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The Edexcel C4 (21/06/12 - PM) Revision Thread

Reply 1580

handsome7654

Original post by Kenan and Kel

Integration > Differentiation and Vectors

Vectors >^100 all other topics :frown:

Reply 1581

King_Arthur

6

SINCE SO MANY PEOPLE ARE HAVING PROBLEMS WITH VECTORS I'L MAKE A SUMMARY OF THE POSSIBILITIES AND APPROACHES

cause in reality it's a very easy topic just need to know the theory properly and the instances when to apply it .....

Reply 1582

tornzy

PLEASE PAY ATTENTION:

Integrating by parts is generally easy if you know what you're doing, however I'm worried that this sort of question may come up, which would throw people off:

∫ e^(2x)sin(3x) dx

This integral is very tricky, as you have to use integration by parts twice and do a little trick. Let me explain how to do it:

let u = e^2x, dv = sin(3x) dx
du = 2e^2x dx, v = (-1/3)cos(3x)

uv - ∫ vdu
= (-(e^(2x)/3)cos(3x) + 2/3 ∫ (e^2x)(cos(3x) dx

let u = e^2x, dv = cos(3x) dx
du = 2e^2x dx, v = (1/3)sin(3x)

= (-(e^(2x)/3)cos(3x) + (2/3) [((e^2x)/3)sin(3x) + 2/3 ∫ (e^2x) sin(3x) dx]
= (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

So we now have:
∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

Now subtract 4/9 ∫ (e^2x) sin(3x) dx from both sides:
5/9 ∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x)

Multiply both sides by 9/5:
∫ (e^2x) sin(3x) dx = (9/5)(-(e^(2x)/3)cos(3x) + (18/15)((e^2x)/3)sin(3x)

= (-3/5)(e^(2x)(cos(3x)) + (2/5)(e^2x)(sin(3x)) + C

PLEASE PASS THIS SOLUTION ON TO AS MANY OF YOUR FRIENDS AS YOU CAN - JUST EMAIL ME AT [email protected]. I WILL BE HAPPY TO HELP PEOPLE WITH OTHER ASPECTS OF C4 WHICH THEY MAY BE STRUGGLING WITH.

Reply 1583

King_Arthur

6

Original post by tornzy

PLEASE PAY ATTENTION:

Integrating by parts is generally easy if you know what you're doing, however I'm worried that this sort of question may come up, which would throw people off:

∫ e^(2x)sin(3x) dx

This integral is very tricky, as you have to use integration by parts twice and do a little trick. Let me explain how to do it:

let u = e^2x, dv = sin(3x) dx
du = 2e^2x dx, v = (-1/3)cos(3x)

uv - ∫ vdu
= (-(e^(2x)/3)cos(3x) + 2/3 ∫ (e^2x)(cos(3x) dx

let u = e^2x, dv = cos(3x) dx
du = 2e^2x dx, v = (1/3)sin(3x)

= (-(e^(2x)/3)cos(3x) + (2/3) [((e^2x)/3)sin(3x) + 2/3 ∫ (e^2x) sin(3x) dx]
= (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

So we now have:
∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

Now subtract 4/9 ∫ (e^2x) sin(3x) dx from both sides:
5/9 ∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x)

Multiply both sides by 9/5:
∫ (e^2x) sin(3x) dx = (9/5)(-(e^(2x)/3)cos(3x) + (18/15)((e^2x)/3)sin(3x)

= (-3/5)(e^(2x)(cos(3x)) + (2/5)(e^2x)(sin(3x)) + C

PLEASE PASS THIS SOLUTION ON TO AS MANY OF YOUR FRIENDS AS YOU CAN - JUST EMAIL ME AT [email protected]. I WILL BE HAPPY TO HELP PEOPLE WITH OTHER ASPECTS OF C4 WHICH THEY MAY BE STRUGGLING WITH.

A really good question .... :smile:

Reply 1584

raheem94

13

Original post by tornzy

...

Original post by King_Arthur

A really good question .... :smile:

I agree its a good question, but i believe it is way beyond C4.

If such a question comes in C4 then they will probably guide through the question. It is more suited to AEA if given without any hints.

Reply 1585

Jack_Smith

8

Original post by tornzy

PLEASE PAY ATTENTION:

Integrating by parts is generally easy if you know what you're doing, however I'm worried that this sort of question may come up, which would throw people off:

∫ e^(2x)sin(3x) dx

This integral is very tricky, as you have to use integration by parts twice and do a little trick. Let me explain how to do it:

let u = e^2x, dv = sin(3x) dx
du = 2e^2x dx, v = (-1/3)cos(3x)

uv - ∫ vdu
= (-(e^(2x)/3)cos(3x) + 2/3 ∫ (e^2x)(cos(3x) dx

let u = e^2x, dv = cos(3x) dx
du = 2e^2x dx, v = (1/3)sin(3x)

= (-(e^(2x)/3)cos(3x) + (2/3) [((e^2x)/3)sin(3x) + 2/3 ∫ (e^2x) sin(3x) dx]
= (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

So we now have:
∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

Now subtract 4/9 ∫ (e^2x) sin(3x) dx from both sides:
5/9 ∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x)

Multiply both sides by 9/5:
∫ (e^2x) sin(3x) dx = (9/5)(-(e^(2x)/3)cos(3x) + (18/15)((e^2x)/3)sin(3x)

= (-3/5)(e^(2x)(cos(3x)) + (2/5)(e^2x)(sin(3x)) + C

PLEASE PASS THIS SOLUTION ON TO AS MANY OF YOUR FRIENDS AS YOU CAN - JUST EMAIL ME AT [email protected]. I WILL BE HAPPY TO HELP PEOPLE WITH OTHER ASPECTS OF C4 WHICH THEY MAY BE STRUGGLING WITH.

wooah that is a really long question :/

Reply 1586

john_bishop1

6

Original post by tornzy

PLEASE PAY ATTENTION:

Integrating by parts is generally easy if you know what you're doing, however I'm worried that this sort of question may come up, which would throw people off:

∫ e^(2x)sin(3x) dx

This integral is very tricky, as you have to use integration by parts twice and do a little trick. Let me explain how to do it:

let u = e^2x, dv = sin(3x) dx
du = 2e^2x dx, v = (-1/3)cos(3x)

uv - ∫ vdu
= (-(e^(2x)/3)cos(3x) + 2/3 ∫ (e^2x)(cos(3x) dx

let u = e^2x, dv = cos(3x) dx
du = 2e^2x dx, v = (1/3)sin(3x)

= (-(e^(2x)/3)cos(3x) + (2/3) [((e^2x)/3)sin(3x) + 2/3 ∫ (e^2x) sin(3x) dx]
= (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

So we now have:
∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x) + 4/9 ∫ (e^2x) sin(3x) dx

Now subtract 4/9 ∫ (e^2x) sin(3x) dx from both sides:
5/9 ∫ (e^2x) sin(3x) dx = (-(e^(2x)/3)cos(3x) + (2/3)((e^2x)/3)sin(3x)

Multiply both sides by 9/5:
∫ (e^2x) sin(3x) dx = (9/5)(-(e^(2x)/3)cos(3x) + (18/15)((e^2x)/3)sin(3x)

= (-3/5)(e^(2x)(cos(3x)) + (2/5)(e^2x)(sin(3x)) + C

PLEASE PASS THIS SOLUTION ON TO AS MANY OF YOUR FRIENDS AS YOU CAN - JUST EMAIL ME AT [email protected]. I WILL BE HAPPY TO HELP PEOPLE WITH OTHER ASPECTS OF C4 WHICH THEY MAY BE STRUGGLING WITH.

Wow that's a very long process. Is it possible for something like this to come up?!

Reply 1587

Yosemite80

yeah it might, i remember my maths teacher saying something about this when we did it as an example in class

Reply 1588

exe

Original post by raheem94

I agree its a good question, but i believe it is way beyond C4.

If such a question comes in C4 then they will probably guide through the question. It is more suited to AEA if given without any hints.

Not to mention he did it incorrectly...

Reply 1589

otrivine

14

guys when you have x squaed = u-1/2
can i do u power 1/2 -1/ 2power 1/2 or is that wrong

Reply 1590

Beresford George

Original post by exe

Not to mention he did it incorrectly...

I agree he missed a minus sign, which mucked it all up.

Should the answer not be.

∫ (e^2x) sin(3x) dx = ((e^2x)/13)[2sin(3x) - 3cos(3x)] + c

(edited 11 years ago)

Reply 1591

exe

Original post by Beresford George

I agree he missed a minus sign, which mucked it all up.

Should the answer not be.

∫ (e^2x) sin(3x) dx = ((e^2x)/13)[2sin(3x) - 3cos(3x)]

yep

but like raheem said, highly unlikely to come up

Reply 1592

otrivine

14

Original post by Beresford George

I agree he missed a minus sign, which mucked it all up.

Should the answer not be.

∫ (e^2x) sin(3x) dx = ((e^2x)/13)[2sin(3x) - 3cos(3x)]

guys when you have x squaed = u-1/2
can i do u power 1/2 -1/ 2power 1/2 or is that wrong

Reply 1593

shahruk

can anyone look at question 4 for from the examination style paper at the back of the book for me ?

0<t<90

why doesn't this correspond to a<x<b

as in when t=0 why isnt x=a ?

I substituted values for t and got x=2 and 0

I understand 2<x<0 is not possible and therefore i switch it around 0<x<2. But is there any other way I can arrive at this conclusion ?

Reply 1594

Beresford George

Original post by otrivine

guys when you have x squaed = u-1/2
can i do u power 1/2 -1/ 2power 1/2 or is that wrong

So..

x^2 = u - 1/2

x = (u - 1/2)^1/2

You have to expand it properly, in order to do it your way I think.

Reply 1595

exe

Original post by shahruk

can anyone look at question 4 for from the examination style paper at the back of the book for me ?

0<t<90

why doesn't this correspond to a<x<b

as in when t=0 why isnt x=a ?

I substituted values for t and got x=2 and 0

I understand 2<x<0 is not possible and therefore i switch it around 0<x<2. But is there any other way I can arrive at this conclusion ?

it is 2<x<0 unless u times the cartesian eqn by -1

Reply 1596

otrivine

14

Original post by Beresford George

So..

x^2 = u - 1/2

x = (u - 1/2)^1/2

You have to expand it properly, in order to do it your way I think.

from that equation u gave me do i have to expand but how thanks?

Reply 1597

Beresford George

Original post by shahruk

can anyone look at question 4 for from the examination style paper at the back of the book for me ?

0<t<90

why doesn't this correspond to a<x<b

as in when t=0 why isnt x=a ?

I substituted values for t and got x=2 and 0

I understand 2<x<0 is not possible and therefore i switch it around 0<x<2. But is there any other way I can arrive at this conclusion ?

x = 2(1 - sint^2)

y/2 = sint

x = 2 - 2(y/2)^2

x = 2 - 2(y^2 / 4)

2x = 4 - y^2

y^2 = 4 - 2x

y = (4 - 2x)^1/2

can't square root a negative number therefore:

x < 2 (less than or equal to)

also from 0 < t < pi/2

when t = pi/2, x = 0

therefore the domain = 0 < x < 2 (thats equal to as well for both sides)