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$$f_n(x) = \begin{cases} 1, & \quad 0\le x\le\frac{1}{n} \\ \frac{1-x^2}{1-1/n^2}, & \quad \frac{1}{n}\le x\le1 \\ \end{cases}$$ is a sequence of functions with $f_n: [0, 1]\longrightarrow\mathbb{R}$

at $x=0$ is $f(x)=1$,what is it for $x \in (0,1)$?

Does $ \int_0^{1 }f_n(x) dx =\int_0^{1 }f(x)dx $?

I am unsure how to aprouch this as im not sure if $ \int_0^{1 }f_n(x) dx= \int_0^{1/n }f_n(x) dx$ is allowed or i use should another method

and does $f_n: \longrightarrow f$ uniformly?

Help on any of these questions would be much appreciated.

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    $\begingroup$ What have you tried? What is $f$? Is that the pointwise limit of the sequence $\{f_n \}$? $\endgroup$ – Theoretical Economist Nov 1 '17 at 23:20
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For $x\in\left(\frac1n,1\right]$ we have $$ \lim_{n\to\infty}\frac{1-x^2}{1-\frac1{n^2}} = 1-x^2, $$ so $f_n$ converges pointwise to $f(x)=1-x^2$. For each $n$, we have $$ |f_n(x)-f(x)| = \left|\frac{1-x^2}{1-\frac1{n^2}}-(1-x^2)\right| = \frac{1-x^2}{n^2-1}, $$ and thus $$ \sup_n|f_n(x)-f(x)| = \frac1{n^2-1}\stackrel{n\to\infty}\longrightarrow0. $$ It follows that $f_n$ converges uniformly to $f$, and so we may compute $$ \lim_{n\to\infty} \int_0^1 f_n(x)\ \mathsf dx = \int_0^1 f(x)\ \mathsf dx = \int_0^1 (1-x)^2\ \mathsf dx = \frac23. $$

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