4U Revising Game (1 Viewer)

[conics2008]

use this :

S between a and -a f(x) dx = S between a and 0 f(x) + f(-x)

when you do that

you will get 2 fractions, then times e^x / e^x with one of them... to remove that e^-x

when you do all that you will get S 1 to 0 of 1 dx

therefore (x)1 and 0

is equal to 1

therefore answer is 1 u^2

 

[conics2008]

hmm ok, different working out mine to yours.. ohh well who cares.. these types of question never come in the HSC.. soo i wont bother with my answer.

 

[independantz]

try:

1
/
|
| 1 / (1 + e^-x) dx
|
/
-1

I(e^x/(1+e^x).dx=I du/u=in/u/=ln(1+e^x)+C

new q: In sec^3x.dx

 

[conics2008]

i already gave my answer to your intergal.. its just 1 by inspection !

 

[conics2008]

hey im not going to finish it ill just tel you what you gota do, see the base , you can write it like this

4S1/ (x-1)^2 -2

and then just integrate 1/(x-1)^2 -2 and then your done..

ok then ill tell you use u=x-1

when your done , break up your fraction.. then integrate one by one. you should get ln in there..

 

[tommykins]

new q: In sec^3x.dx

int sec.sec^2x u = sec u' = secxtanx , v' = sec^2x, v = tanx

int by parts
Let I = int sec^3xdx
I = secxtanx - int secxtan^2x dx
= secxtanx - [int sec^3x - int secx]
= secxtanx - I + int secx -> multiply secx by (tanx+secx)
2I = secxtanx + ln|secx + tanx|

.:. I = (secxtanx + ln|secx + tanx|) / 2



Conics
The normal to the parabola (4x^2)/3 - 4y^2 = 1 at the point (3sqrt2/4,sqrt2/4) is also a tangent to the circle x^2 + y^2 = 1 at the point (x1,y1). Find x1 and y1.

 

[conics2008]

do you assume that alpha is one of the roots and then you sub it in,

a^101+a^51+a=1

factor out a gives you a^100+a^50+1 = 1/a

then i dont know whats going on ??

 

[conics2008]

Tommy conics.

dy/dx = x^2/2y

therefore gradient = 9root2/8

find the stupid line..

this is the equation.

36xroot2 - 32y -54+8root2 =0

then use this and solve it together with the circle equation.. am i on the right track or did i go off track.

 

[conics2008]

3units did you factor out x and made it look like this

x(x^100+x^50+1-1/x)

 

[conics2008]

ok i give up, i had enough of that question, my mind doesn't work for easy quesitons... soo i think ill pass.. because that equation has no integer roots or what so ever..

anyways, im out.. ive been playing too much games here.. take care. thanks for the questions =)

 

[undalay]

pi/4
/
| 1
|----------------dx
|2sin(2x) + cosx
/
0

 

[YannY]

pi/4
/
| 1
|----------------dx
|2sin(2x) + cosx
/
0

=1/[cosx(4sinx+cosx)] x sec^2x/sec^2x
=sec^2x/(4tanx+1)
=1/4ln(4tanx+1)
=1/4ln5

 

[undalay]

=1/[cosx(4sinx+cosx)]

= 1/[4cosxsinx + cos^2 x]
= 1/[2sin(2x) + cos^2 x]
not equal the question!.

You made a small mistakeeee

 

[YannY]

Oh blimsey i did so.

 

[ngogiathuan]

(5) evaluate:

I{-1 -> 1} sqrt [(1+x) / (1-x)] dx

by using the substitution (1+x) / (1-x) = z^2

Not sure if this correct

 

[ronnknee]

Haha man I need to maintain this thread more
Boredofstudies was down for me and friends the past two days for some reason T_T

 

[pLuvia]

OH WON'T SOMEBODY PLEASE PROVIDE A WORKED SOLUTION?


:sadface:

You have to think about it case by case i.e. if there are 4 repeated letters, 3 repeated letters etc etc..

Answer is 1645

A bit tedious but in the 4u HSC they may ask something similar but less tedious

I'll let you guys ponder on that

 

[pLuvia]

MOAR:

(6)

i) prove: 1 - r^2 + r^4 - r^6 + r^8 - ... = 1/(1+r^2), for |r|<1

ii) hence find a series for tan inverse

iii) hence show that pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 ...

(7) by considering a geometric series prove:

I [r^n / (1-r)] dr = ln|1- r| - (r + r^2/2 + r^3/3 + ... + r^n/n)

6

i) Geometric series, where r=-r² and n approaches infinity hence its just 1/(1+r²)

ii) Int both sides wrtr, RHS becomes tan^-1x, LHS becomes r-r³/3+r⁵/5...
hence
tan^-1r=r-r³/3+r⁵/5-...+(-1)ⁿr^2n-1/(2n-1)

ii) Using the series for tan^-1r, consider letting r=1
RHS=tan^-11=pi/4
LHS=1-1/3+1/5...
Hence
pi/4=1-1/3+1/5...
pi=4-4/3+4/5...

7
Consider the geo series (1-rⁿ)/(1-r)
(1-rⁿ)/(1-r)=1+r+r²+r³+...+r^n-1
rⁿ/(1-r)=1/(1-r)-(1+r+r²+r³+...+r^n-1)
int. [rⁿ/(1-r)]dr = int. [1/(1-r)-(1+r+r²+r³+...+r^n-1)]
=ln|1-r|-(r+r²/2+r³/3+....+rⁿ/n

as reqd

 

[pLuvia]

The point P(a*cosθ, b*sinθ ) is any point on the ellipse: x^2/a^2 + y^2/b^2 = 1, with focus S.
The point M is the midpoint of the interval SP.
Show that as P moves on the ellipse, M lies on another ellipse whose centre is midway between the origin O and the focus S.

Edit. Hows your degree exphate?

 

[pLuvia]

It's a bit tedious at the moment, especially "multivariate" which isn't really what they proclaim it to be. Hopefully I can murder the end of session tests and get good finaly marks in both the math subjects.

And then I can start teaching myself extension 2 >_<

We did some bivariate stats, which was alright I think. Not sure how different it is in the actual maths field And anyway I rather like stats then maths

You hardly touch on any of the MX2 topics in uni, I think the only thing we did was complex and some integration. I'm sure you can teach yourself the mx2 topics, since you now have the mathematics mind