What is the integral of e and how does it relate to acceleration and velocity?

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Discussion Overview

The discussion revolves around the integration of the exponential function e^(5t) in the context of relating it to acceleration and velocity. Participants explore the correct application of integration techniques, particularly in the presence of constants and derivatives.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant claims that the integral of e is simply e, suggesting that integrating 24 + e^(5t) results in 24t + e^(5t) for velocity.
  • Another participant counters that the derivative of the exponent must be considered, stating that the derivative of e^(5t) is 5e^(5t).
  • A participant expresses confusion about the integration process, questioning how to apply the derivative in the context of integration and suggesting a substitution method.
  • Further clarification is provided that the integral of 5e^(5t) is e^(5t) + C, emphasizing that 5e^(5t) is not a constant but a function of t.
  • One participant critiques another for misunderstanding the integral of e, asserting that the integral of e (as a constant) is ex, not e itself.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the interpretation of the integral of e and the proper method for integrating e^(5t). There is no consensus on the correct approach to the problem, and confusion persists about the application of integration techniques.

Contextual Notes

Participants express uncertainty about the integration process, particularly in relation to constants and derivatives. The discussion highlights the need for clarity in defining terms and understanding the relationship between differentiation and integration.

Dae
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The integral of e is e right? So if you were to take the integral of 24+e^(5t) (acceleration), it would be 24t+e^(5t) (velocity)?
 
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No. Don't forget to include the derivative of the exponent.

Remember that

[itex] \frac{d}<br /> {{dt}}e^{5t} = e^{5t} \cdot 5[/itex]

- Warren
 
Alright, thanks for the help. :)

I see what you're saying there. However, that's when taking the derivative, and I don't understand how to apply that. If I were taking the integral of e^(5t)*5, the 5 would be a constant, so since the constant of just plain e^(5t) is 1.. how could the integral change? Does it become like Ce^(5t) after integrated? When you say I have to include the derivative, do you mean (24+e^u du)dt where u = 5t, du = 5. For some reason I'm leaning towards du = 5 dt, dt = 1/5 du, so its (1/5)*24 du + e^u * (1/5) du, which when integrated becomes (24t + e^(5t))/5.

I'm obviously confused. Maybe you could please show how to solve it for me? Integral of e
 
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Dae said:
If I were taking the integral of e^(5t)*5, the 5 would be a constant, so since the constant of just plain e^(5t) is 1..

Whoops! That's not right. Just because 5 is a constant doesn't mean 5 e^(5t) is a constant. It's not; it's still a function of t.

If you agree that the derivative of e^(5t) is 5 e^(5t), then you also have to agree that the integral of 5 e^(5t) is e^(5t) + C.

What I mean when I say "don't forget to include the derivative of 5t," I mean this:

You cannot directly integrate e^(5t). What you'd have to do is multiply it by 5 inside the integral sign, and divide it by 5 outside the integral sign, like this:

[itex] \int {e^{5t} dt = \frac{1}<br /> {5}} \int {5e^{5t} dt} [/itex]

Thus, the integral of e^(5t) is (1/5) e^(5t) + C.

Maybe you could please solve it for me?

We don't solve homework problems. We help you understand how to solve them.

- Warren
 
A lot of your problem is just sloppy thinking. You started by saying "The integral of e is e right?" No, the "integral of e" is NOT e. The integral of e (with respect to x) is ex since e is just a constant. If you MEANT that the integral of ex is ex (plus a constant NOT times a constant as you say) say that!
 

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