Limit of a Sequence of Definite Integrals This was on a mock test for an examination that grants admission to an undergraduate course in mathematics. So, in theory, a school-going $17$ year old with a bit of extra knowledge should be able to solve it. As such, I'm looking for "elementary" answers.
Q. Suppose $$I_n=\int_0^2(2x-x^2)^ndx.$$ Show that $$\lim_{n\to\infty}I_n=0.$$
I would give my attempt, but I do not know where to begin. Perhaps we set up some sort of recurrence relation?
 A: Here's how to solve this by forming a recurrence relation. First, complete the square and perform a trigonometric substitution $x-1=\sin\phi,\text{d}x=\cos\phi\text{ d}\phi$.
$$\begin{aligned}
I_n&=\int_0^2\left(2x-x^2\right)^n\text{ d}x\\
&=\int_0^2\left(1-(x-1)^2\right)^n\text{ d}x\\
&=\int_{-\pi/2}^{\pi/2}\left(1-\sin^2\phi\right)^n\cos\phi\text{ d}\phi\\
&=\int_{-\pi/2}^{\pi/2}\cos^{2n+1}\phi\text{ d}\phi
\end{aligned}$$
Now, integrate by parts, integrating $\cos\phi$ and differentiating $\cos^{2n}\phi$.
$$\begin{aligned}
I_n&=\int_{-\pi/2}^{\pi/2}\cos^{2n+1}\phi\text{ d}\phi\\
&=\left[\sin\phi\cos^{2n}\phi\right]_{-\pi/2}^{\pi/2}-\int_{-\pi/2}^{\pi/2}2n\cos^{2n-1}\phi\cdot\sin\phi\cdot\sin\phi\text{ d}\phi\\
&=-2n\int_{-\pi/2}^{\pi/2}\cos^{2n-1}\phi\cdot(1-\cos^2\phi)\text{ d}\phi\\
&=-2n\int_{-\pi/2}^{\pi/2}\cos^{2n-1}\phi\text{ d}\phi+2n\int_{-\pi/2}^{\pi/2}\cos^{2n+1}\phi\text{ d}\phi\\
&=-2n\ I_{n-1}+2n\ I_n
\end{aligned}$$
Solving this for $I_n$ gives
$$I_n=\frac{2n-1}{2n}I_{n-1}$$
Can you prove that $I_n$ decreases to $0$ now?
A: Take $\delta\in(0,1)$, set $I_\delta=(1-\delta,1+\delta)$, and set
$$M_\delta:=\max_{x\in[0,2]\setminus I_\delta}\lvert 2x-x^2\rvert.$$
Notice that then $M_\delta<1$. In particular we have that
$$\left\lvert\int_{[0,2]\setminus I_\delta} (2x-x^2)^n~\mathrm{d}x\right\rvert\leq\int_{[0,2]\setminus I_\delta} M_\delta^n~\mathrm{d}x=(2-2\delta)M_\delta^n\to0$$
as $n\to\infty$. Similarly we also have that
$$\left\lvert\int_{I_\delta} (2x-x^2)^n~\mathrm{d}x\right\rvert\leq\int_{I_\delta} 1^n~\mathrm{d}x=2\delta\to0$$
as $\delta\to0$. Now fix $\varepsilon>0$. Choose $\delta$ such that
$$\left\lvert\int_{I_\delta} (2x-x^2)^n~\mathrm{d}x\right\rvert<\frac{\varepsilon}{2}$$
and choose $N$ such that
$$\left\lvert\int_{[0,2]\setminus I_\delta} (2x-x^2)^n~\mathrm{d}x\right\rvert\leq\frac{\varepsilon}{2}$$
for all $n\geq N$. Then, for all $n\geq N$ we have that
$$\lvert I_n\rvert\leq\left\lvert\int_{[0,2]\setminus I_\delta} (2x-x^2)^n~\mathrm{d}x\right\rvert+\left\lvert\int_{I_\delta} (2x-x^2)^n~\mathrm{d}x\right\rvert<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$
which proves that
$$\lim_{n\to\infty}I_n=0.$$
