well?

here is my thinking

Ok, so let's start with the easy part.
x+x+x=30 <=> x=10
30+y+y=20 <=> y=5
5+2*z+2*z=9 <=> 4*z=4 <=> z=1

And the fomula is:
Which means, with the values from above, it spells out:
The integral of (10*sin(a))/(2*a) from 0 to infinity.
So here we go:
Let's simplify the inner part of this a bit.
(10*sin(a))/(2*a) <=> (5*sin(a))/a

which means our new formula is:
the integral (5*sin(a))/a from 0 to infinity <=>
5 * the integral (sin(a))/a from 0 to infinity <=>
5* PI/2 ≈ 7.85 If I've done everything properly.

>>8676765
21

>>8676804

no.

>>8676765
beer=9.999999...

does this change anything?

[math]\int_0^\infty \frac{10 sin(x)}{2x} dx[/math]

[math]5 \int_0^\infty \frac{sin(x)}{x} dx[/math]

do you see the dirichlet integral?

[math]5 \frac{pi}{2}[/math]

[math]2.5 \pi[/math]

>>8676842

9.9999999999 etc = 10

>>8676870
5* Pi/2 is the answer

[math]\int_{}^{} \frac_{sin x}{x}[/math] isn't an elementary function. Therefore, the problem is impossible to solve

>>8677260

>says problem is impossible
>doesnt even do math right

>>8677260
dirichlets integral

>>8677260
[math] \int e^{-x^2} dx [/math]
is nonelementary as well, but
[math] \int_0^\infty e^{-x^2} dx [/math]
is known exactly.

>>8676870
What's the primitive of sin(x)/x ?
That's where I got stuck

>>8677487

are you from the other thread?

>>8677461
So what is it then, Mr. Smartass?

>>8676765
you fucktards you cant take the integral of sinx/x.
well you can but its theoretical

Fucking brainlets. ∫sin(t)/t dt = Si(t) in non-elementary functions.

so the exact answer is:
-10pi+20Si(30)

approximate:
-0.0808

>>8677739
Holy shit i guess i'm just as retarded at you guys.

I thought the lower boundary was burger+burger+bottle.

exact answer is:
-10pi

approximate:
-31.416

>>8676770

this is the only answer

other posts are brainlets

t. CERN scientist

>>8677760
^is correct if the beer mugs are treated as one variable

>>8677763

I think OP trolled us all into solving his homework.

>>8677821

nah i made the pic

>tfw planetoidbrain

>>8677570
I forget, fucking google it brainlet. it's that one trick where you square it and evaluate it as a double integral after transforming to polar coordinates.

>>8677980

>brainlet
>brainlet
>planetoidbrain
>brainlet

why is /sci/ so weird?

>>8677487
>>8677260
>>8677668
[math]\int_0^\infty \frac{\sin(u)}{u} du[/math]

Write [math]C_R: z = Re^{ia} \Rightarrow dz = iRe^{ia}da ,\;\; 0 \leq a \leq \pi[/math]
Then [math] C_r: z = \frac{1}{R} e^{ia} \Rightarrow dz = \frac{1}{R} i e^{ia}da ,\;\; -\pi \leq a \leq 0[/math]
Write [math] R_1: -R \leq x \leq -\frac{1}{R}[/math]
And [math] R_2: \frac{1}{R} \leq x \leq R[/math]

Consider the integral over the complex plane of the closed region [math]C = C_R \cup C_r \cup R_1 \cup R_2[/math]: [math]\oint_C \frac{e^{iz}}{z} dz[/math], which equals 0 by cauchy's theorem. Expanding the integral:

[math]\displaystyle 0 = \oint_C \frac{e^{iz}}{z} dz = \int_{C_R} \frac{e^{iRe^{ia}}}{Re^{ia}} iRe^{ia}da + \int_{C_r} \frac{e^{(1/R) ie^{ia}}}{(1/R) e^{ia}} i \frac{1}{R} e^{ia} da + \int_{R_1} \frac{e^{ix}}{x} dx + \int_{R_2} \frac{e^{ix}}{x} dx = i \int_{C_R} e^{iRe^{ia}} da + i \int_{C_r} e^{1/R i e^{ia}} da + \int_{R_1} \frac{e^{ix}}{x} dx + \int_{R_2} \frac{e^{ix}}{x} dx[/math]
[math]\displaystyle = i \int_0^\pi e^{iRe^{ia}} da + i \int_{-\pi}^0 e^{(1/R) i e^{ia}} da + \int_{-R}^{-1/R} \frac{e^{ix}}{x} dx + \int_{1/R}^R \frac{e^{ix}}{x} dx[/math]

>>8678339
Let's look at the first integral:
[math]i \int_0^\pi e^{iRe^{ia}} da = i \int_0^\pi e^{iR \cos(a)} e^{-R\sin(a)} da[/math], as [math]R \rightarrow \infty,\, e^{-R\sin(a)} \rightarrow 0[/math], so [math]i \int_0^\pi e^{iR \cos(a)} e^{-R\sin(a)} da \rightarrow 0[/math]

Now for the second:
[math]i \int_{-\pi}^0 e^{(1/R) i e^{ia}} da = -i \int_0^\pi e^{(1/R) i e^{ia}} da[/math], which becomes [math]- i \int_0^\pi 1 da = - i \pi[/math] as [math]R \rightarrow \infty[/math]

Taking the limit [math]R \rightarrow \infty[/math]
[math]0 = 0 -i\pi + \int_{-\infty}^0 \frac{e^{ix}}{x} dx + \int_0^\infty \frac{e^{ix}}{x} dx = \int_{-\infty}^\infty \frac{e^{ix}}{x} dx -i\pi \Rightarrow \int_{-\infty}^\infty \frac{e^{ix}}{x} dx = i\pi[/math]
[math]\Rightarrow \int_{-\infty}^\infty \frac{\cos(x)}{x}dx + i\int_{-\infty}^\infty \frac{\sin(x)}{x}dx = i\pi \Rightarrow \int_{-\infty}^\infty \frac{\sin(x)}{x}dx = \pi[/math]

By symmetry: [math]\int_0^\infty \frac{\sin(x)}{x}dx = \frac{\pi}{2}[/math]

>>8678027
We're all insecure about our intelligence

>>8678339
>>8678341
These both just look like autistic ramblings, I think you're full of shit

>>8678339
>>8678341
complex analysis is so beautiful
thanks friendo

File: Untitled.png (16KB, 800x600px) Image search: [Google]

16KB, 800x600px

>>8678346
It's just countour integration over this region.

https://en.wikipedia.org/wiki/Methods_of_contour_integration

>>8678339
>>8678341

nice psuedo rambling bullshit

using alot of symbols to explain intro level shit, hope you got a kick of of it

File: IMG_9921.jpg (2MB, 3024x4032px) Image search: [Google]

2MB, 3024x4032px

>>8678423
Come on guys, that’s too complicated. Given
[eqn]\int\ \frac{\sin x}{x}\ dx=\int\left (\int e^{-xy}\sin(x)\ dy\right )dx=\int \int e^{-xy}\sin(x)\ dx\ dy=\int \left (\frac{-ye^{-xy}\sin(x)\ -\ e^{-xy}\cos(x)}{1+y^{2}} \right )dy=\int \frac{1}{1+y^{2}}\ dy=\arctan y+C[/eqn]then [eqn]\int_{0}^{\infty}\frac{5\sin x}{x}\ dx=5\lim_{n\rightarrow \infty}\int_{0}^{n}\frac{\sin x}{x}\ dx=5\lim_{n\rightarrow \infty}\arctan y\ |_{0}^{n}=5\lim_{n\rightarrow \infty}\arctan n\ -\ \arctan 0=5\left [ \left ( \frac{\pi}{2} \right )\ -\ \left ( 0 \right ) \right ]=\frac{5\pi}{2} [/eqn] Now THAT, imho, is as simple as intro level shit gets. I got a picture for you too. I cleaned it up while typing.

>>8678509

wtf did you just spend your whole night working on an autism picture math problem?

hahh ahahahahah

>>8678521
Yes lolololol now do you think you could help me with this? >>8678534

All these plebs falling for the elaborate homework thread

>>8678597

real, non memed, 100%, literal, in the flesh - Autism

AND STILL WRONG

>>8678521
It's almost like people browsing /sci/ would be interested in solving Science and Math problems. Go figure.

you are all clearly wrong
>absence of operator between two variables indicates addition
it's clearly
burger + 2(beer)^2=9
hence beer=sqrt(2)
brainlets

>>8678638

thank you for you informative post. it was well received and im a better person after reading it.

>>8678652
True.

However it doesn't actually change the final answer of the problem, since z^2 = 2 if you consider that and with the other assumption, 2z = 2, if you consider z = 1 from assuming it's 2z+2z.

>>8676871
dude no

>>8676804
you stupid

>>8679082
lrn2convergence fgt pls

>>8676765
The answer is nasty hangover and diarrhoea

>>8679103
>2017
>trying to heaven a thread
Ha!

>>8676765
Bottle * infinity / sgn(beer)

File: fmathe1.png (104KB, 478x897px) Image search: [Google]

104KB, 478x897px

>>8676765

File: fmathe2.png (60KB, 466x889px) Image search: [Google]

60KB, 466x889px

>>8680341

File: fmathe3.png (53KB, 628x883px) Image search: [Google]

53KB, 628x883px

>>8680344

>>8678423
>question is a parody of those Facebook math question images
>uses food and beverages as variables
An unnecessarily complex answer is the correct way to respond to this question. The only way to improve the answer or make it any more appropriate would be to apply even more difficult or excessive methods.

>>8678652
>beer=sqrt(2)
So beer is a completely artificial construct, devised by mathematicians who reject constructivism (another way of spelling "common sense") and whose only purpose is pacifying the masses who can't, won't and shan't learn rational trigonometry and the concept of quadrance? That seems about right.

>>8679103
>convergence
more like convirgin
that is you

>>8676765
Do we need to answer in fast food notation?

File: 1469987411089.jpg (249KB, 2000x1088px) Image search: [Google]

249KB, 2000x1088px

>trying to integrate sine of hotdog