A recent Math SE question
asked for help computing the value of
\$\$int_0^{2000} e^{x/2-leftlfloor x/2rightrfloor};
dx.tag{\$star\$}\$\$

(!!leftlfloor frac x2 rightrfloor!! means !!frac x2!! rounded
down to the nearest integer.)

Often when I see someone’s homework problems I exclaim “what blockhead
TA assigned this?” But I think this is a really good exercise.
Here’s why.

In a calculus class, some people will have learned to integrate common
functions by rote manipulatation of the expressions. They have
learned a set of rules for converting \$\$int_a^b x^k; dx\$\$ to
\$\$left.frac{x^{k+1}}{k+1}rightrvert_a^b\$\$ and then to
\$\$frac{b^{k+1}}{k+1}- frac{a^{k+1}}{k+1}\$\$ and such like, and they grind
through the algebra. If this is all someone knows how to do, they are
going to have a lot of trouble with !!(star)!!. They might say “But
nobody ever taught us how to integrate functions with !!leftlfloor
frac x2rightrfloor!!”.

A calculus tyro trying to deal with this analytically might also try
rewriting \$\$e^{x/2-leftlfloor x/2rightrfloor}\$\$ as
\$\$frac{e^{x/2}}{e^{leftlfloor x/2rightrfloor}}\$\$ but that makes
the problem harder, not easier.

To solve this, the student has to actually understand what the
integral is computing, and if they don’t they will have to learn
something about it. The integral is computing the area under a curve.
if you graph the function \$\$frac x2-leftlfloor frac
x2rightrfloor\$\$

you find that it looks like this:

If the interval of integration in !!(star)!! were only !!(0,2)!!
instead of !!(0, 2000)!!, the problem would be
very easy because, on this interval, the
complicated exponent is identically equal to !!frac
x2!!:

\$\$begin{align}
int_0^2 e^{x/2-leftlfloor x/2rightrfloor}; dx
& = int_0^2 e^{x/2}; dx \
& = left. 2e^{x/2} rightrvert_0^2 \
& = 2e-2
end{align}
\$\$

Since the function is completely periodic, integrating over any of
the !!1000!! intervals of length !!2!! will produce the same value, so
the final answer is simply \$\$1000cdot (2e-2).\$\$

But just pushing around the symbols won’t get you there, to solve this
problem you have to actually know something about calculus.

The student who overcomes this problem might learn the following
useful techniques:

• If some expression looks complicated, try graphing it and see if you
get any insight into how it behaves.

• Some complicated functions can be understood by breaking them into
simple parts and dealing with the parts separately.

• Piecewise-continuous functions can be integrated by breaking them
into continuous intervals and integrating the intervals separately.

• You can exploit symmetry to reduce the amount of calculation
required.

None of this is deep stuff, but it’s all valuable technique. Also
they might make the valuable observation that not every problem should
be solved by pushing around the symbols.