How do I solve this integral $\int_{e}^{4+e} (3x- \lfloor 3x \rfloor)dx$? I haven't come across this yet, but the question is to solve this definite integral: $$\int_{e}^{4+e} (3x- \lfloor 3x \rfloor)dx$$
What is obviously causing problems is the whole part of the number function. Any suggestions, help on how to solve this? It's for highschool level, so have that in mind..

Break the integral into smaller pieces. Forgive my abuse of notation but it must be absolutely clear
$$ \int_e^{4+e} = \int_e^3 + \int_3^{3+1/3} + \int_{3+1/3}^{3+2/3} + \int_{3+2/3}^{4} + ... + \int_{4+2/3}^{4+e}. $$
On each of these integrals, $\lfloor 3x \rfloor$ takes the values $(8,9,10,...,19,20)$; to see this, check that $\lfloor 3e \rfloor=8$, $\lfloor 3(4+e) \rfloor=20$ and basically $\lfloor 3x \rfloor$ changes value every time $3x$ increase by an integer, which is everytime $x$ increases by $1/3$. 
So you have to solve
$$ \int_e^3 (3x-8)dx+ ... + \int_{4+2/3}^{4+e} (3x-20) dx $$
and you can easily finish it from here. To do this, I recommend to write to solve the general problem 
$$ \int_a^b (3x-C) dx$$
for any constant $C$, this must give you 
$$(3/2)(b^2-a^2)-(b-a)C$$
and then, sum the values of all the integrals. Try to simplify and cancel terms. The good news is that many terms will cancel. Here, $b-a=1/3$.
Th key point of all of this here is that if you want to integrate from $a$ to $b$, and $c$ is another number, then the integral from $a$ to $b$ is the sum of the integral from $a$ to $c$ and the integral from $c$ to $b$ (there are some restriction, but these will never occur in high school; basically, if the integrals do exist, then this must be true). So the spirit was to break in subintervals where you know the result.
I will give you a last hint, although you can add up all the number from $8$ to $20$ by hand (this will be needed in simplifying), use the following instead
$$ 8 + 9 + ... + 20 = \frac{20*21}{2}-\frac{7*8}{2} = 210-28=182$$
I will let you think why this is true as high school exercise.
