Let $S=[0,1]\times[0,1]\subset \mathbb R^2$. Evaluate :$\iint_s max(x,y)\,\mathbb dx\mathbb dy$

Question

Let $S=[0,1]\times[0,1]\subset \mathbb R^2$.

Evaluate:$$\iint_s max(x,y)\,\mathbb dx\mathbb dy$$

MY TRY:$\iint_s max(x,y)\,\mathbb dx\mathbb dy=\iint_{s_1} x\,\mathbb dx \mathbb dy+\iint_{s_2} y\,\mathbb dx \mathbb dy$

where $s_1$ is the triangle obtained by joining the point $(0,0),(1,0)$ and the line $y=x$ and similarly $s_2$ is the rest half.$$\iint_{s_1} x\,\mathbb dx \mathbb dy=\int_{y=0}^{y=1}\int_{x=y}^{x=1} x\,\mathbb dx \mathbb dy=\frac 1 3$$ and similarly $$\iint_{s_2} y\,\mathbb dx \mathbb dy=\int_{y=0}^{y=1}\int_{x=0}^{x=y} y\,\mathbb dx \mathbb dy=\frac 1 3$$ So,the answer is $\frac 2 3$.

I don't know whether it is correct or not.So,please check it and if you have any better tricks/result please share. Thank you.

https://math.stackexchange.com/q/1874340/321264 – StubbornAtom May 19 '20 at 15:13 — StubbornAtom, May 19 '20 at 15:13

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

Your calculations are correct.

There is also a different approach to calculate the result using probability theory. You try to calculate the expectation of the maximum of two independent random variables, that are distributed uniformly on $[0, 1]$. This can be calculated like here.

More generally, the approach in the linked thread will easily yield the following result: $$\int_{[0, 1]^n} \max(x_1, \ldots, x_n) \, dx_1 \ldots dx_n = \int_0^1 1 - t^n \, dt = \frac{n}{n + 1}.$$

edited Apr 13 '17 at 12:21

Community

1

answered Dec 24 '16 at 19:32

Dominik

19,963

So,what will be the answer if it is "min" by using probability theory approach?@Dominik – MatheMagic Dec 24 '16 at 19:35
In the "min"-case we get $\frac{1}{n + 1}$. – Dominik Dec 24 '16 at 19:40
Nice answer -- I should have pointed out that the min case involves slightly easier integrals/limits in mine. – John Hughes Dec 25 '16 at 14:30

John Hughes · Answer 2 · 2016-12-24T19:24:59.390

1

That looks fine to me.

You might ask the same question about "min", and note that the sum of the two answers must be the integral of $x+y$ over the whole square. :)

edited Dec 24 '16 at 19:24

answered Dec 24 '16 at 19:18

John Hughes

93,729

If we change "max" to "min" to this question the answer according to you is 1 (@John Hughes) but it is $\frac 1 3$ according to Dominik.then which one is correct? – MatheMagic Dec 25 '16 at 04:28
@SubratKumarJena What John says is that $\int \max + \min = \int x + y = 1$, i.e. $\int \min = 1 - \int \max = \frac{1}{3}$. – Dominik Dec 25 '16 at 08:28
Thanks @Dominik,now i understood... – MatheMagic Dec 25 '16 at 08:57

Let $S=[0,1]\times[0,1]\subset \mathbb R^2$. Evaluate :$\iint_s max(x,y)\,\mathbb dx\mathbb dy$

2 Answers2