Calculating Area Using Integration

[helpm3pl3ase]

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

 

[hage567]

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.

 

[helpm3pl3ase]

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.

 

[hage567]

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.

 

[helpm3pl3ase]

alright so integrate from 2 to -2 which i would get:

16/3?

 

[hage567]

I agree with that answer.

 

[helpm3pl3ase]

alright man.. Thanks..

 

[hage567]

You're welcome.

 

[helpm3pl3ase]

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

 

[helpm3pl3ase]

Anyone?

 

[hage567]

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.

 

[helpm3pl3ase]

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??

 

[hage567]

Yeah, I would try that.

 

[helpm3pl3ase]

All added together i get 481/4??

is the any definite way to know the right answer..

 

[hage567]

I can't think of one of off the top of my head. Is there someone you can clarify the question with?

 

[helpm3pl3ase]

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 ****

 

[hage567]

I'll give it a try. You also might see more action if you post your questions in the Homework Forums.

 

[helpm3pl3ase]

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??

 

[arildno]

Integration by parts works nicely here.

 

[helpm3pl3ase]

i know but Iam unsure of where to begin..

 

[helpm3pl3ase]

u= t-1 du = dt

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

= u^2 * u^(-1/2) then antideriv??

 

[arildno]

Set $u(t)=t^{2},\frac{dv}{dt}=\frac{1}{\sqrt{t-1}}$