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The value of $$\int_{0}^{1} \int_{0}^{1}\{\operatorname{Min}(x, y)-x y\} d x d y$$ is

In the solution I checked

$\int_{0}^{1} \int_{0}^{1}\{\operatorname{Min}(x, y)-x y\} d x d y$

$\boxed{=\int_{0}^{1} \int_{0}^{y} x d x d y+\int_{0}^{1} \int_{0}^{x} y d x dy =\frac{1}{3}}$

And $\int_{0}^{1} \int_{0}^{1} x y d x d y=\frac{1}{4}$

$\therefore \mathrm{I}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$

How did the boxed part come into the picture? Can anybody explain

  • The first integral, you are integrating $x$ from $0$ to $y$ signifying $x<y$ over all $y$. Similar is the case with the second integral – learner Jul 07 '20 at 19:10
  • @an4s yes totally! brilliant work.thanks indeed. –  Jul 07 '20 at 19:12

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