I think my problem solving is absurd.

  • Thread starter Thread starter stanton
  • Start date Start date
  • Tags Tags
    Problem solving
Click For Summary

Homework Help Overview

The discussion revolves around evaluating a definite integral involving an exponential function and a transformation using substitution. The original poster expresses confusion regarding the application of integration rules and the handling of a potential division by zero.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the substitution method, with one suggesting a specific substitution for the integral. Questions arise about the implications of the substitution and the correctness of the approach taken by the original poster.

Discussion Status

There is an ongoing exploration of the integral's transformation through substitution. Some participants provide feedback on the original poster's attempts, indicating that the approach may be on the right track, while others seek clarification on the integral's setup.

Contextual Notes

The original poster mentions encountering a division by zero issue, which raises questions about the assumptions made in the problem setup. There is also a reference to the need for absolute value signs in the solution, which is debated among participants.

stanton
Messages
74
Reaction score
0

Homework Statement



Evaluate the definite integral:
200911281932436339503356392025007792.jpg


Homework Equations



\int u^n du/dx = u^(n+1)/(n+1) (n not equal to -1)


The Attempt at a Solution



(-1) muliplied by \int (e^-x +2)^-1 (-e^-x) dx = ?

and if I follow the equation above, I got denominator zero, for I broke the rules 'n is not equal to zero'
I think I did as the equation. but now what should I do now?

200911281926416339503320196712503694.jpg
 
Physics news on Phys.org
Do you know how to integrate

<br /> \int \frac 1 u \, du<br />
?
 
Let u = e^(-x) + 2
the du = ...
 
-e^(-x)
 
So with u = e^{-x} + 2 and du = -e^{-x}, what happens to your integral?
 
So with LaTeX Code: u = e^{-x} + 2 and LaTeX Code: du = -e^{-x} , what happens to your integral?

Thank you for your help. I did like this:

cramster-equation-20091130231296339514508975162501709.gif


cramster-equation-20091130234126339514525212662509813.gif


cramster-equation-20091130236176339514537720475008948.gif


Can you check is this the right answer?
 
Looks good. You are correct that the absolute value signs are not needed in the answer, since e^{-x} + 2 is never negative.
 

Similar threads

Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K