Can someone give me the solution

Please provide the step by step solution for the integration. Thank You.

\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ |x-y|(6{ x }^{ 2 }y)dxdy } }

|.| = Absolute value.

#Calculus

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

Is this what you mean? : $\huge \displaystyle \int_{0}^{1} \int_{0}^{1} |x-y|(6x^2 y) dx \ dy$

Mohammad Farhat - 2 years, 8 months ago

Ya. I want the solution for this

Ankush Gogoi - 2 years, 8 months ago

@Aaghaz Mahajan, I have sorted out the $\LaTeX$ . May you help @Ankush Gogoi

Mohammad Farhat - 2 years, 8 months ago

@Mohammad Farhat – Hey!!! Thanks a lot!!! @Ankush Gogoi Well wait then, I'll send the solution shortly....!!

Aaghaz Mahajan - 2 years, 8 months ago

You can get rid of the absolute value by dividing the integration region into two parts: One above the line $y = x$ and one below it:

$\large{\int_0^1 \int_0^1 | x - y | (6 x^2 y) \, dx \, dy = \int_0^x \int_0^1 (x - y) (6 x^2 y) \, dx \, dy + \int_x^1 \int_0^1 (y - x) (6 x^2 y) \, dx \, dy }$

The rest is tedious but trivial

Steven Chase - 2 years, 8 months ago

@Steven Chase Yeah that also works......but still, the integral is quite long and monotonous to evaluate....!!!

Aaghaz Mahajan - 2 years, 8 months ago

Indeed. I would just integrate it numerically

Steven Chase - 2 years, 8 months ago

@Steven Chase – Yeah.but maybe he was practicing multivariable calculus for the first time........then this sort of question works for improving one's understanding......

Aaghaz Mahajan - 2 years, 8 months ago

Thank You @Steven Chase for the approach. I got the answer.

Ankush Gogoi - 2 years, 7 months ago

Would you please post this as a problem in the calculus section?

Steven Chase - 2 years, 7 months ago

@Steven Chase – I'm not good at writing using Latex. But I will try to post it.

Ankush Gogoi - 2 years, 7 months ago

Well, LATEX is not clear........I don't know what are you asking for.......

Aaghaz Mahajan - 2 years, 8 months ago

Sorry. Not good with latex for writing.

Ankush Gogoi - 2 years, 8 months ago

@Ankush Gogoi Ok......if you are familiar with methods of Multiple Integration, then this is fairly easy by a change of variables.........
Firstly, map x and y to (u+v) and (u-v) respectively and then find the Jacobian.......The motivation behind this step was to remove both variables from inside the modulus operator and replace them with one variable...........Now, after change of variables, apply the desired limits and then open the modulus accordingly.......and thats that!!!! You are done............!!!

Aaghaz Mahajan - 2 years, 8 months ago

Can you please provide the solution.

Ankush Gogoi - 2 years, 8 months ago

That is what I have written.....!! What do you need??

Aaghaz Mahajan - 2 years, 8 months ago

@Aaghaz Mahajan – I assume he wants you to write out the whole thing for him?

A Former Brilliant Member - 2 years, 8 months ago

@A Former Brilliant Member – Well.......I really can't and this is the thing I regret........I mean I have ZERO knowledge of programming and LATEX and stuff........and also, I dont have the interest to learn that......so that is why on this site, most of the things that I do are based more on writing....(except solving problems of course!!!).....

Aaghaz Mahajan - 2 years, 8 months ago

@Aaghaz Mahajan – It's not that you CAN'T; you SHOULDN'T. I replied to the previous post directly; let's end it on that note.

A Former Brilliant Member - 2 years, 8 months ago

Have you made a decent attempt at the problem? If so, please show. If not, then you should at least try before asking others on this site to give you worked answers.

A Former Brilliant Member - 2 years, 8 months ago

Sounds like a good plan... Just a little confused as to the limits of integration once you put in $x=u+v$ and $y=u-v$ . Kinda feel like I haven't touched this stuff for a good 6-7 years, but I only remember briefly (like the change of variables & the Jacobian associated to it).

A Former Brilliant Member - 2 years, 8 months ago

Sir, well the limits are inside the square on the u-v plane.......i.e. .......the region is a square with vertices on (0,0),(0.5,0.5),(1,0) and (0.5,-0.5) on the u-v plane.........

Aaghaz Mahajan - 2 years, 8 months ago

@Aaghaz Mahajan – Ahk. Got it.

A Former Brilliant Member - 2 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$