- #1

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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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- Thread starter helpm3pl3ase
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- #1

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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

- #2

hage567

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- #3

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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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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

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

16/3?

16/3?

- #6

hage567

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I agree with that answer.

- #7

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alright man.. Thanks..

- #8

hage567

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You're welcome.

- #9

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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

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

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Anyone????

- #11

hage567

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- #12

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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

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All added together i get 481/4??

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

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

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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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- #18

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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

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i know but Iam unsure of where to begin..

- #21

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u= t-1 du = dt

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

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

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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