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In polar coordinates, the circle is $r=2$, the line $y=\sqrt{3}x$ is $\theta =\pi /3$, and $y=1$ is $r\sin \theta =1$. The inequalities selectThe surface height isThus the required volume is

For fixed $x>0$, $x\ne 1$,with limiting value one at $x=1$. Integrating over $0<x<1$ in the opposite order givesOn the other hand, after setting $x=e^{-u}$,Consequently

LetThenUsing polar coordinates on the first quadrant,Since $I>0$, the Gaussian integral is

Because $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous and bounded; write $|f|\le M$. Given $ϵ>0$, choose $\delta >0$ such thatThenThe bound is independent of $p$. Taking the supremum, then $n\to \infty$, gives a limit superior at most $ϵ$. Since $ϵ$ is arbitrary,This is the Bernstein polynomial proof of the Weierstrass approximation theorem.

The sample mean $S_n/n$ has expectation $p$ and varianceThe sum in the question is exactlyChebyshev's inequality therefore gives, uniformly in $p$,

The Chebyshev inequality states that for a random variable $Y$ with finite variance and every $\delta >0$,Indeed, Markov inequality applied to the nonnegative variable $(Y-\mathbb{E}Y{)}^2$ gives

The sum $S_n=X_1+\cdots +X_n$ has the binomial distribution, soFor fixed $n$ this is a finite sum of polynomial functions of $p$, hence is a polynomial.

Ignore rolls showing $2,3,4,5$ and consider the first roll among $\{1,6\}$. By symmetry, the event $A$ that a 1 occurs before a 6 has probability $1/2$. The event that the very first roll is 1 has probability $1/6$ and is contained in $A$. Hence

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then $f(c)=y$ for some $c\in [a,b]$.

Let $x_1<x_2<x_3$. If, for example, $f(x_2)$ exceeded both $f(x_1)$ and $f(x_3)$, choose $y$ strictly between $f(x_2)$ and $\max \{f(x_1),f(x_3)\}$. Applying the theorem on both $[x_1,x_2]$ and $[x_2,x_3]$ would give two distinct preimages of $y$, contradicting injectivity. The analogous argument excludes $f(x_2)$ below both endpoint values. Since the three values are distinct,

Fix $a<b$. If $f(a)<f(b)$, applying the displayed betweenness property to triples containing $a,b$ forces the same increasing order for every pair $x<y$; an order reversal would create a triple whose middle value is not between the other two. Thus $f$ is strictly increasing. If $f(a)>f(b)$, the same argument shows that it is strictly decreasing. This proves that every continuous bijection of the real line is monotone.