fracvarepsilona-b, forall x,y in[x_i,x_i+1], |x_i+1-x_i|<delta$ (see the definition of uniform continuity). This makes $x^m$ integrable (see Riemann's criterion, theorem 3 here, for more details).
answered Jul 28 at 15:42
rtybase
8,77221333
I will assume $m>0$. Using MVT, for a partition $P:a=x_1<x_2<...<x_n+1=b$, we have $exists xi_i in (x_i,x_i+1)$ s.t. $xi_i^m(x_i+1-x_i)=fracx_i+1^m+1m+1-fracx_i^m+1m+1$. As a result
$$S^*(P)=sumlimits_i=1^n xi_i^m(x_i+1-x_i)=
sumlimits_i=1^n left(fracx_i+1^m+1m+1-fracx_i^m+1m+1right)=\
fracx_n+1^m+1m+1-fracx_1^m+1m+1=fracb^m+1-a^m+1m+1$$
Because $inflimits_xin[x_i,x_i+1]leftx^mright leq xi_i^m leq suplimits_xin[x_i,x_i+1]x^m$, we also have
$$L(x^m,P)leq S^*(P)leq U(x^m,P)$$
But $x^m$ is continuous on $[a,b]$, thus uniform continuous, so $L(x^m,P)$ and $U(x^m,P)$ will squeeze to $S^*(P)$, because
$$0< U(x^m, P) - L(x^m, P)=sumlimits_i=1^n left(suplimits_xin[x_i,x_i+1]x^m-inflimits_xin[x_i,x_i+1]x^mright)(x_i+1-x_i)leq \
fracvarepsilonb-a (b-a) = varepsilon$$
assuming we choose the partition such that $|y^m-x^m|<fracvarepsilona-b, forall x,y in[x_i,x_i+1], |x_i+1-x_i|<delta$ (see the definition of uniform continuity). This makes $x^m$ integrable (see Riemann's criterion, theorem 3 here, for more details).
answered Jul 28 at 15:42
rtybase
8,77221333
edited Jul 28 at 16:30
answered Jul 28 at 15:42
rtybase
8,77221333
answered Jul 28 at 15:42
rtybase
8,77221333
answered Jul 28 at 15:42
rtybase
8,77221333
8,77221333
I think you would have to explain the notations you have used a bit more explicitly as I am not familiar with them
– tatan
Jul 28 at 15:43
I have provided many links explaining those notations. $S^*(P)$ is just a particular sum I constructed based on MVT, i.e. just an arbitrary notation. However $L$ (lower) and $U$ (upper) are very well known (see the link coming with "... we also have").
– rtybase
Jul 28 at 16:20
add a comment |Â
I think you would have to explain the notations you have used a bit more explicitly as I am not familiar with them
– tatan
Jul 28 at 15:43
I have provided many links explaining those notations. $S^*(P)$ is just a particular sum I constructed based on MVT, i.e. just an arbitrary notation. However $L$ (lower) and $U$ (upper) are very well known (see the link coming with "... we also have").
– rtybase
Jul 28 at 16:20
I think you would have to explain the notations you have used a bit more explicitly as I am not familiar with them
– tatan
Jul 28 at 15:43
I think you would have to explain the notations you have used a bit more explicitly as I am not familiar with them
– tatan
Jul 28 at 15:43
I have provided many links explaining those notations. $S^*(P)$ is just a particular sum I constructed based on MVT, i.e. just an arbitrary notation. However $L$ (lower) and $U$ (upper) are very well known (see the link coming with "... we also have").
– rtybase
Jul 28 at 16:20
I have provided many links explaining those notations. $S^*(P)$ is just a particular sum I constructed based on MVT, i.e. just an arbitrary notation. However $L$ (lower) and $U$ (upper) are very well known (see the link coming with "... we also have").
– rtybase
Jul 28 at 16:20
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865286%2fhow-to-find-int-abxmdx-using-the-limit-sum-definition-of-definite-integral%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
The sum doesn't appear correct, at least not without restrictions. Is $n$ intended to be $(b-a)/h$?
– B. Mehta
Jul 28 at 14:32
@B.Mehta Yep.... Edited that
– tatan
Jul 28 at 14:33
In that case, is the limit as $n to infty$ also?
– B. Mehta
Jul 28 at 14:34
@B.Mehta $hto 0$ directly implies $ntoinfty$... doesn't it?
– tatan
Jul 28 at 14:35
More likely $h=(b-a)/n$ and the limit should be taken as $ntoinfty,$ I would think. It's not clear what the sum means if $n$ isn't an integer.
– saulspatz
Jul 28 at 14:35