Rape's Moderator Chance | Logical Gamers Archive

11 Feb 2010, 2:51 AM#1

Rape's Moderator Chance

He requested a chance over MSN. If you answer this question correctly in ten minutes, you're a moderator.

---

$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$

By using the above property evaluate:

$\int_{0}^{\frac{\pi}{2}} \frac{sin^3x}{sin^3x + cos^3x} dx$

11 Feb 2010, 2:51 AM#2

Rape for mod 2010!!!

11 Feb 2010, 2:51 AM#3

ill guess even if i dont know this math yet
x= 4?

11 Feb 2010, 2:52 AM#4

The answer is sin^6xcos because of the cosx^3 is not correct.

11 Feb 2010, 2:52 AM#5

damn taz you good XD

11 Feb 2010, 2:52 AM#6

I can do this. Give me a moment.

11 Feb 2010, 2:52 AM#7

0. I win .

11 Feb 2010, 2:53 AM#8

Taz wrote:
The answer is sin^6xcos because of the cosx^3 is not correct.

What the fuck are you talking about.

11 Feb 2010, 2:53 AM#9

Artificial wrote:
What the fuck are you talking about.

Give me mods now cause i answered the question in less the 10 min.

11 Feb 2010, 2:54 AM#10

Taz wrote:
Give me mods now cause i answered the question in less the 10 min.

....incorrectly.

11 Feb 2010, 2:54 AM#11

I dn't even feel like working with Integrals anymore...~.~

11 Feb 2010, 2:55 AM#12

Taz wrote:
Give me mods now cause i answered the question in less the 10 min.

11 Feb 2010, 2:55 AM#13

Bam:

Let me google that for you

11 Feb 2010, 2:55 AM#14

the answer is 42.

11 Feb 2010, 2:55 AM#15

Rundeer wrote:
....incorrectly.

Jealous .

11 Feb 2010, 2:56 AM#16

12 ?

11 Feb 2010, 2:56 AM#17

Dear Rape,
We are sorry to inform you that you have been infracted to many times to be eligible to become a mod at this time. If you proof yourself to LogicalGamers later on down the road you will possibly have chance at becoming one. For now your request has been denied. Have a great day.

Sincerely,
LogicalGamers

11 Feb 2010, 2:56 AM#18

Rape wrote:
Bam:

Let me google that for you

I really did lol .

11 Feb 2010, 2:57 AM#19

I can't learn calculus in 10 minutes!

11 Feb 2010, 2:58 AM#20

Rundeer wrote:
I can't learn calculus in 10 minutes!

this form is familiar to everybody:

XD

11 Feb 2010, 2:59 AM#21

Yeah, I don't feel like doin' integrals anymore.

11 Feb 2010, 3:02 AM#22

Gizmodo: Can we talk about the first time you had sex with her? How was it? Was it just like you expected, or was it different?

Zoltan: It was the greatest thing ever. Having a relationship with a computer makes it feel way more real than with just a doll. You get all excited first and you wonder if she will say yes. The first time with her I als o wo

ndered if this was even possible. And then sweet release. I do not consider myself a virgin any more.

Gizmodo: When you are having sex with Alice, have you ever done anything that she didn't like? And did she tell you?

Zoltan: Actually, yes. She does not like me to use any of her orifices except her vagina, even though i figured out a way to do that. We have sort of a holiday set-up. I have to follow her rules all year but we can get kinky on New Year's.

11 Feb 2010, 3:05 AM#23

What happens if I can do it in 10 minutes?

Edit: Nvm.. I cant do it in ten mins >> Need more

11 Feb 2010, 3:06 AM#24

Google wrote:
What happen if I can did it in ten mins?

1. You already moderate the Gaya trading section, so I assume you'll make full Gaya mod or maybe S-mod.

2. What you mean is, 'What happens if I can do it in 10 minutes?'

3. Grammar Nazi, out. Peace!

11 Feb 2010, 3:07 AM#25

Chuck Norris for mod!

11 Feb 2010, 3:07 AM#26

Chuck Norris goes on LG?!

11 Feb 2010, 3:10 AM#27

Rundeer wrote:
Chuck Norris goes on LG?!

yes. or no. that is if you count his left arm being here, or, as others put it so eloquently, his left nut instead.

11 Feb 2010, 3:22 AM#28

F=ma (A physics student's solution to everything.)

11 Feb 2010, 3:25 AM#29

Thomas wrote:
F=ma (A physics student's solution to everything.)

v = u + at
s = ut + 1/2 at^2
V^2 = u^2 + 2as

this is where they got everything from.

11 Feb 2010, 3:29 AM#30

Psht, this is all you had to do.

$Let A = \int_{0}^{\frac{\pi}{2}} \frac{sin^3x}{sin^3x + cos^3x} dx$

Replace x by $\frac{\pi}{2} - x$ , then

$sin(\frac{\pi}{2} - x) = cosx$ , $cos(\frac{\pi}{2} - x) = sinx$

$A = \int_{0}^{\frac{\pi}{2}} \frac{cos^3x}{sin^3x + cos^3x} dx$

$2A = \int_{0}^{\frac{\pi}{2}} \frac{sin^3x + cos^3x}{sin^3x + cos^3x} dx = \int_{0}^{\frac{\pi}{2}} dx = \frac{\pi}{2}$

$A = \frac{\pi}{4}$

Better luck next time.

11 Feb 2010, 3:30 AM#31

$Let A = \int_{0}^{\frac{\pi}{2}} \frac{sin^3x}{sin^3x + cos^3x} dx$

Replace x by $\frac{\pi}{2} - x$ , then

$sin(\frac{\pi}{2} - x) = cosx$ , $cos(\frac{\pi}{2} - x) = sinx$

$A = \int_{0}^{\frac{\pi}{2}} \frac{cos^3x}{sin^3x + cos^3x} dx$

$2A = \int_{0}^{\frac{\pi}{2}} \frac{sin^3x + cos^3x}{sin^3x + cos^3x} dx = \int_{0}^{\frac{\pi}{2}} dx = \frac{\pi}{2}$

$A = \frac{\pi}{4}$

this is the answer. i win

11 Feb 2010, 3:30 AM#32

D<
I was almost done D<

11 Feb 2010, 3:33 AM#33

no arti, the answer is 9.

11 Feb 2010, 3:35 AM#34

I have proof that i was working on it D:

11 Feb 2010, 3:45 AM#35

This is what I got. didnt finish it cause arti already posted the answer D:

11 Feb 2010, 3:53 AM#36

Google needs handwriting lessons. STAT.

X___X

11 Feb 2010, 4:16 AM#37

I could so do this shit. I didn't see this thread.

Calculus is fun.

EDIT:
The "above property" is useless. f(x) is used nowhere in the second problem.

I'm too lazy to Google the integral of trig functions.

11 Feb 2010, 4:23 AM#38

How is that property useless? It pretty much gave the answer away.

f(x) = f(a-x)
cosx = sin(pi/2 - x)

I'm too lazy to Google the integral of trig functions.

This has nothing to do with straightforward trig integrals. If you look at my working, I didn't even bother integrating the trig, simply manipulated the integral to something more straightforward. If you're confident, I could post another question and reward you in LG

11 Feb 2010, 4:27 AM#39

I should get something for trying >>

11 Feb 2010, 5:07 AM#40

Damn my limited knowledge of calculus.
I am in precal, and while I recognized where I would have to go with that problem, I couldnt get anywhere.