How to calculate the integral as a limit of sum?

Click For Summary

Homework Help Overview

The problem involves calculating the integral of the function \(\frac{1}{x}\) using the definition of the integral as the area under the curve expressed as a limit of a sum.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss dividing the interval \([a,b]\) into \(n\) parts and express the sum in terms of \(h = \frac{b-a}{n}\). There is mention of obtaining a vague expression for the sum and questions about how to proceed from there.

Discussion Status

Some participants are exploring the relationship between the limit and the logarithmic function, while others are questioning the accuracy of the expressions used in the limit process. There is an ongoing examination of the setup and notation used in the attempts.

Contextual Notes

There are indications of potential typos in the expressions, particularly regarding the powers of \(h\) in the sum, which participants are addressing. The discussion reflects a focus on ensuring clarity in the mathematical formulation.

PrakashPhy
Messages
35
Reaction score
0

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
 
Physics news on Phys.org
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]
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
4
Views
2K