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.
Best Answer
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.