How to calculate the integral as a limit of sum?

PrakashPhy
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Homework Statement



The problem is to find the integral of function \frac{1}{x} using the definition i.e the area under the curve as a limit of a sum?



Homework Equations


\int_a^b \frac{1}{x} dx = ln(\frac{b}{a})



The Attempt at a Solution


I tried with dividing the interval [a,b] into n parts such that h=\frac{b-a}{n} but what I got was a too vague expression of the sum.

\lim_{n \rightarrow ∞} \frac{b-a}{n} = h
now
\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg]

how do I proceed now
 
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PrakashPhy said:

Homework Statement



The problem is to find the integral of function \frac{1}{x} using the definition i.e the area under the curve as a limit of a sum?



Homework Equations


\int_a^b \frac{1}{x} dx = ln(\frac{b}{a})



The Attempt at a Solution


I tried with dividing the interval [a,b] into n parts such that h=\frac{b-a}{n} but what I got was a too vague expression of the sum.

\lim_{n \rightarrow ∞} \frac{b-a}{n} = h
now
\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg]

how do I proceed now

So is the idea that you should end up with the power series for ln(x)?
 
Something like that; probably some limit of ln(x) because \lim_{h\rightarrow 0} is there intact throughout.
 
Last edited:
PrakashPhy said:

Homework Statement



The problem is to find the integral of function \frac{1}{x} using the definition i.e the area under the curve as a limit of a sum?

Homework Equations


\int_a^b \frac{1}{x} dx = ln(\frac{b}{a})

The Attempt at a Solution


I tried with dividing the interval [a,b] into n parts such that h=\frac{b-a}{n} but what I got was a too vague expression of the sum.

\lim_{n \rightarrow ∞} \frac{b-a}{n} = h
now
\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg]

how do I proceed now
Maybe that's a typo, but that h should not be squared in your last line.

\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg]

should be

\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h ...\bigg]
 
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