if $\lim _{n \rightarrow \infty} \int_{0}^{a_{n}} x^{n} d x=2,a_n>0,$find $ \lim _{n \rightarrow \infty}a_n$ if $$\lim _{n \rightarrow \infty} \int_{0}^{a_{n}} x^{n} d x=2,a_n>0,$$find $$\lim _{n \rightarrow \infty}a_n$$
what I did is this :
$$\begin{aligned}
\lim _{n \rightarrow \infty} \int_{0}^{a_{n}} x^{n} d x&=\lim _{n \rightarrow \infty} \frac{1}{n+1} a_{n}^{n+1}=2 \Rightarrow a_{n}^{n+1} \rightarrow \infty\\
\lim _{n \rightarrow \infty} a_{n}&=\lim _{n \rightarrow \infty}(2(n+1))^{\frac{1}{n+1}}=e^{\frac{1}{n+1} \ln (2(n+1))}=1?
\end{aligned}$$
Am I right? I don't have answer. 
I think I was wrong , because  my method always get $\lim_{n \rightarrow \infty}a_n=1$, no matter the integral equals any other number ,
 A: Suppose that $\lim_{n\to\infty}a_n=L\gt1$ then there exists $N\gt0$ such that $\forall n\gt N$, $a_n\gt \frac{L+1}{2}\gt1$. Hence we have that
$$\int_0^{a_n}x^n\mathrm{d}x\gt\int_0^{(L+1)/2}x^n\mathrm{d}x=\frac1{n+1}\left(\frac{L+1}{2}\right)^{n+1}\to\infty$$
Similarly suppose that $\lim_{n\to\infty}a_n=M\lt1$ then there exists $N\gt0$ such that $\forall n\gt N$, $a_n\lt \frac{M+1}{2}\lt1$. Hence we have that
$$\int_0^{a_n}x^n\mathrm{d}x\lt\int_0^{(M+1)/2}x^n\mathrm{d}x=\frac1{n+1}\left(\frac{M+1}{2}\right)^{n+1}\to0$$
So if the limit in question tends to $2$ then we must have $\lim_{n\to\infty}a_n=1$ otherwise the limit is either $0$ or diverges.
A: You have seen that $\frac {a_n^{n+1}} {n+1} \to 2$. This gives $(n+1) \ln a_n-\ln (n+1) \to ln \, 2$. Divide by $n+1$ and take limit. You get $\ln a_n \to 0$ and hence $a_n \to 1$. 
[$\frac {ln (n+1)} {n+1} \to 0$ by L'Hopital's Rule]. 
A: Your answer is correct 
Note that $$\lim _{n\to \infty}a^{1/{(n+1)}}=1 $$ for any positive value of $a$
Thus if you change the value of the integral from $2$ to any other positive integer the limit does not change. 
