= Solution
The integral is finite and nonnegative because $|f|$ is <continuous> on a compact interval. If it vanishes and $f(x_*)\ne0$, continuity gives a relative interval of positive length on which $|f|\geq|f(x_*)|/2$, a contradiction. Thus the integral is zero only for $f=0$. For a real scalar $a$, $\|af\|=|a|\|f\|$, and the pointwise <triangle inequality> $|f+g|\leq|f|+|g|$ integrates to $\|f+g\|\leq\|f\|+\|g\|$. These are all the <norm> axioms.
The given sequence satisfies
$$
\boxed{\|f_n-0\|=\int_0^1e^{-nx}\,dx=\frac{1-e^{-n}}n\longrightarrow0.}
$$
Hence \b[it converges to the zero function in this integral <norm>]. Although $f_n(0)=1$ for all $n$, an isolated endpoint contributes no integral. The pointwise limit is discontinuous, and the sequence does not converge in the <uniform norm>; neither fact obstructs convergence in the stated <norm>.
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