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(x^2-4x+4)(x^2-4x+4), i think the question should be why you took the square of (x^2-4x+4)This might be a senseless question but how did you take square of (x^2-4x+4)?
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(x^2-4x+4)(x^2-4x+4), i think the question should be why you took the square of (x^2-4x+4)This might be a senseless question but how did you take square of (x^2-4x+4)?
Wrote it twice and multiplied... There must be something that we are not applying here...This might be a senseless question but how did you take square of (x^2-4x+4)?
To get volume we have to square the given equation... If I take it, shouldn't we end up with the same answer?(x^2-4x+4)(x^2-4x+4), i think the question should be why you took the square of (x^2-4x+4)
the answer comes the same this way too, i.e 32π/5Wrote it twice and multiplied... There must be something that we are not applying here...
Oh ye. Well, intregating (x-2)^4 gives me the same result as mubarka got.To get volume we have to square the given equation... If I take it, shouldn't we end up with the same answer?
Assalamoalaikum wr wb!9709_s11_qp_11.pdf
9709_s11_ms_11.pdf
Q3 ii) I do get the answer by just squaring the equation (x-2)^4 and integrating it...
But what if I open the equation y= (x-2)^2, take its square (x^2-4x+4)^2 and now integrate it? I don't end up getting the same answer. WHY?
If I further open it (x^2-4x+4)^2 and then integrate it I do get the answer... But why not without opening it?
the answer comes the same this way too, i.e 32π/5
writing it twice n multiplying gives
x^4 - 8x^3 + 24x^2 - 32x + 16
and integrating it we get, [(1x^5)/5 - 2x^4 +8x^3 - 16x^2 + 16x] and the limits are 2 and 0, putting these in the integrated form,
we get (32/5-32 +64 -64 +32)π
32π/5
Oh ye. Well, intregating (x-2)^4 gives me the same result as mubarka got.
Assalamoalaikum wr wb!
u can integrate the equation directly if it's in the form (ax + b)^n and NOT otherwise unlike differentiation!
so u don't need to open the brackets!
integrate (x-2)^4 directly
=[(x-2)^5 ]/(5 x 1 )
and simplify...
Use the double angle formula in (i) and (ii)http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s10_qp_11.pdf
q1,
asalam alikum, help me please
and what are asending powers of x?
thanks alot
marking scheme
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s10_ms_11.pdf
That is because there is no formula applicable for integration of (ax^2 + bx +c)^n9709_s11_qp_11.pdf
9709_s11_ms_11.pdf
Q3 ii) I do get the answer by just squaring the equation (x-2)^4 and integrating it...
But what if I open the equation y= (x-2)^2, take its square (x^2-4x+4)^2 and now integrate it? I don't end up getting the same answer. WHY?
If I further open it (x^2-4x+4)^2 and then integrate it I do get the answer... But why not without opening the second equation?
Thank you!That is because there is no formula applicable for integration of (ax^2 + bx +c)^n
U r welcome!Thank you!
yup...that's what i said UNLIKE differentiationYes. But when you open the first equation... Square it and then integrate this (x^2-4x+4)^2 without opening... The answer isn't same... Where as it should come that way too?
First list the partial fractions:I need help with part (ii), I don't understand how he got the differentaition to
"0.1 ln(a)-0.1 ln(10 -a) " where did the negative sign come from?
I attached the question and the markscheme below
Assalamu alaikum,
November 2003, Paper 3, Q.4, 6 (2) , 7 (3) , 8 and 10.
Assalamu alaikum,
November 2003, Paper 3, Q.4, 6 (2) , 7 (3) , 8 and 10.
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