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.

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## Comments

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

Ya. I want the solution for this

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

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

Hey!!! Thanks a lot!!!Log in to reply

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

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

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Indeed. I would just integrate it numerically

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

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

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

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Sorry. Not good with latex for writing.

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@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............!!!

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Can you please provide the solution.

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

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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.

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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).

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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.........

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