Mathematics: Post your doubts here!

[iKhaled]

(-1/2;0) is the point of intersection of the graph with x-axis hence the lower limit is -1/2
it's given by:

pi* integral from -1/2 to 0 of y^2

duuuhhh -_- i know that but how can i integrate that ? i know we have to use integration by parts but it is not working for me !

 

[Kandinsky]

why not. take exponent as dv and the polynomial as u:

integral u dv = uv - integral v du

 

[iKhaled]

why not. take exponent as dv and the polynomial as u:

integral u dv = uv - integral v du

yeah i tried to but its failing..can u do it please ?

 

[Kandinsky]

so your integral is given by
I = integral(-0.5 ; 0) of e^(-x) * (1+2x) dx
u = 2x+1, dv = e^(-x) dx. hence du = 2dx and v = -e^(-x)

I = (2x+1)*(-e^(-x)) +2 integral (e^(-x) dx)
I = -e^(-x) * (2x+1+2) = -e^(-x)*(2x+3)
inserting limits, obtain:

I = -3 + e^(0.5)*2;
hence the volume is pi*(2 sqrt(e) - 3)

 

[iKhaled]

can someone please explain to me the first question in this paper ?
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

 

[Tharindu]

question 5) (ii) May June, paper 3 (M1) ....how do we find the tension of S2 ?

 

[whitecorp]

can someone please explain to me the first question in this paper ?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

ln( 2+ e^-x )=2
2+e^-x = e^2
e^-x = e^2 - 2
-x = ln (e^2 - 2)
Hence, x = - ln (e^2 - 2)= -1.68 (shown)

Hope this helps. Peace.

 

[whitecorp]

question 5) (ii) May June, paper 3 (M1) ....how do we find the tension of S2 ?

Posting the question paper would be nice, you know?

 

[iKhaled]

ln( 2+ e^-x )=2
2+e^-x = e^2
e^-x = e^2 - 2
-x = ln (e^2 - 2)
Hence, x = - ln (e^2 - 2)= -1.68 (shown)

Hope this helps. Peace.

how did uk that 2+e^-x = e^2

 

[Yousif Mukkhtar]

Guys for this question in the attached file why cant we used this method:
S(x-x bar)^2
= Sx^2-x bar^2

Mean of x= Sx/n=645/150=4.3
x bar^2= 4.3^2=18.49
S(x-x bar)^2=8287.5/18.49
=8269.01

 

[whitecorp]

how did uk that 2+e^-x = e^2

You simply take the exponential base on both sides.

Hope this helps. Peace.

 

[iKhaled]

how this stupid question can be solved!!!

(2^x + 1 ) / (2^x - 1 ) = 5

first question in m/j 2010 32

 

[VelaneDeBeaute]

how this stupid question can be solved!!!

(2^x + 1 ) / (2^x - 1 ) = 5

first question in m/j 2010 32

Studied logs? Well, try it that way.

 

[iKhaled]

Studied logs? Well, try it that way.

yeah nevermind..i got it now i just didn't know how to multiply 5(2^x-1)

 

[Alice123]

how do i differentiate ln(xy) with respect to x???

 

[whitecorp]

how do i differentiate ln(xy) with respect to x???

This is done implicitly. d/dx [ ln(xy)] = 1/(xy) * [ x (dy/dx) +y ]

Hope this helps. Peace.

 

[Alice123]

This is done implicitly. d/dx [ ln(xy)] = 1/(xy) * [ x (dy/dx) +y ]

Hope this helps. Peace.

thanks
x=cos^3t y=sin^3t.....find the cartesian equation with the parametric equations.... pls help with this too..

 

[whitecorp]

thanks
x=cos^3t y=sin^3t.....find the cartesian equation with the parametric equations.... pls help with this too..

x= cos^3 t ======> cos^2 t = x^ (2/3) ---------------(1)

y= sin^3 t ======> sin^2 t = y^ (2/3) --------------(2)

(1)+(2): x^(2/3) +y ^(2/3) =1 (shown)

Hope this helps. Peace.

 

[kronix6]

Hi can anyone solve this plz.

use the substitution x=tanQ to show that

( f(1-x^2)/(1+x^2)^2 dx )is equal to (fcos2Q dQ)

i dont know how to make integration sign so i used f instead. so f= integration sign

 

[iKhaled]

Hi can anyone solve this plz.

use the substitution x=tanQ to show that

( f(1-x^2)/(1+x^2)^2 dx )is equal to (fcos2Q dQ)

i dont know how to make integration sign so i used f instead. so f= integration sign

ok bro for this question u need to know trig identities and integration by substitution so here is the solution for it..

x = tan Q

∫ (1-x^2)/ (1+x^2)^2 dx = ∫cos 2Q dQ

dx/dQ = sec^2 Q

dx = sec^2 Q dQ

∫(1-tan^2 Q)/ (1+tan^2 Q)^2 sec^2 dQ

ok i wont add the Q now cuz its a bit annoying will leave that to the end..

we know that 1+tan^2 = sec^2 so
∫ [(1-tan^2)/(sec^2)^2] x sec^2 dQ ( here we substituted the dx with sec^2 Q dQ )
∫(1-tan^2)/sec^2 dQ
∫(1- sin^2/cos^2)/ (1/cos^2) dQ
∫ cos^2 Q - sin^2 Q dQ

we have a rule that says cos^2 Q - sin^2 Q = cos 2Q

therefore.. ∫ cos 2Q dQ

i hope this is clear to u and u understood it now!