once again (1 Viewer)

[crammy90]

appears i have the last 2 threads aha i feel needy :O
anwho aha
when we are doing volume
V = pi(integral) y^2
Just say the equation is y=(x^2 - 4)
and then we do the volume
V=pi(integral) (x^2 + 4)^2
do we have to expand this before integrating
i.e. x^4 + 8x + 16 ?? and then wed integrate
or can we do an integral on the equation not in separate terms. i.e.
(x^2 + 4)^2
= [(x^2 + 4)^3]/3
yeh just a bit confused :S


Area sector = 8(theta-sintheta)
ii) using the expression, discuss the limiting value of (sine theta)/ theta as theta gets closer and closer to zero :S
i know its one of the small angles but i dont know how to discuss it :S
and limiting values, do they have anything to do with limiting sums lol

 

[syriangabsta]

for the volume question, personally i'd expand it. i find it much easier to work with, and that way i dont do any stupid mistakes.

But im pretty sure the other way is possible too...

but theres a certian method of integrating it..like..

(x^2 -4)^2 = [(x^2 -4)^3]/3

but if it was

(2x^2 -4)^2 ...it would become (i think) ..


[(2x^2 -4)^3]/6

:S

 

[dolbinau]

You don't have to expand if you can integrate it correctly,

But this (x^2 + 4)^2
= [(x^2 + 4)^3]/3

Is not correct. I'm having a mind blank but I don't think it you can actually integrate this without expanding

(ax+b)^K = (ax+b)^k+1/(a*(K+1)

So f'(x) of x. But I don't know if it works with x^2, but I'm sure your integration is wrong so just expand it's easier .

regarding second. When theta approaches zero area approaches 0 so I think the answer is just 0. I sub values for this, I don't know if it is sufficient.

 

[crammy90]

(2x^2 -4)^2 ...it would become (i think) ..


[(2x^2 -4)^3]/6

:S

yeh im pretty sure because its in the form:
[(a^n+1)/n+1*a] i think its times a or maybe even the derivative of the bracket :S
EDIT: just saw dolbinau's post after posting

 

[crammy90]

But this (x^2 + 4)^2
= [(x^2 + 4)^3]/3

Is not correct.
(ax+b)^K = (ax+b)^k+1/(a*(K+1)

[(x^2 + 4)^3]/3*a which is 1 in this case yeh? so its just 3 i think

 

[dolbinau]

The form of that case is when x is to the power of 1 only, I think. Besides you can test it by expanding/integrating sub x=1 I'm sure you'll find they are different answers.

 

[crammy90]

The form of that case is when x is to the power of 1 only, I think. Besides you can test it by expanding/integrating sub x=1 I'm sure you'll find they are different answers.

so basically if its a sum of squares or whatever
i.e. (x-3)^2 then we could do ^3 all over 3
but if its (x^2 -3) then we have to expand and what not yeh?

 

[dolbinau]

so basically if its a sum of squares or whatever
i.e. (x-3)^2 then we could do ^3 all over 3
but if its (x^2 -3) then we have to expand and what not yeh?

I thought it was function of function or something.

But Yes. That's my understanding.


Can someone check their text book though for this integral to make sure? I have maths in focus which is crap and doesn't have this.