The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Edexcel C3,C4 June 2013 Thread

Scroll to see replies

Original post by posthumus
What answer did you get ?


-3.5
Original post by otrivine
wow! wait, can i now just use substitution, my integration by parts was the same as urs but do not get then how you or why you considered I?:confused:


You don't have to use I, it just makes it easier.

You let I = ∫e3xcos2x dx\displaystyle\int e^{3x} cos 2x \ dx

You have:

∫e3xcos2x dx=13e3xcos⁡2x+29e3xsin⁡2x−49∫e3xcos⁡2xdx\displaystyle\int e^{3x} cos 2x \ dx = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x - \dfrac{4}{9} \displaystyle\int e^{3x} \cos 2x dx

If you notice, you have the same function to be integrated on both sides. So you can replace them both with "I":

I=13e3xcos⁡2x+29e3xsin⁡2x−49II = \dfrac{1}{3}e^{3x} \cos 2x + \dfrac{2}{9}e^{3x} \sin 2x -\dfrac{4}{9}I

Then, because I = ∫e3xcos2x dx\displaystyle\int e^{3x} cos 2x \ dx (i.e. the question) you just make "I" the subject of the formula. :smile:
Original post by otrivine
they got 12x/y but should there not be a y with 12x?


If it's 12x/y, then it's 12x/y... the question is still doable.
Original post by posthumus
Its quite unlikely to come up if you ask me - I don't think it has ever before... where did you get the question from ?


It's never come up, no, and a question like that probably won't to be honest. I got it from my maths teacher who got it from a really old (like 10-15 years old) A-Level maths textbook. :tongue:
Original post by usycool1
It's never come up, no, and a question like that probably won't to be honest. I got it from my maths teacher who got it from a really old (like 10-15 years old) A-Level maths textbook. :tongue:


I just realised that the textbook I have has a whole section on integrals of the form eaxcos(bx)e^{ax}cos(bx) and eaxsin(bx)e^{ax}sin(bx) :tongue:.
Original post by brittanna
I just realised that the textbook I have has a whole section on integrals of the form eaxcos(bx)e^{ax}cos(bx) and eaxsin(bx)e^{ax}sin(bx) :tongue:.


I see! I don't have any questions like that in my textbook :frown:
Reply 346
Avatar for samfreak
samfreak
someone please help me? maths.jpgmaths.jpg
Reply 347
Avatar for QwertyG
QwertyG
9
Original post by samfreak
someone please help me? maths.jpgmaths.jpg


Nah solutionbank is right

Original post by brittanna
I just realised that the textbook I have has a whole section on integrals of the form eaxcos(bx)e^{ax}cos(bx) and eaxsin(bx)e^{ax}sin(bx) :tongue:.


which book you using i dont see it in the green book?
Original post by samfreak
someone please help me? maths.jpgmaths.jpg


No.

Notice that the limits of the integral changed.

∫abu dx=−∫bau dx\displaystyle \int_a^b u \mathrm{ \ } dx = -\int_b^a u \mathrm{ \ } dx
Original post by justinawe
No.

Notice that the limits of the integral changed.

∫abu dx=−∫bau dx\displaystyle \int_a^b u \mathrm{ \ } dx = -\int_b^a u \mathrm{ \ } dx


Hi, in connecting rates, what is r/h in triangle? is this a formula for finiding what ? thanks
Reply 350
Avatar for samfreak
samfreak
Original post by justinawe
No.

Notice that the limits of the integral changed.

∫abu dx=−∫bau dx\displaystyle \int_a^b u \mathrm{ \ } dx = -\int_b^a u \mathrm{ \ } dx


where have they chnaged? :colondollar:
Original post by QwertyG
Nah solutionbank is right



which book you using i dont see it in the green book?


I have an older one I think. It says:
Advanced Maths A2 core for edexcel.

It's this one here:

a2-core-mathematics-for-edexcel-2076-p.jpg
Original post by samfreak
where have they chnaged? :colondollar:


On the second line (the line beginning with A_1=) it was ∫pi40ydx\int^0_\frac{pi}{4} y dx whereas on the line after it was ∫0pi4ydx\int^\frac{pi}{4}_0 y dx
Original post by brittanna
On the second line (the line beginning with A_1=) it was ∫pi40ydx\int^0_\frac{pi}{4} y dx whereas on the line after it was ∫0pi4ydx\int^\frac{pi}{4}_0 y dx


Hi :smile: In cone why do we use r/h ? and what is it in triangle
Original post by otrivine
Hi :smile: In cone why do we use r/h ? and what is it in triangle


I don't think I know what you mean. Have you get an example of a question that uses it?
*cries*
had to do a C4 paper as a test in class, i knew what paper it was gona be and got 51% :colondollar:

ha hahahalgnsrlkgnsrin :frown:
Original post by brittanna
I don't think I know what you mean. Have you get an example of a question that uses it?


http://www.examsolutions.net/maths-revision/core-maths/differentiation/methods/chain-rule/applications/connected-rates-of-change/tutorial-3.php

this tutor uses that r/h formula? thanks
Original post by otrivine


It's just the way similar shapes are defined (from GCSE). So in this case, the ratio of the height to the radius is the same in each of the triangles he looked at. I guess triangles would be the same as it was a triangle he was looking at in the cone question.
Original post by brittanna
It's just the way similar shapes are defined (from GCSE). So in this case, the ratio of the height to the radius is the same in each of the triangles he looked at. I guess triangles would be the same as it was a triangle he was looking at in the cone question.


oh so what does r/h calculate ?
Original post by otrivine
oh so what does r/h calculate ?


It's just the ratio of of the length and height of the triangle, it doesn't represent any physical quantity. But because the two triangles are mathematically similar, it gives us an extra equation we can use.

Do you remember doing similar triangles at GCSE?

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you