Question about Integration

  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
hage567
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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.
 
  • #3
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.
 
  • #4
hage567
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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.
 
  • #5
alright so integrate from 2 to -2 which i would get:

16/3?
 
  • #6
hage567
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I agree with that answer.
 
  • #7
alright man.. Thanks..
 
  • #8
hage567
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You're welcome. :smile:
 
  • #9
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
hage567
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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
hage567
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Yeah, I would try that.
 
  • #14
All added together i get 481/4??

is the any definite way to know the right answer..
 
  • #15
hage567
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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 ya 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
hage567
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I'll give it a try. You also might see more action if you post your questions in the Homework Forums.
 
  • #18
Awesome.. heres 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
arildno
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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
arildno
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Set [itex]u(t)=t^{2},\frac{dv}{dt}=\frac{1}{\sqrt{t-1}}[/itex]
 

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