Prove $\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=1$ I was messing around on my calulator and I noticed that for large values of $n$, $\int_0^1\sqrt{x^n+1}~dx\to1$. I would like to prove this. So, this is my question:
Prove $$\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=1$$
Geometrically this seems to make sense; for all values of $x$ in the interval $[0,1)$ it is obvious that as as $n$ increases then $x^n+1$ is very close to $1$, but I would like a formal proof.
Many thanks for your help.
 A: Note that $\sqrt{1+x^n} \le \sqrt{1+x^n+x^{2n}/4}=1+x^n/2$ for $x \ge 0$; so
$$
1=\int_{0}^{1}dx\le \int_{0}^{1}\sqrt{1+x^n}\;dx \le \int_{0}^{1}\left(1+x^n/2\right)dx=x+\frac{x^{n+1}}{2(n+1)}\bigg\vert_{0}^{1}=1+\frac{1}{2(n+1)}.
$$
Since the endpoints of this interval $\rightarrow 1$ as $n\rightarrow \infty$, the value of the integral must also approach $1$ in this limit.
A: The point is we can permute intergral and limit (see below):
$$\lim_{n\to\infty}\int_0^1\sqrt{x^n+1}~dx=\int_0^1\lim_{n\to\infty}\sqrt{x^n+1}~dx = \int_0^1 1 dx = 1$$
Justification of the permutation limit - integral :
The dominated convergence theorem states that :
If $(f_n)$ are functions such that :

*

*$f_n$ are piecewise continuous functions

*$(f_n)$ simply converges to $f$ which is piecewise continuous

*$\exists \varphi : I\mapsto \mathbb{R}^{+}$ piecewise continuous and integrable such that $\forall n\in \mathbb{N}, |f_n|\leqslant \varphi$
Then :
The $f_n$ and $f$ are integrable on $I$ and $$\int_I f_n \rightarrow \int_I f$$
Here you can use $\varphi :x\mapsto 2$, which is of course integrable.
A: Since this tools may be advanced for you (as you wrote in the comment) I just would like to give you a counterexample that rises the main problem here: it is not true that you can, in general, exchange limit and integral, that formally means: $$\lim _{n \rightarrow \infty} \int_{a}^{b} f_{n}(x) d x \neq \int_{a}^{b} \lim_{n \rightarrow \infty}f_n(x) d x$$
Just as an example try to compute the followings: $$\lim _{n \rightarrow \infty} \int_{-n}^{n} \frac{1}{2n} d x =1$$ and $$\int_{-n}^{n} \lim _{n \rightarrow \infty} \frac{1}{2n} d x =0$$ Clearly you get they are not equal.
In general, to have such limit theorems either you need uniform convergence of a sequence of functions if you are dealing with Riemann integrable functions, or you need more powerful tools like Lebesgue integral and the Lebesgue theory of integration.
What you are applying is what is known as the Lebesgue Dominated Convergence Theorem. All what you need is to consider the sequence of functions (with values in $[0,1]$) given by $$f_n(x)=\sqrt{x^n+1}$$ and noticing that we have the following pointwise convergence where $f(x)=0$: $$f_n(x) \rightarrow f(x) $$ and the following integrability condition $$|f_n| \le g \text{  for all  } n \in \mathbb{N}$$ where $g$ is an integrable function (in this case just pick $g$ equal the constant function 2 on $[0,1]$, that has clearly finite integral and dominates all the $|f_n|$). Then, since all this holds in your example, you can exchange limit ans integral and conclude the result.
You could also apply what is known as the Monotone Convergence Theorem (a bit modified) in this case for monotone decreasing sequences.
