# Urgent Maths Help NeededWatch

This discussion is closed.
#1
Got a maths exam tomorrow.. so can someone pleeeeeeassseeee help.

1. Use integration by parts to evaluate 01 In (1 + x) dx.

And the other questions I'm stuck on is...

2. Functions x(t) and y(t) satisfy

dx/dt = -x2y, dy/dt = -xy2

When t = 0, x = 1 and y = 2.

(a) Express dy/dx in terms of x and y and hence obtain y as a function of x

I get this one... the answer being y = 2x but get stuck on the next bit

(b) Deduce that dx/dt = -2x3 and obtain x as a function of t for t>=0.

plz help.
0
13 years ago
#2
2.
b) dx/dt = -x2y = -x(2x) = -2x3

(1/x3)(dx/dt) = -2

INT x-3 dx = INT (-2) dt

x-2/(-2) = -2t + c

-1/(2x2) = -2t + c

t=0, x=1 => -1/2 = c

=> -1/(2x2) = -2t - 1/2
1/(2x2) = 2t + 1/2

x = +/- rt[1/(1+4t)]
0
13 years ago
#3
1.

∫ ln (1 + x) dx =∫1.ln (1 + x) dx

u = ln(1+x) ; v' =1
u' = 1/(1+x) ; v = x

∫1.ln (1 + x) dx = xln(1 + x) - ∫x/(1+x) dx

x/(1+x) = (x+1-1)/(x+1) = 1 - 1/(x+1)

∫1.ln (1 + x) dx = xln(1 + x) - ∫ [1 -1/(1+x) ] dx
∫1.ln (1 + x) dx = [ xln(1 + x) - x + ln(1+x) ] {1,0}
0
#4
Cheers!
0
13 years ago
#5
Int ln(1+x) dx

Int u'v = uv - Int uv'

u' = 1
u = x
v = ln(1+x)
v' = 1/(1+x)

Int ln(1+x) = xln(1+x) - Int x/1+x

x/(1+x) = 1 - 1/(x+1)

Int ln(1+x) = xln(1+x) - Int 1 - 1/(x+1)
Int ln(1+x) = xln(1+x) - x + ln(x+1)
Int ln(1+x) = (x+1)ln(1+x) - x

from 0 to 1

[1ln1 - 0] - [2ln2 - 1]
= 1 - 2ln2
0
#6
Cheers for the help everyone... I always forget the '1' in the integration by parts... will have to remember it for tomorrow though.
0
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