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I was reading a text and it had the following listed as immediate:

$\int_{0}^{1}\int_{0}^{1-x_{1}}\cdots \int_{0}^{1-x_{1}-\cdots -x_{n-1}} dx_{n}\cdots dx_{1}=\frac{1}{n!}$

I haven't actually evaluated an integral and done calculus in a while so I'm sorry if this is apparent. But why is this immediate?

Amview QMage
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    http://math.stackexchange.com/questions/769545/volume-of-t-n-x-i-ge0x-1-cdotsx-n-le1 ; http://math.stackexchange.com/questions/1151465/what-is-the-size-of-x-1-ldots-x-n-in-mathbbrnx-1-cdotsx-na-text-a?noredirect=1&lq=1 ; http://math.stackexchange.com/questions/1116322/find-the-volume-of-the-set-s-x-x-1-x-2-cdots-x-n-in-bbbrn0-le-x-1-le?noredirect=1&lq=1 – Winther Nov 13 '16 at 00:23
  • Can you do it when $n=2?$ – zhw. Nov 13 '16 at 02:15

1 Answers1

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You can also prove this by induction by replacing $1$ by $r$ in the equation:

$$\int_{0}^{r}\int_{0}^{r-x_{1}}\cdots \int_{0}^{r-x_{1}-\cdots -x_{n-1}} dx_{n}\cdots dx_{1}=\frac{r^n}{n!}$$

Let $f_n(r)$ be the integral.

Then $$f_{n+1}(r)=\int_0^r f_{n}(r-x_1)\,dx_1=\int_0^rf_n(x_1)\,dx_1$$

Then apply the induction hypothesis.

Thomas Andrews
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