Find the value of : $\lim_{n\to\infty}\int _0^1 nx ({1 - x^2})^{n} dx$

Question

Find the value of : $$\lim_{n\to\infty}\int _0^1 nx ({1 - x^2})^{n} dx$$ $1)0$

$2)\frac{1}{2}$

$3)1$

$4) \infty $

I think we have : $\lim _{n\to\infty}\int _0^1 nx ({1 - x^2})^{n} dx =\lim _{n\to\infty} n \int _0^1 x ({1 - x^2})^{n} dx =\lim _{n\to\infty} \frac{n}{-2(n+1)} \int _0^1 -2(n+1) x ({1 - x^2})^{n} dx=\lim _{n\to\infty} \frac{n}{-2(n+1)} \int _0^1 ( ({1 - x^2})^{n+1})^\prime dx= \lim _{n\to\infty} \frac{n}{-2(n+1)}(({1 - x^2})^{n+1}|_0^1)=\frac{1}{2}$

Answer 1

Yes. With $1-x^2=t$,

$$\int_0^1nx(1-x^2)^ndx=-\frac12\int_1^0 nt^ndt=\frac12\frac n{n+1}\to\frac12.$$