Maths H P2 Disaster Prevention.

OTliddy

BraveheartGal wrote:

can anybody help me out with the option question?

2000 paper 2, b (i)
its integration by parts but its the trickier integration by parts

youve to integrate e ^2xCosxdx
i integrated it, then integrated it again but im completely stuck

if someone cld post up the solution id love ya forever cos i reckon that type of int. by parts will come up because it hasnt come up since 2000

thanks!

note: let Y =the integration sign and the first(given) sum be I'
u=cosx dv=e^2x dx
Ydv=_le^2x dx
du = -sinx v=1/2.e^2x
dx

du = -sinxdx

I' = cosx.1/2.e^2x - Y1/2.e^2x.-sinxdx
= cosx.1/2.e^2x + 1/2.Ye^2x.sinxdx
then treat the italisized bit as whole new sum. call it I'':
it works out as sinx.1/2.e^2x - Y1/2e^2x.cosx
italisized bit is I'
so: I' = cosx. 1/2.e^2x +1/2.I''
I' = cosx.1/2e^2x + sinx.1/4.e^2x - 1/4.I'
5/4.I' = cosx.1/2e^2x + sinx.1/4.e^2x...........x across by 4
5.I' = 2cosx.e^2x + sinx.e^2x
I' = e^2x{2cosx + sinx} + c
5

Fortinbras'

question on probability to get you going

what is the prob that three people have birtday on same day of the week?
three different days of the week?

mrs. hamlet

Okay!

|=integral sign thingy!

|e^2xcosxdx

Let u = cosx
Du=-sinxdx

Let dv = e^2x
V=1/2 e^2x

|e^2xcosxdx = 1/2e^2xcosx + |1/2e^2x.-sinxdx
= 1/2e^2xcosx - |1/2e^2xsinxdx

|1/2e^2xsinxdx

Let u = sinx
Du = cosxdx

Dv = 1/2e^2xdx
V = ¼ e^2x

¼ e^2xsinx - |1/4e^2xcosxdx
¼ e^2xsinx – ¼ |e^2xcosxdx

|e^2xcosxdx = ½ e^2xcosx+ ¼ e^2xsinx – ¼ |e^2xcosxdx
5/4 |e^2xcosxdx = ½ e^2xcosx + ¼ e^2xsinx
|e^2xcosxdx = 2/5 e^2xcosx + 1/5 e^2xsinx
= 1/5 (2e^2xcosx + e^2xsinx)

Then sub in the values and u should get 1/5 (e^pie – 2) (n=5)

K hope you can understand my crappy longwinded way of doing it!

Fortinbras'

Fortinbras' wrote:

question on probability to get you going

what is the prob that three people have birtday on same day of the week?
three different days of the week?

bump

Beno

i surprised that alot of people find Q8 difficult, it one of the few q's i actually know...dont like the circle, dont like Vectors, dont like the line, hate trig with a vengence and seem to always get prob wrong.

*Angel*

So you really don't like the rest of the paper then.

I'm doing coord geom the line/the circle, trig, vectors and further calculus.

I can't see why people don't like coord geom, but anyway I don't like probability however I'm quite good at it so 6/7 will be my last questions to do.

Gileadi

hoping a straight foward enough Q 8 comes,if it does i can see myself doing 7 questions as the 6/7 never take more than 25 mins for the 2 of them and some of the trig questions can be very straight forward not gonna bank on doing 7 though,

probably gonna do Q1,2,3,6,7,8 a trig one with the cosine rule proof as a b part would be beautiful

Rredwell

The 12 proofs we need to know aren't actually that bad. They all basically stem from about 3 or 4 different situations. Cos (A-B), and thus cos (A+B), start off with a circle with radius 1 and points p(cos A, sinA) and q(cos B, sin (I think, feel free to correct me); the cos 2A ones start off with cos^2 A + sin^2 A = 1, and so on.

Personally I would really like tan(A+B) to come up. It is such an easy proof!

mrs. hamlet

oh my head is bangin' from crammin' maths all day!
Think i will be fine on the circle and the line! But i'm absolutely no better at trig, it's just not gonna happen!
Q8 is fine as long as we don't get tan inverse for maclaurin series!
Fingers crossed for a hard 20marks integration by parts!!

rainbow kirby

Just to interrupt at this point... Good luck tomorrow!

BraveheartGal

Crumbs wrote:

int(e^2x.cosx)dx

Let u = cosx and dv = e^2x.dx

Should get it out as (cosx.e^2x)/2 + 1/2[int(e^2x.sinx)dx]


Integrate that last bit by parts again to get (sinx.e^2x)/2 - 1/2[int(e^2x.cosx)dx]


Jam that back into the first part, multiply out the brackets and you should get:

int(e^2x.cosx)dx = (cosx.e^2x)/2 + (sinx.e^2x)/4 - 1/4[int(e^2x.cosx)dx]


Rearrange to get:

5/4[int(e^2x.cosx)dx] = (2cosx.e^2x + sinx.e^2x)/4


Some more manipulation:

int(e^2x.cosx)dx = (e^2x/5)(sinx + 2cosx) + c


Hmm, that's messy and I tried to summarise it as short as I possibly could.
Just ask if you need more clarification on any part of it.

E&OE

Where*did*u*get*5/4*from?*tanx*again

int(e^2x.cosx)dx = (cosx.e^2x)/2 + (sinx.e^2x)/4 - 1/4[int(e^2x.cosx)dx]

If you look at the above line you will see that the thing on the left of the equals is similar to the thing that is furthest to the right, so you can bring it over and add them.

Gileadi

doesnt matter anymore ,,wahaaaaay!

JackKelly

woohay. Thats the worst of it over for me.

Was worried about English and Irish (ordinary-lol) and now they are done and dusted. Wasn't worried about Maths until i did p1! Now with a good P2 im happy.Week to study for my last 4 now!

*Angel*

No more maths , wish I could sit maths again instead of damn french.

Agreed.

jacksie

oh holy god. i never want to see maths again. wait i wont. unless i failed. that would be ****ed