Prove sequence (of integrals) is bounded.

Question

I've got the function $f(x)=x^2-6x+8$ and the sequence of integrals
$I_n=\int_3^4 (x^2-6x+8)^n dx$
$ n>0 $
I have to prove that $I_n$ is convergent. We are taught that a sequence converges provided it is monotone and bounded.

My idea would be as follows: $f(x)$ is decreasing on $(3,4)$ and because n>0 we can say $f(x)^n$ is decreasing as well. To find the bounds I plug the values of the integral bounds, and since it is decreasing it would mean $f(4)<f(x)<f(3)$ and therefore $-1<x<0$. Is that right? And if so, is the answer complete?

Answer 1

It is false (since $f(x)$ is increasing in $(3,4)$), and so the answer is not complete!

A hint can be:

Show that $-1 \leq f(x) \leq 0$ for any $x \in (3,4)$. And then use Dominated Convergence Theorem to show that $I_n \to 0$.

Alternatively, you can show that (since $f$ has a constant sign) $$ |I_n| = \int_3^4 |x^2-6x+8|^n \leq \int_3^4 |x^2-6x+8|^{n-1} = |I_{n-1}|$$ So $|I_n|$ is decreasing (in $n$) and thus (since $\geq 0$) converges. Therefore, to show that $I_n$ converges it's enough to show that this limit is $0$.

Suppose the limit is $l > 0$. Then for $n$ even and big enough, we have $0 \leq f(x)^n \leq \frac{l}{2}$ for any $x \in (3+ \frac{l}{4} , 4)$

So $|I_n| \leq \frac{l}{2} \cdot (1-\frac{l}{4}) + \frac{l}{4} < l$.

Therefore, we must have $l=0$ and so $I_n \to 0$.

Answer 2

We have that

$$-1\le (x^2-6x+8)=(x-2)(x-4)\le 0$$

and therefore by MVT $\forall n \, \exists \, c_n \in (3,4)$

$$|I_n|=\left| \int_3^4 (x^2-6x+8)^n dx \right|=\frac1{4-3}\int_3^4 (8+6x-x^2)^n dx=[f(c_n)]^n\le 1$$

Answer 3

Using Riemann integral only.

Note that $|f(x)| = 1$ only at $x = 3$, we split the interval into two parts. For every $c \in (3,4)$, note that $|f|$ is decreasing, we have \begin{align*} \varlimsup_n |I_n| &\leqslant \varlimsup_n \int_3^c + \int_c^4 |f(x)|^n\mathrm d x \\ &\leqslant \varlimsup_n \int_3^c 1\mathrm d x +\varlimsup_n \int_c^4 |f(c)|^n\mathrm dx\\ &= c-3 +\varlimsup_n (4-c) |f(c)|^n \\ &= c-3 \quad [|f(c)| < 1]\\ &\to 0 \quad [c \to 3^+]. \end{align*}

Answer 4

Let $x=\sqrt{t}+3$ then $$I_n=\int_3^4 (x^2-6x+8)^n dx=\dfrac{(-1)^n}{2}\int_0^1(1-t)^nt^{-\frac12}\ dt<\dfrac{1}{2}\int_0^1t^{-\frac12}\ dt<1$$ Indeed $$I_n=\dfrac{(-1)^n}{2}\int_0^1(1-t)^nt^{-\frac12}\ dt=\dfrac{(-1)^n\sqrt{\pi}}{2}\dfrac{\Gamma(n+1)}{\Gamma(n+\frac32)}$$ then $$|I_n|<\dfrac{\sqrt{\pi}}{2}$$