Tricks for Saving Money on Groceries

  • Context: MHB 
  • Thread starter Thread starter bearn
  • Start date Start date
  • Tags Tags
    Money
Click For Summary

Discussion Overview

The discussion revolves around the concepts of integration and differentiation in calculus, specifically focusing on the antiderivative and its relationship to the derivative. Participants are examining a mathematical expression and clarifying the correct identification of derivatives versus antiderivatives.

Discussion Character

  • Technical explanation

Main Points Raised

  • Some participants assert that a previous response incorrectly identified an antiderivative as a derivative, emphasizing the need to distinguish between the two processes.
  • One participant suggests an alternative expression for the integral, proposing that it should be $\displaystyle \int 3x+5 \, dx = \dfrac{3}{2}x^2 + 5x + C$.
  • Another participant reinforces the idea that the original claim was incorrect by stating that the derivative of the proposed antiderivative yields the original function.

Areas of Agreement / Disagreement

Participants generally agree that there was a misunderstanding regarding the identification of derivatives and antiderivatives, but the specific details and interpretations remain contested.

Contextual Notes

There are unresolved assumptions regarding the definitions of the functions involved and the rules applied in the differentiation and integration processes.

bearn
Messages
11
Reaction score
0
Screenshot (45).png
 
Physics news on Phys.org
no ... you found the derivative, not the antiderivative.

$\displaystyle \int \dfrac{3}{2}x^2 + 5x + C \, dx = \dfrac{x^3}{2} + \dfrac{5}{2}x^2 + Cx + K$
 
skeeter said:
no ... you found the derivative, not the antiderivative.

$\displaystyle \int \dfrac{3}{2}x^2 + 5x + C \, dx = \dfrac{x^3}{2} + \dfrac{5}{2}x^2 + Cx + K$
Where is your answer based from? What rule, if there is?
 
Maybe you meant ...

$\displaystyle \int 3x+5 \, dx = \dfrac{3}{2}x^2 + 5x + C$ ?

in any case, watch the video

 
Last edited by a moderator:
As skeeter first said, you went "the wrong way"- you found the derivative, not the integral. $\frac{d(\frac{3}{2}x^2+ 5x+ C)}{dx}= 3x+ 5$
 

Similar threads

  • · Replies 179 ·
6
Replies
179
Views
14K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 28 ·
Replies
28
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
2
Views
5K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K