Approximating Integrate Using Series Methods

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Homework Help Overview

The discussion revolves around approximating the integral of the function (1 - e^x)/x from 0 to 1 using series methods, specifically through the Taylor series expansion of e^x. The original poster expresses uncertainty about their approach and seeks assistance in the context of a homework assignment.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers using the Taylor series for e^x and integrating the resulting power series. Some participants confirm this approach and suggest substituting the limits of integration after obtaining the series.

Discussion Status

Participants are actively engaging with the problem, with some providing feedback on the original poster's calculations. There is a recognition of the need to account for the sign of the integrand, and adjustments are being made based on this feedback. The discussion reflects a collaborative effort to clarify the approach without reaching a definitive conclusion.

Contextual Notes

The original poster mentions a time constraint due to the assignment's deadline and expresses feelings of being challenged after a break from school, which may influence their confidence in tackling the problem.

blue mango
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I need help solving the following (it is due tomorrow :frown: and it just got assigned yesterday ):

use series methods to obtain the approximate value of integral(from 0 to1) of (1-e^x)/x dx

What I have thought of so far is to use the Taylor series approx for e^x and carry that out to a few terms then solve for 1 and 0 and plug it into the original equation but I'm not sure that I'm headed in the right direction. I took a year off from school and now I'm having to relearn a bunch of things that were once simple but now seem quite challenging. Any help is appreciated. Thanks.
 
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Looks good to me. You should get an easy power series to integrate and then you just substitute in 1 and 0.

Mat
 
I plugged in the Taylor series for e^x and took that out to 6 terms then integrated it from 0 to 1 and got 1.262...does that sound about right?
 
The integrand is negative so you should end up with a negative answer.

As an example just using two terms, ex=1+x you would end up with

[tex]\int_0^1 \frac{1-(1+x)}{x}dx = \int_0^1 \frac{-x}{x}dx=-1[/tex]

As you add more terms the integral should become more negative
 
Thanks for pointing that out Office Shredder. There was a negative that I had forgotten to carry over. So I got -1.262 with 6 terms in the e^x Taylor series.
 

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