Author Topic: Couple Q's (Read 1943 times) Tweet Share

sxcalexc · « **on:** July 04, 2008, 10:10:10 pm »

Hey guys I know how to get the rule and I think I understand the basics of finding out the domain, however I don't understand their answer completely. I get how they arrived that the composition has to exclude -5/2. (Because 0 doesn't work in f, and if you make g = to 0, you get -5/2). I don't get why the answer is from negative infinity to 0. Don't you take the domain of g (as it fits into f). Shouldnt it be from negative infinity to -5/2 but excluding -5/2) ... hope that makes sense lol

With the next question I can get a couple of the answers but i'm not completely sure of it. (By no means am I asking anyone to answer all of the parts; just a little push in the right direction would be nice)

Glockmeister · « **Reply #1 on:** July 04, 2008, 11:21:33 pm »

I'll do the probability one

a. This one you should be able to do, so I won't do it
b. k must equal $\frac{1}{2}$ as a probability distribution function must equal one be definition. Adding a half in the equation at a., we will get the number 1 (this explaination might not be clear, I'm not good with words)
c. Should be easy enough

That b question after c that's just sticking there confused me for a bit.

d. If you differentiated corrected, you should get a derivitive. What you need to do to integrate both sides and you should get an answer that looks like this

$\int xcos(\frac{x}{2}) dx = 2x\sin(\frac{x}{2}) - \frac{x^{2}}{2}$

Mao · « **Reply #2 on:** July 05, 2008, 04:02:58 pm »

question 4a

you are right. it should be $\left( -\infty, -\frac{5}{2}\right)$ , or $R^-\backslash \left[ -\frac{5}{2},0\right)$ , even though the second one is really unconventional.

/0 · « **Reply #3 on:** July 05, 2008, 06:21:25 pm »

waaah you're doing density functions already

sxcalexc · « **Reply #4 on:** July 05, 2008, 09:07:30 pm »

Thx for your replies guys!

Quote from: DivideBy0 on July 05, 2008, 06:21:25 pm

waaah you're doing density functions already

Haha I didn't know these were a separate topic. Our teacher just gave us a few practise exams. I just answered it using the knowledge I already had about anti differentiation; deriving circular functions and probability.

Also I still don't quite get part d, why does it start with Hence? What information are you using? lol thx

Mao · « **Reply #5 on:** July 05, 2008, 09:23:35 pm »

looking at that, if you photoshopped this you've copied part b of the question twice.

c)

$\frac{d}{dx}\left(x\sin\left(\frac{x}{2}\right) +2 \cos\left(\frac{x}{2}\right)\right)=\sin\left(\frac{x}{2}\right) + \frac{x}{2}\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)\right)=\frac{1}{2}x\cos\left(\frac{x}{2}\right)$

hence:

d)

$\int \frac{1}{2}x\cos\left(\frac{x}{2}\right)\; dx = x\sin\left(\frac{x}{2}\right) +2 \cos\left(\frac{x}{2}\right) + C$

$\implies \frac{1}{2} \int x\cos\left(\frac{x}{2}\right)\; dx = x\sin\left(\frac{x}{2}\right) +2 \cos\left(\frac{x}{2}\right) + C$

$\implies \int x\cos\left(\frac{x}{2}\right)\; dx = 2 x\sin\left(\frac{x}{2}\right) +4 \cos\left(\frac{x}{2}\right) + C$

sxcalexc · « **Reply #6 on:** July 05, 2008, 09:42:59 pm »

All good!! I figured it out I think... The answer to part c is half of question d. So the answer to d should be 2 multiplied by the question for c... right? lol

EDIT: Oh you typed it out.. even better

Cheers Mao!!
Oops.. i did put question b twice.
Also Mao, the C at the end wouldn't be necessary because you are antidiffing between pi and 0 right? Like the C's kinda cancel out. And it says evaluate for part d, so you rekon they would want pi to be subbed in?

2nd EDIT: Lol you may be right with the C... i'm subbing in pi and i'm getting 2pi for d. and pi for e. Their answer is 2pi - 4 and pi - 2, respectively. I can't work it out :S

Mao · « **Reply #7 on:** July 07, 2008, 08:40:41 pm »

actually, the C doesn't matter

$\implies \int x\cos\left(\frac{x}{2}\right)\; dx = 2 x\sin\left(\frac{x}{2}\right) +4 \cos\left(\frac{x}{2}\right) + C$

$\implies \int_0^{\pi} x\cos\left(\frac{x}{2}\right)\; dx = \left[2 x\sin\left(\frac{x}{2}\right) +4 \cos\left(\frac{x}{2}\right)\right]_0^{\pi}=(2\pi \cdot 1 + 0)-(0+4) = 2\pi-4$

e)

$E(X)=\int_a^b x\cdot f(x)\; dx = \int_0^{\pi}x\cdot k\cdot \cos\left(\frac{x}{2}\right)\; dx=k\int_0^{\pi} x\cos\left(\frac{x}{2}\right)\; dx=\frac{1}{2}\cdot (2\pi-4) = \pi-2$

sxcalexc · « **Reply #8 on:** July 08, 2008, 12:46:13 pm »

Ah cause we are integrating, of course! <3 Mao

Mao · « **Reply #9 on:** July 08, 2008, 08:03:37 pm »

Quote from: sxcalexc on July 08, 2008, 12:46:13 pm

Ah cause we are integrating, of course! <3 Mao

assuming you referred to the "C"
it's because its a definite integral [there's numbers on top and bottom]

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