Where am I wrong in finding area of this triangle?

I was self-reading Mathematics for Economists by Simon and Blume. On page 815, Section 29.4, he has discussed "Norms on Function Space". And here I am stuck:

Let $$f_n = \begin{cases} 2n^2-2n^3x, & \text{0\leq x\leq\frac1n,} \\ 0, & \text{\frac1n\leq x\leq1.} \\ \end{cases}$$ The graph of $f_n$ is a line segment of slope $-2n^3$ from $(0,2n^2)$ to $(\frac1n,0)$ and then runs along $x$-axis from $(\frac1n,0)$ to $(1,0)$. The area under the graph of $f_n$ is $\frac1n$ and thus $$||f_n||_{L^1}=\int_0^1|f_n(x)|dx\text{ (x\in[0,1])}\longrightarrow0.$$

But I think the corresponding area should be $$\frac12\times \text{Base}\times\text{Height}=\frac12\times\frac1n\times2n^2=n.$$ Please let me know where I am wrong.

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You seem to be right. – egreg Nov 27 '13 at 9:27
Thank you. But if I am right, then, I can't go further on this topic in the book. The book says, since the area under the graph of $f_n$ is $\frac1n$, $$||f_n||_{L^1}=\int_0^1|f_n(x)|dx\text{ (x\in[0,1])}\longrightarrow0.$$ Again I am stuck! – Silent Nov 27 '13 at 9:38
As $\int_{0}^{\frac{1}{n}}2n^2-2n^3x dx$=n, the book seems to contain a typo. – Peter Nov 27 '13 at 9:48
@Peter, So does this mean that $$||f_n||_{L^1}\longrightarrow\infty?$$ A bigger typo? – Silent Nov 27 '13 at 9:51
No, this typo follows from the wrong calculated integral. – Peter Nov 27 '13 at 9:53

You're right. The consequence derived from this wrong computation is of course false too. Indeed $$\lim_{n\to\infty}\|f_n\|_{L^1}=\infty.$$
A “correct” example might be $$f(x)=\begin{cases} n-n^3x & \text{for 0\le x\le\frac{1}{n^2}}\\ 0 & \text{for \frac{1}{n^2}<x\le 1} \end{cases}$$ Then $$\int_0^1 f(x)\,dx=\int_0^{1/n^2}(n-n^3x)\,dx= n\frac{1}{n^2}-\frac{1}{2}n^3\frac{1}{n^4}=\frac{1}{2n}$$ so $$\lim_{n\to\infty}\|f_n\|_{L^1}=\lim_{n\to\infty}\frac{1}{2n}=0$$ but the sequence of functions $(f_n)$ is not pointwise convergent, because $$\lim_{n\to\infty}f_n(0)=\infty.$$
@Sush I'll try supplying an example later. But the idea is to take a triangle with area $1/n$ so that the vertex on the $y$-axis goes far away. – egreg Nov 27 '13 at 17:18