Find sequence of function that has several conditions

Question

$\require{cancel}$I have to find a sequence of function $f_n(x)$ that would have these conditions:

1. $f_n(x)$ is continious in the range $[0;1]$

2. $f_n(x)\rightarrow 0$, $x\in[0,1]\; \forall{x}$

3. $\int_0^1 f_n(x)\, dx \cancel{\rightarrow}0$

Answer 1

Inspired by the answer by @Kelenner above, here is a family of functions that have the same non-zero integral for all $n$:

$$f_n(x) = \begin{cases} 4n^2x, &0\le x < \frac1{2n},\\ 2n, &x = \frac1{2n},\\ 4n^2\left(\frac1n-x\right), &\frac1{2n}<x<\frac1n,\\ 0, &\frac1n \le x \le 1. \end{cases}$$

Condition 2 is satisfied, because $\forall x\in(0,1]$, $$n\ge\frac1x\implies f_n(x) = 0,$$

and $\forall n\in\mathbb N$, $f_n(0) = 0$.

Condition 3 is satisfied: $$ \begin{align*} \int_0^1 f_n(x)\ dx &= \int_0^\frac1{2n}4n^2x\ dx + \int_{\frac1{2n}}^{\frac1n} 4n^2\left(\frac1n-x\right)dx\\ &= 2\int_0^\frac1{2n}4n^2x\ dx\\ &= 2\cdot4n^2\cdot\frac12\cdot\frac{1}{(2n)^2}\\ &= 1\\ \lim_{n\to \infty}\int_0^1 f_n(x)\ dx &\ne 0 \end{align*}$$