Calculate $int_{a}^{b}ln x ~dx$ using the definition of integral
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I am being asked to calculate the integral $int_{a}^{b}ln x~dx$ using the definition of integral (i.e. expressing it as limit of Riemann sums)
Here's what I did:
Let's divide the interval $[a,b]$ into $n$ subintervals of the form $[x_{i},x_{i+1}]$.
Let $x_i=a+frac{(b-a)i}{n}$ where $i=0,1,2...,n-1. $ be the start points of each subinterval such that $x_0=a$ and $x_n=b$
Let $Delta x_i=x_i-x_{i-1}=frac{b-a}{n}$
I defined the integral like this:
$$begin{align}int_{a}^{b}ln x , dx&=lim_{nrightarrow infty}sum_{i=o}^{n-1}f(x_{i})Delta x_{i}\
&=lim_{nrightarrow infty}frac{b-a}{n}sum_{i=0}^{n-1}lnleft(a+frac{(b-a)i}{n}right)\
&=lim_{nrightarrow infty}frac{b-a}{n}lnleft(prod_{i=0}^{n-1}left(a+frac{(b-a)i}{n}right)right)end{align}$$
And I'm stuck here, I don't know how to find this product, any hint would be appreciated , thank you !
real-analysis integration definite-integrals
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up vote
2
down vote
favorite
I am being asked to calculate the integral $int_{a}^{b}ln x~dx$ using the definition of integral (i.e. expressing it as limit of Riemann sums)
Here's what I did:
Let's divide the interval $[a,b]$ into $n$ subintervals of the form $[x_{i},x_{i+1}]$.
Let $x_i=a+frac{(b-a)i}{n}$ where $i=0,1,2...,n-1. $ be the start points of each subinterval such that $x_0=a$ and $x_n=b$
Let $Delta x_i=x_i-x_{i-1}=frac{b-a}{n}$
I defined the integral like this:
$$begin{align}int_{a}^{b}ln x , dx&=lim_{nrightarrow infty}sum_{i=o}^{n-1}f(x_{i})Delta x_{i}\
&=lim_{nrightarrow infty}frac{b-a}{n}sum_{i=0}^{n-1}lnleft(a+frac{(b-a)i}{n}right)\
&=lim_{nrightarrow infty}frac{b-a}{n}lnleft(prod_{i=0}^{n-1}left(a+frac{(b-a)i}{n}right)right)end{align}$$
And I'm stuck here, I don't know how to find this product, any hint would be appreciated , thank you !
real-analysis integration definite-integrals
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am being asked to calculate the integral $int_{a}^{b}ln x~dx$ using the definition of integral (i.e. expressing it as limit of Riemann sums)
Here's what I did:
Let's divide the interval $[a,b]$ into $n$ subintervals of the form $[x_{i},x_{i+1}]$.
Let $x_i=a+frac{(b-a)i}{n}$ where $i=0,1,2...,n-1. $ be the start points of each subinterval such that $x_0=a$ and $x_n=b$
Let $Delta x_i=x_i-x_{i-1}=frac{b-a}{n}$
I defined the integral like this:
$$begin{align}int_{a}^{b}ln x , dx&=lim_{nrightarrow infty}sum_{i=o}^{n-1}f(x_{i})Delta x_{i}\
&=lim_{nrightarrow infty}frac{b-a}{n}sum_{i=0}^{n-1}lnleft(a+frac{(b-a)i}{n}right)\
&=lim_{nrightarrow infty}frac{b-a}{n}lnleft(prod_{i=0}^{n-1}left(a+frac{(b-a)i}{n}right)right)end{align}$$
And I'm stuck here, I don't know how to find this product, any hint would be appreciated , thank you !
real-analysis integration definite-integrals
I am being asked to calculate the integral $int_{a}^{b}ln x~dx$ using the definition of integral (i.e. expressing it as limit of Riemann sums)
Here's what I did:
Let's divide the interval $[a,b]$ into $n$ subintervals of the form $[x_{i},x_{i+1}]$.
Let $x_i=a+frac{(b-a)i}{n}$ where $i=0,1,2...,n-1. $ be the start points of each subinterval such that $x_0=a$ and $x_n=b$
Let $Delta x_i=x_i-x_{i-1}=frac{b-a}{n}$
I defined the integral like this:
$$begin{align}int_{a}^{b}ln x , dx&=lim_{nrightarrow infty}sum_{i=o}^{n-1}f(x_{i})Delta x_{i}\
&=lim_{nrightarrow infty}frac{b-a}{n}sum_{i=0}^{n-1}lnleft(a+frac{(b-a)i}{n}right)\
&=lim_{nrightarrow infty}frac{b-a}{n}lnleft(prod_{i=0}^{n-1}left(a+frac{(b-a)i}{n}right)right)end{align}$$
And I'm stuck here, I don't know how to find this product, any hint would be appreciated , thank you !
real-analysis integration definite-integrals
real-analysis integration definite-integrals
edited Nov 17 at 14:57
Masacroso
12.6k41746
12.6k41746
asked Nov 17 at 14:18
Maths Survivor
441219
441219
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1 Answer
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Some hints:
Let ${bover a}=:rho>1$, and choose an $Ngg1$. Then use the partition $a=x_0<x_1<ldots<x_N=b$ given by
$$x_k:=a,rho^{k/N}qquad(0leq kleq N) .$$
Then
$$R_N:=sum_{k=0}^{N-1}log(x_k)(x_{k+1}-x_k)$$
is an admissible Riemann sum for the given integral, i.e., we can be sure that $$lim_{Ntoinfty} R_N=int_a^blog x>dx .$$
Now $R_N$ can be computed in an elementary way; you only need a formula for finite sums of the type $sum_{k=0}^{N-1} k, q^k$.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Some hints:
Let ${bover a}=:rho>1$, and choose an $Ngg1$. Then use the partition $a=x_0<x_1<ldots<x_N=b$ given by
$$x_k:=a,rho^{k/N}qquad(0leq kleq N) .$$
Then
$$R_N:=sum_{k=0}^{N-1}log(x_k)(x_{k+1}-x_k)$$
is an admissible Riemann sum for the given integral, i.e., we can be sure that $$lim_{Ntoinfty} R_N=int_a^blog x>dx .$$
Now $R_N$ can be computed in an elementary way; you only need a formula for finite sums of the type $sum_{k=0}^{N-1} k, q^k$.
add a comment |
up vote
5
down vote
Some hints:
Let ${bover a}=:rho>1$, and choose an $Ngg1$. Then use the partition $a=x_0<x_1<ldots<x_N=b$ given by
$$x_k:=a,rho^{k/N}qquad(0leq kleq N) .$$
Then
$$R_N:=sum_{k=0}^{N-1}log(x_k)(x_{k+1}-x_k)$$
is an admissible Riemann sum for the given integral, i.e., we can be sure that $$lim_{Ntoinfty} R_N=int_a^blog x>dx .$$
Now $R_N$ can be computed in an elementary way; you only need a formula for finite sums of the type $sum_{k=0}^{N-1} k, q^k$.
add a comment |
up vote
5
down vote
up vote
5
down vote
Some hints:
Let ${bover a}=:rho>1$, and choose an $Ngg1$. Then use the partition $a=x_0<x_1<ldots<x_N=b$ given by
$$x_k:=a,rho^{k/N}qquad(0leq kleq N) .$$
Then
$$R_N:=sum_{k=0}^{N-1}log(x_k)(x_{k+1}-x_k)$$
is an admissible Riemann sum for the given integral, i.e., we can be sure that $$lim_{Ntoinfty} R_N=int_a^blog x>dx .$$
Now $R_N$ can be computed in an elementary way; you only need a formula for finite sums of the type $sum_{k=0}^{N-1} k, q^k$.
Some hints:
Let ${bover a}=:rho>1$, and choose an $Ngg1$. Then use the partition $a=x_0<x_1<ldots<x_N=b$ given by
$$x_k:=a,rho^{k/N}qquad(0leq kleq N) .$$
Then
$$R_N:=sum_{k=0}^{N-1}log(x_k)(x_{k+1}-x_k)$$
is an admissible Riemann sum for the given integral, i.e., we can be sure that $$lim_{Ntoinfty} R_N=int_a^blog x>dx .$$
Now $R_N$ can be computed in an elementary way; you only need a formula for finite sums of the type $sum_{k=0}^{N-1} k, q^k$.
answered Nov 17 at 14:45
Christian Blatter
171k7111325
171k7111325
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