Riemann Sum for 1/x

Mar 2012
565
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How to find the integral 1/x from 1 to 2 using riemann sum?

∫ 1/x dx = lim as n goes infinity Σ(i = 1 to n)f(xi)Δx

Δx = 2-1/n = 1/n
xi = 1 + iΔx = 1 + (i/n)

= lim as n goes infinity Σ(i = 1 to n)f(1 + (i/n))(1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (1 + (i/n)) ] (1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (n + i) ]

NOW, how to get rid of i?
 

HallsofIvy

MHF Helper
Apr 2005
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How to find the integral 1/x from 1 to 2 using riemann sum?

∫ 1/x dx = lim as n goes infinity Σ(i = 1 to n)f(xi)Δx

Δx = 2-1/n = 1/n
You mean Δx= (2- 1)/n. That had me puzzled for a minute!

xi = 1 + iΔx = 1 + (i/n)

= lim as n goes infinity Σ(i = 1 to n)f(1 + (i/n))(1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (1 + (i/n)) ] (1/n) = lim as n goes infinity Σ(i = 1 to n)[ 1 / (n + i) ]

NOW, how to get rid of i?
Your do the sum! But there is no "simple" formula for that sum, if that is what you are asking. Do you know the McLaurin series for ln(x)?
 
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Mar 2012
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I don't know the McLaurin series. It seems a tough problem because I don't know how to use Mclaurin series or Euler's constant.
 

HallsofIvy

MHF Helper
Apr 2005
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Have you not taken a Calculus course? A "McLaurin" series is a Taylor's series about a= 0. Specifically, the McLaurin Series for f(x) is
\(\displaystyle f(0)+ f'(0)x+ (f''(0)/2)x^2+ \cdot\cdot\cdot+ f^{(n)}(0)x^n+ \cdot\cdot\cdot\).

(\(\displaystyle f^{(n)}\) is the nth derivative of f.)
 
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Mar 2012
565
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but zero is undefined in the function 1/x
 

Plato

MHF Helper
Aug 2006
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but zero is undefined in the function 1/x
Hallsofivy in reply told #2 told you the truth. The fact is that you nor anyone else is going to find a neat way to use simple Riemann Sums to find $\int_1^2 {\frac{{dx}}{x}}$. All of the other replies rely on advanced facts and/or techniques.

Some calculus textbooks develop the integral using an area function and never mention Riemann sums.
In those texts, the logarithm is defined as $\displaystyle\log (x) = \int_1^x {\frac{{dt}}{t},\;x > 0}$, again no sums!
 
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