Rape's Moderator Chance

**Artificial** · 11 Feb. 2010 02:51am

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 02:51am

Rape for mod 2010!!!

**Sam** · 11 Feb. 2010 02:51am

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

**Taz** · 11 Feb. 2010 02:52am

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

**Sam** · 11 Feb. 2010 02:52am

damn taz you good XD

**Tatsuo** · 11 Feb. 2010 02:52am

I can do this. Give me a moment.

**Enjoi** · 11 Feb. 2010 02:52am

0. I win .

**Artificial** · 11 Feb. 2010 02:53am

Originally Posted by Taz

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

What the fuck are you talking about.

**Taz** · 11 Feb. 2010 02:53am

Originally Posted by Artificial

What the fuck are you talking about.

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

**Rundeer** · 11 Feb. 2010 02:54am

Originally Posted by Taz

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

....incorrectly.

**Tatsuo** · 11 Feb. 2010 02:54am

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

**noob** · 11 Feb. 2010 02:55am

Originally Posted by Taz

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

11 Feb. 2010 02:55am

Bam:

Let me google that for you

11 Feb. 2010 02:55am

the answer is 42.

**Taz** · 11 Feb. 2010 02:55am

Originally Posted by Rundeer

....incorrectly.

Jealous .

**Stapled** · 11 Feb. 2010 02:56am

12 ?

**Maverick** · 11 Feb. 2010 02:56am

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 02:56am

Originally Posted by Rape

Bam:

Let me google that for you

I really did lol .

**Rundeer** · 11 Feb. 2010 02:57am

I can't learn calculus in 10 minutes!

**noob** · 11 Feb. 2010 02:58am

Originally Posted by Rundeer

I can't learn calculus in 10 minutes!

this form is familiar to everybody:

XD

**Tatsuo** · 11 Feb. 2010 02:59am

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

11 Feb. 2010 03:02am

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

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.

**Google** · 11 Feb. 2010 03:05am

What happens if I can do it in 10 minutes?

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

**noob** · 11 Feb. 2010 03:06am

Originally Posted by Google

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!

**C0FFINCASE** · 11 Feb. 2010 03:07am

Chuck Norris for mod!

**Rundeer** · 11 Feb. 2010 03:07am

Chuck Norris goes on LG?!

**noob** · 11 Feb. 2010 03:10am

Originally Posted by Rundeer

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.

**Elliot** · 11 Feb. 2010 03:22am

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

**noob** · 11 Feb. 2010 03:25am

Originally Posted by Thomas

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.

**Artificial** · 11 Feb. 2010 03:29am

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 03:30am

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

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

**Google** · 11 Feb. 2010 03:30am

D<
I was almost done D<

11 Feb. 2010 03:33am

no arti, the answer is 9.

**Google** · 11 Feb. 2010 03:35am

I have proof that i was working on it D:

**Google** · 11 Feb. 2010 03:45am

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

**noob** · 11 Feb. 2010 03:53am

Google needs handwriting lessons. STAT.

X___X

11 Feb. 2010 04:16am

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.

**Artificial** · 11 Feb. 2010 04:23am

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

**Google** · 11 Feb. 2010 04:27am

I should get something for trying >>

**Smithno13** · 11 Feb. 2010 05:07am

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.

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