Edexcel A2 C4 Mathematics June 2016 - Official Thread

oops should be a negative will change other post

Nope. TM's solution is correct:

$\displaystyle V = \pi \int_{0}^{\frac{\pi}{8}}(\tan(2x)-1)^2 \mathrm{d}x$

$\displaystyle V = \pi \int_{0}^{\frac{\pi}{8}}1 - 2\tan(2x) + \tan^2(2x) \mathrm{d}x[br]$

$\displaystyle V = \pi \int_{0}^{\frac{\pi}{8}}\sec^2(2x) - 2\tan(2x) \mathrm{d}x$

$\displaystyle V = \pi \left[\frac{1}{2}\tan(2x) - \ln|\sec(2x)|\right]_{0}^{\frac{\pi}{8}}$

$\displaystyle V = \pi \left[\frac{1}{2} - \ln|\sqrt{2}|\right]$

$\displaystyle V =\frac{\pi}{2} \left[1 - \ln(2)\right]$

My point was that it integrates to (2/3)sin^3

$(1-\tan(2x)) = y$ works for a flip

But $y^2 = (1-\tan(2x))^2 \neq (2-\sec^2(2x))$

$y^2 = (1-\tan(2x))^2 = 1 - 2\tan(2x) + \tan^2(2x)$

Can someone help me with my understanding of integrating to get area? So on the left example if you integrate all together the are under the graph has a negative sign and the area above a positive sign so they cancel to get zero.

So why on the question on the right when I integrate the curve on the right side of the y axis do I obtain the area of the upper right quadrant? Surely the area above should cancel with the area below resulting in zero?

I am so confused

Thanks for the response.
I used:
$[br]V = \pi\int_a^b{(y_1^2-y_2^2)}dx [br]$
For the volume of revolution between two curves or is formula not valid in this case?
I also tried to verify by taking way the said integral from the volume of a cylinder and still ended up with the same answer.

Oh yeah hahaha I can't believe I forgot that, thanks for your help!

Don't normally post here but can someone have a look at C4 IYGB Paper U question 8. I seem to be getting a different answer to the mark scheme:



The mark-scheme says:
$[br]= \dfrac{1}{2}\pi(1-\sqrt{2})units^3[br]$

$\displaystyle V = \pi\int_a^b{(y_1^2-y_2^2)}dx$

This integral should work. I'd always do it separately though.

Not really sure what you're asking? Do you mean, for example, an odd function will cancel if you integrate from $-a$ to $a$? I don't really get what you're saying in your second paragraph?

Integration gives the net signed area! So any function with $y< 0$ will give a negative area and a positive one if $y>0$, so you need to evaluate these separately and apply modulus where necessary!

The shaded area isn't below the x, I think you have to do volume of rev's about the y axis, not sure how you actually do that though

Please can someone explain question 4 of jan 2009???? Thanks in advance

Sorry so the area on the left question under the graph is -A and the area above the graph is A. So when you integrate together you get zero

The are on the left question the area under the graph is -B and above the graph is B so when you integrate why do you obtain the correct answer and not 0? As I integrated and got a non zero value which was correct

Does that help?

can someone explain how your meant to know to put an a in front as didnt think you had to

The shaded area isn't below the x, I think you have to do volume of rev's about the y axis, not sure how you actually do that though

can someone explain how your meant to know to put an a in front as didnt think you had to

Please can someone explain question 4 of jan 2009???? Thanks in advance