Calculating Area Using Integration

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Discussion Overview

The discussion revolves around calculating areas using integration, specifically focusing on the area enclosed by the curves y = x and y^2 = x + 2, as well as determining the total distance traveled by a particle given its velocity function v(t) = t^3 - 3t. The scope includes mathematical reasoning and problem-solving related to integration techniques.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Exploratory

Main Points Raised

  • One participant initially calculated the area enclosed by the curves and arrived at an area of 10/3, but another participant pointed out a potential mistake in evaluating the integral at x = 0.
  • There was a discussion about the correct limits of integration, with some participants suggesting that the integration should be from x = -2 to x = 2 instead of from 0 to 2.
  • One participant proposed that the area calculated from the new limits would yield an area of 16/3, and another participant agreed with this result.
  • In a separate question, a participant sought to calculate the total distance traveled by a particle, but another participant cautioned that simply evaluating the integral from 0 to 5 would give displacement rather than total distance.
  • Participants discussed breaking the interval into segments to account for changes in direction to accurately compute total distance.
  • Another participant asked about integrating a function involving a square root, expressing uncertainty about the substitution method and integration by parts.

Areas of Agreement / Disagreement

There is no clear consensus on the correct limits for the area calculation, as some participants suggest different approaches. The discussion on total distance also remains unresolved, with differing opinions on how to approach the problem.

Contextual Notes

Participants express uncertainty about specific steps in their calculations and the implications of their results, indicating a reliance on assumptions that may not be fully articulated.

Who May Find This Useful

Students or individuals interested in calculus, particularly in integration techniques and applications related to area and distance calculations.

helpm3pl3ase
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I was wondering if I did this correctly:

Sketch the region enclosed by y = x and y^2=x+2 and then determine its area.

I got the area enclosed to be x = 0, x = 2.

Then I used Integral from 0 to 2 (x+2)^(1/2) - (x) = [(2/3)(x+2)^(3/2) - (x^2/2)] evaluated at 2 and 0 which = 10/3
 
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Are you sure? Your method looks fine but I think you may have made a mistake when you put x=0 in your result. (2/3)(0+2)^(3/2) is not zero.
 
oo your right.. goddamn.. lol.. so does this look right

10/3 - (4*(2)^(1/2))/3

I just wasnt sure if I went from the right points (0,2) because the graph y = (x+2)^(1/2) ends at x = -2.
 
helpm3pl3ase said:
oo your right.. goddamn.. lol.. so does this look right

10/3 - (4*(2)^(1/2))/3

I just wasnt sure if I went from the right points (0,2) because the graph y = (x+2)^(1/2) ends at x = -2.

3, you mean? not that it's going to matter...

I just graphed it and I see what you mean. Yeah, I guess you should take it from x=-2, since there is nothing in your question about x=0.
 
alright so integrate from 2 to -2 which i would get:

16/3?
 
I agree with that answer.
 
alright man.. Thanks..
 
You're welcome. :smile:
 
one more question:

I need total distance of a particle traveling at v(t) = t^3 - 3t for [0,5]

t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 5 and x = 0

= 475/4?? Did i do this correctly for the TOTAL distance
 
Last edited:
  • #10
Anyone?
 
  • #11
Hmmm, yeah I would be careful with that. If you put in just x=0 and x=5 you would get the displacement, which is not necessarily the total distance. So you have to know which they are really asking for. Try adding up the distance between t=0 to t=1, then t=1 to t=2 up to t=5 to get a better idea of the total distance.
 
  • #12
so

t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 1 and x = 0
+
t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 2 and x = 1
+
t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 3 and x = 2

etc?? Like that??
 
  • #13
Yeah, I would try that.
 
  • #14
All added together i get 481/4??

is the any definite way to know the right answer..
 
  • #15
I can't think of one of off the top of my head. Is there someone you can clarify the question with?
 
  • #16
not really.. but i do have another question for you if your up for it.. Yes I know i have a lot, but I am being tested quite soon and I need to know my ****
 
  • #17
I'll give it a try. You also might see more action if you post your questions in the Homework Forums.
 
  • #18
Awesome.. here's one..

find Integral of (t^2)/(t-1)^(1/2) dt

I know we need to use the substitution rule.. but I am not sure where to begin:

u = t^2, du = 2t dt, dt = du/2t??
 
  • #19
Integration by parts works nicely here.
 
  • #20
i know but Iam unsure of where to begin..
 
  • #21
u= t-1 du = dt

Integral (u)^2/u^(1/2)

= u^2 * u^(-1/2) then antideriv??
 
  • #22
Set [itex]u(t)=t^{2},\frac{dv}{dt}=\frac{1}{\sqrt{t-1}}[/itex]
 

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