• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Messages
304
Reaction score
193
Points
53
let u^2 =x
that means dx/du =2u that means differentiating u^2 woth respect to u
dx= 2u du
next thing is to change limits...
if u^2 =x
when upper limit of x=p^2 then replace x with p^2
u^2 = p^2
new upper limit =p
and since o will remain o
the new limits are p and o

then u have to integrate cos of root of u^2
root will cut the sqr u will get
integral of cos u dx
now look at the third line of this solution u will find dx= 2u du
integral of cosu 2u du for the limit p and 0
integrating by parts
u= 2u dv= cos u
u'=2 v= sinu
2u sin u- integral of 2sin u
2usinu -( -2cosu)
2usinu + 2cos u... apply limits p and 0
2p sinp + 2cosp - ( o +2 cos0)
2psin p +2cosp -2(1)=1 as this is also equal to the area of the shaded region...
2p sinp= 3-2cosp
sinp= 3-2cosp / 2p
Thanks A lot!!!
I wasn't integrating it by parts :/
Anyways,thanks again :)
 
Messages
4,493
Reaction score
15,418
Points
523
Part 5 ii again Rutzaba (sorry for tagging, but due to quick response I am really really really glad)
Yess More prayers coming!View attachment 42627
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_31.pdf

Can anyone explain to me wat the ms is saying for Q10 part iv??! >.<
these cie people cud try making a marking scheme when they are not drunk for a change >_<

this doesnt make any sense

I got the answer 0.906 but all these other values given in the ms seem to b mumbo jumbo to me
hogaye ye wale or shall i do em?
 
Messages
147
Reaction score
743
Points
103
how the gradient of AC is -1/2 plz help :(upload_2014-5-16_17-43-47.png
 
Top