Verification of an integral reduction for $I_n=\int_{0}^{2}(2x-x^2)^n\,dx$

Question

Find the reduction formula for the integral $$I_n=\int_{0}^{2}(2x-x^2)^n\,dx.$$

I wanted to verify my formula for this integral. Please check for mistakes.

My Solution:

Since $2x-x^2$ is symmetric about $x=1$ in interval $[0,2]$, we can write the integral as $$I_n=2\int_{0}^{1}x^n(2-x)^n\, dx.$$

Using $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)\,dx$, we get:

$$I_n=2\int_{0}^{1}(1-x)^n(1+x)^ndx=2\int_{0}^{1}(1-x^2)^{n}\,dx.$$

Let \begin{align*} x&=\sin\theta\\ dx&=\cos\theta\, d\theta. \end{align*}

Thus $$I_n=2\int_{0}^{\pi/2}\cos^{2n+1}\theta\, d\theta.$$

Now we know that $$2k\int_{0}^{\pi/2}\cos^k\theta\,d\theta=2(k-1)\int_{0}^{\pi/2}\cos^{k-2}\theta\, d\theta.$$

Putting $k=2n+1$, we get:

$$(2n+1)I_n=2nI_{n-1}\implies I_n=\dfrac{2nI_{n-1}}{2n+1}.$$

Please check this solution for any mistakes. Also please help in finding $\lim_{n\to \infty}I_n.$

Thank you

Answer 1

As for the $\lim_{n\to\infty}I_{n}$, we can use the Dominated Convergence Theorem.

In this case, it suffices to check that

1. The sequence of functions $\{f_n\}$ given by $f_{n}(x) = (2x-x^2)^n$ converges pointwise to some function $f$ on the interval $[0,2]$.
2. There exists some integrable function $g$ such that for all $x \in [0,2]$ and for all $n$ we have $|f_n(x)| \le g(x)$.

If both conditions are met we can conclude that $$\lim_{n\to\infty}I_{n} = \lim_{n\to\infty}\int_0^2 f_n(x)\,dx = \int_0^2\lim_{n\to\infty}f_n(x)\,dx = \int_0^2 f(x)\,dx.$$

Considering the pointwise limit, we have that for $x\in [0,2]$, \begin{align*} \lim_{n\to\infty}(2x- x^2)^n &= \begin{cases} 1, &x = 1\\ 0, &\text{else} \end{cases}\\ &= 0\quad \text{a.e.} \end{align*} which means the first condition is satisfied. Furthermore, for $x\in [0,2]$, $|(2x-x^2)^{n}| \leq 1,$ meaning the second condition is also satisfied. Therefore, we have that $$\lim_{n\to\infty}I_{n} = \int_{0}^{2} 0\,dx = 0.$$