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..
 A: 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.
