# 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
3 weeks, 5 days ago

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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

- 3 weeks, 3 days ago

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

- 3 weeks, 3 days ago

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

- 3 weeks, 3 days ago

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

- 3 weeks, 2 days ago

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

- 3 weeks, 3 days ago

Indeed. I would just integrate it numerically

- 3 weeks, 3 days ago

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

- 3 weeks, 3 days ago

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

- 3 weeks, 5 days ago

Ya. I want the solution for this

- 3 weeks, 5 days ago

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

- 3 weeks, 5 days ago

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

- 3 weeks, 5 days 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............!!!

- 3 weeks, 5 days 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).

- 3 weeks, 4 days 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.........

- 3 weeks, 4 days ago

Ahk. Got it.

- 3 weeks, 4 days ago

Can you please provide the solution.

- 3 weeks, 5 days 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.

- 3 weeks, 4 days ago

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

- 3 weeks, 5 days ago

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

- 3 weeks, 4 days 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!!!).....

- 3 weeks, 4 days 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.

- 3 weeks, 4 days ago

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

- 3 weeks, 5 days ago

Sorry. Not good with latex for writing.

- 3 weeks, 5 days ago