# 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.

Note by Ankush Gogoi
2 years, 9 months ago

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

- 2 years, 9 months ago

Ya. I want the solution for this

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

Indeed. I would just integrate it numerically

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

Sorry. Not good with latex for writing.

- 2 years, 9 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............!!!

- 2 years, 9 months ago

Can you please provide the solution.

- 2 years, 9 months ago

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

- 2 years, 9 months ago

I assume he wants you to write out the whole thing for him?

- 2 years, 9 months ago

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!!!).....

- 2 years, 9 months ago

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

- 2 years, 9 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.

- 2 years, 9 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).

- 2 years, 9 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.........

- 2 years, 9 months ago

Ahk. Got it.

- 2 years, 9 months ago