$\lim_{n \rightarrow \infty}\int_0^1f_n(x)$

Question

For $n=1, 2,...,$ let $f_n(x)=\frac{2nx^{n-1}}{x+1}, x\in [0, 1]$. Then $\lim_{n \rightarrow \infty}\int_0^1f_n(x)$

Function is unbounded at $1$, How do I solve?

Answer 1

\begin{align*} \int_{0}^{1}f_{n}(x)dx&=\int_{0}^{1}\dfrac{2nx^{n-1}}{x+1}dx\\ &=\dfrac{2}{x+1}\cdot x^{n}\bigg|_{x=0}^{x=1}-\int_{0}^{1}-\dfrac{2}{(x+1)^{2}}\cdot x^{n}dx\\ &=1+2\int_{0}^{1}\dfrac{x^{n}}{(x+1)^{2}}dx, \end{align*} now \begin{align*} \dfrac{x^{n}}{(x+1)^{2}}\leq\dfrac{1}{(x+1)^{2}},~~~~x\in[0,1], \end{align*} and \begin{align*} \int_{0}^{1}\dfrac{1}{(x+1)^{2}}=-\dfrac{1}{x+1}\bigg|_{x=0}^{x=1}=\dfrac{1}{2}<\infty, \end{align*} and \begin{align*} \dfrac{x^{n}}{(x+1)^{2}}\rightarrow 0,~~~~x\in[0,1), \end{align*} so by Lebesgue Dominated Convergence Theorem, \begin{align*} \int_{0}^{1}f_{n}(x)dx\rightarrow 1. \end{align*}

Another way: \begin{align*} \int_{0}^{1}\dfrac{x^{n}}{(x+1)^{2}}dx\leq\int_{0}^{1}x^{n}dx=\dfrac{1}{n+1}\rightarrow 0. \end{align*}

Answer 2

Since $\int_0^1nx^{n-1}dx=1$, we have

$$\begin{align} 0 &\le\left|\int_0^1f_n(x)dx-1 \right|\\ &=\left|\int_0^1{2nx^{n-1}\over1+x}dx-\int_0^1nx^{n-1}dx \right|\\ &=\left|\int_0^1\left({2nx^{n-1}\over1+x}-{n(1+x)x^{n-1}\over1+x} \right)dx \right|\\ &=\left|\int_0^1{n(x^{n-1}-x^n)\over1+x}dx\right|\\ &=\int_0^1{n(x^{n-1}-x^n)\over1+x}dx\quad\text{(since the integrand is clearly nonnegative)}\\ &\le\int_0^1n(x^{n-1}-x^n)dx\\ &=1-{n\over n+1}\\ &={1\over n+1}\to0 \end{align}$$

So by the Squeeze Theorem,

$$\lim_{n\to\infty}\int_0^1f_n(x)dx=1$$