integration problem

Shox

could someone tell me how to integrate 1/y(squared)

tolosenc

1/y^2 = y^-2

Are there limits? If not: |1/y^2.dx = 1/y, in my reckoning

ZorbaTehZ

1/y^2 = y^-2

Are there limits? If not: |1/y^2.dx = 1/y, in my reckoning

But the problem with that is that you are integrating with respect to x when there is that y inside there.

If the question was integrate 1/x-squared with respect to x or integrate 1/y-squared with respect to y, then your method would be almost correct (you forgot the minus signs, and the constant).

i.e. S 1/y-squared dy = S (y^-2) dy = (y^-1)/-1 + c = -1/y + c

The OP needs to be clearer about what the question is.

Shox

hey sry tez, wasnt clear enough.

ya its from da "fundamental applied maths" book page 289, Q 2.

ive ended up wit:

Integral of 1/v^2 dv = Integral of dt

Dunno where to go from here?

ZorbaTehZ

This question I assume:

dx/dt [2] = (dx/dt)^2 (dx/dt [2] meaning second derivitive)

Looks like you let v=dx/dt
So:
v = dx/dt
dv/dt = dx/dt [2]

Original equation becomes:

dv/dt = v^2

First find v then x:
So manipulation of the above gives:

(1/v^2) dv = dt

Integrate both sides:

S (1/v^2) dv = S dt <-what you posted
-1/v = t + c (how I got this is posted above in my original)

But v = dx/dt, so:

-dt/dx = t + c

dx/dt =1 @ t = 0 (given in the question)

Therefore:

-1 = 0 + c
-1 = c

So therefore:

-dt/dx = t -1
dt/dx = 1-t
dx/dt = 1/(1-t)
dx = dt/(1-t)
S dx = S dt/(1-t)
x = -ln(1-t) + c

x = 0 @ t=0 (given in the question)

Therefore:

0 = -ln(1) + c
c = ln1
c = 0

Therefore:

x = -ln(1-t) + 0
x = -ln(1-t)
x = ln((1-t)^-1)
x = ln(1/(1-t))

QED

Nehpets

Fair play for typing that out! I HATE typing out maths with a keyboard

lauraxo

any1 have dare art history paper done???? and a question on newgrange??????????? i need help fast!!!!!!!!!!

Shox

ya thanks a million tehz, thats a huge help an i no ya prob spent a while typin it. i appreciate it!!!