Understanding the Integral of e^x: Solving for \int_0^1 e^{-3x} dx Explained

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Discussion Overview

The discussion revolves around understanding the integral of the function \( e^{-3x} \) over the interval from 0 to 1. Participants are exploring the steps involved in solving the integral, particularly focusing on the appearance of the constant \(-\frac{1}{3}\) in the solution process.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states the integral of \( e^x \) is \( e^x + C \) and expresses confusion about the integral \( \int_0^1 e^{-3x} dx \) and the origin of the constant \(-\frac{1}{3}\).
  • Another participant requests help, specifically asking for clarification on where the constant \(-\frac{1}{3}\) comes from.
  • A participant explains that the constant \(-\frac{1}{3}\) is derived from the need to counteract the derivative of the exponent in \( e^{-3x} \), noting that the derivative of \( e^{-3x} \) introduces a factor of \(-3\).
  • One participant acknowledges their oversight in forgetting to apply the derivative rule when integrating functions of the form \( e^{kx} \), expressing relief at the clarification provided.
  • Another participant relates to the confusion, indicating that they also sometimes forget this aspect of differentiation when working with exponentials.

Areas of Agreement / Disagreement

Participants generally agree on the reasoning behind the appearance of the constant \(-\frac{1}{3}\) in the integral, but there is no formal consensus on the overall understanding of the integral process as some participants express confusion.

Contextual Notes

Some participants express uncertainty about the application of derivative rules in integration, indicating a potential gap in understanding the relationship between differentiation and integration for exponential functions.

Who May Find This Useful

Students preparing for tests in calculus or those seeking clarification on integration techniques involving exponential functions.

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Ok I know that the [tex]\int e^x = e^x + C[/tex]

now I don't understand this problem.

[tex]\int_0^1 e^{-3x} dx = -(1/3)e^{-3(0)} - -(1/3)e^{-3(1)}[/tex]
[tex]= 1/3(1- e^{-3} )[/tex]

Where does the 1/3 come from?
 
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anyone?? have to study for a test any help would be greatly appreciated.

I just want to know where the constant -1/3 came from.
 
Because...

*I'm using -> to point to the derivative.

e^x -> e^x

e^3x -> 3e^3x

e^-3x -> -3e^-3x

See how the 3 came up? The -1/3 is used to get rid of the -3 because in the integral we don't have a -3.

-1/3e^-3x -> e^-3x

Note: I know it's sloppy, but you should get the idea.
 
JasonRox said:
Because...

*I'm using -> to point to the derivative.

e^x -> e^x

e^3x -> 3e^3x

e^-3x -> -3e^-3x

See how the 3 came up? The -1/3 is used to get rid of the -3 because in the integral we don't have a -3.

-1/3e^-3x -> e^-3x

Note: I know it's sloppy, but you should get the idea.


wow I feel really stupid to forget that lol, I forgot that when [tex]e^x[/tex] is raised to something like 3x you use the derivative of that times the original [tex]e^x[/tex] function.

thanks :D
 
I forget it sometimes to because your so used to copying e^x down as the derivative.

Good Luck!
 

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