Solving integral I need to show that:
$\displaystyle ||f||_1= \int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}} |1-n(x-\frac{1}{2})|dx = 0$ as $n \to \infty$
I know $x-\frac{1}{2}$ will go to $0$ as $n\to\infty$, but I don't understand how to solve this integral.
 A: You can perform this integration directly for any given $n$. Here's the idea.
If the integrand is positive on the interval of integration $(\frac12,\frac12+\frac1n)$, you can just drop the absolute value bars and integrate the linear function (because $|a|=a$ if $a$ is positive).
If it is sometimes positive and sometimes negative, break the interval into sub intervals where the sign doesn't change, and integrate over each one separately, adding the resulting integrals up. As before, where the integrand is positive, just drop the bars; where it is negative, drop the bars and change the sign of the whole integrand (because $|a|=-a$ if $a$ is negative).
Because the expression inside the bars is linear in $x$, it will change at most once (if at all) on the interval of integration. So you just need to find where it is zero and check whether or not this is inside the interval.
Another approach is to find a constant upper bound $M_n$ on the integrand, then bound the integral from above by $M_n/n$, which is the bound on the integrand times the length of the interval of integration.
A: $$0 \leq y \leq \frac{1}{n} \implies 0 \leq ny \leq 1 \implies 1 \geq 1-ny \geq 0 \implies |1-ny|=1-ny $$
You can safely substitute $y=x - \frac{1}{2}$ and the integral becomes:
$$\int_0^{1/n} |1-ny|dy = \int_0^{1/n} (1-ny)dy = \int_0^{1/n}dy - n\int_0^{1/n} ydy = \frac{1}{n} - n \frac{1}{2n^2}= \frac{1}{n} -  \frac{1}{2n}=\frac{1}{2n} \to 0$$
