I have to find the value of $$I = \int_0 ^{10} [x]^3\{x\}dx$$
Where $[x]= $greatest integer less than or equal to$ x $(the greatest integer function or the floor function)
And $\{x\}= $fractional part of $x $
So I began as follows:
We know, $x=[x]+\{x\}$ which gives us $\{x\}=x-[x]$
Now, I get, $$I = \int_0 ^{10} [x]^3 xdx-\int_0 ^{10} [x]^4dx$$, and then I could go on to evaluate it by breaking it down at integral points $1,2,3,....,10$ and then compute the value. But that seems pretty hard to do by hand, atleast by my method. Is there a more easier/less calculation intensive way to do this? The answer given is $\frac{2025}{2}$
EDIT: Computing $$I= \int_0 ^{10}[x]^3\{x\}dx$$ I get, $$I = 0 + 1^3\int_1 ^2\{x\}dx + 2^3\int_2 ^3\{x\}dx+...+9^3\int_9 ^{10}\{x\}dx$$
So can I just write $$\int_1 ^2\{x\}dx=\int_2 ^3\{x\}dx=...=\int_9 ^{10}\{x\}dx=\int_0 ^1\{x\}dx$$ and then go from there?
Write the integral as a sum of integrals over intervals $(n, n+1)$: $$\int_0^{10}[x]^3\{x\}dx=\sum_{n=0}^9\int_{n}^{n+1}[x]^3\{x\}dx$$ Now in these intervals $[x]=n$ and $\{x\}=x-n$. So we have $$\sum_{n=0}^9\int_{n}^{n+1}n^3(x-n)dx=\sum_{n=0}^9\left(n^3\frac{(n+1)^2-n^2}{2}-n^4(n+1-n)\right)=\sum_{n=0}^9\frac{n^3}2=\frac{2025}2$$