Easy As A Sum, But The Sum Of Squares?

\[\begin{array} {} \displaystyle \int _0^{12} {\left( \{ x \}^2 + \lfloor x \rfloor ^2 \right)dx} & = \displaystyle \int _0^{12} { \{ x \}^2 dx} + \int _0^{12} { \lfloor x \rfloor ^2 dx} \\ & = \displaystyle 12 \int _0^{1} { x^2 dx} + \sum_{n=1}^{11} {n^2} \quad \quad \small \color{blue}{\text{See graphs}} \\ & = 12 \left[ \dfrac {x^3}{3} \right]_0^1 + \dfrac {11(12)(23)}{6} \\ & = 4+506 = \boxed{510} \end{array}\]

We want to find $\displaystyle \int_{0}^{12} \Bigl( \{x\}^2+\lfloor x \rfloor ^2\Bigr) \ dx$ First use the fact that $x=\lfloor x \rfloor + \{x\}$ .

$x^2=(\lfloor x \rfloor + \{x\})^2 = \lfloor x \rfloor^2 + \{x\}^2+2 \lfloor x \rfloor \{x\}$

$\lfloor x \rfloor^2 + \{x\}^2=x^2-2 \lfloor x \rfloor \{x\}$ . Make a substitution. $\displaystyle \int_{0}^{12} \Bigl( x^2-2 \lfloor x \rfloor \{x\}\Bigr) \ dx =$ . $\displaystyle \int_{0}^{12} x^2\ dx -2\int_{0}^{12} \lfloor x \rfloor \{x\} \ dx=$ .

$\displaystyle \int_{0}^{12} x^2\ dx -2\Bigl( \int_{0}^{1} 0x \ dx+\int_{0}^{1} 1x \ dx +\int_{0}^{1} 2x \ dx +...+ \int_{0}^{1} 11x \ dx \Bigr)=$ .

$\displaystyle \int_{0}^{12} x^2\ dx -\int_{0}^{1} 132x\ dx =$ .

$\left[ \dfrac {x^3}{3} \right]_0^{12} -\left[ 66x^2 \right]_0^{1} =\dfrac{12^3}{3}-66= \boxed{510}$

Since integration is linear, and $x \mapsto \{x\}^2$ and $x \mapsto \lfloor x \rfloor^2$ behave "well" (there are only countably many non-essential discontinuities, and no essential discontinuities), we can separate the two terms:

$\displaystyle \int_0^{12} \left( \{x\}^2 + \lfloor x \rfloor^2 \right) \,dx = \int_0^{12} \{x\}^2 \,dx + \int_0^{12} \lfloor x \rfloor^2 \,dx$

We begin by computing $\displaystyle \int_0^{12} \{x\}^2 \,dx$ . Note that the function $x \mapsto \{x\}^2$ is periodic with period $1$ (since $x \mapsto \{x\}$ is periodic with period $1$ ). For any periodic function $f$ with period $p$ , we have the identity $\displaystyle \int_a^b f(x) \,dx = \int_{a+kp}^{b+kp} f(x) \,dx$ for any integer $k$ , which we can utilize for this problem:

$\begin{aligned} \displaystyle \int_0^{12} \{x\}^2 \,dx &= \sum_{k=0}^{11} \int_k^{k+1} \{x\}^2 \,dx \\ &= \sum_{k=0}^{11} \int_0^1 \{x\}^2 \,dx \\ &= 12 \int_0^1 \{x\}^2 \,dx \end{aligned}$

Additionally, for $x \in (0,1)$ , $\{x\} = x$ , so we can continue:

$\begin{aligned} \displaystyle 12 \int_0^1 \{x\}^2 \,dx &= 12 \int_0^1 x^2 \,dx \\ &= 12 \cdot \left. \frac{1}{3}x^3 \right|_0^1 \\ &= 12 \cdot \left( \frac{1}{3} - 0 \right) \\ &= 4 \end{aligned}$

We then need to compute $\displaystyle \int_0^{12} \lfloor x \rfloor^2 \,dx$ . Again, we can break this integral into parts:

$\displaystyle \int_0^{12} \lfloor x \rfloor^2 \,dx = \sum_{k=0}^{11} \int_k^{k+1} \lfloor x \rfloor^2 \,dx$

Additionally, when $x \in (k, k+1)$ for integer $k$ , $\lfloor x \rfloor = k$ is constant, so we can simplify further:

$\begin{aligned} \displaystyle \sum_{k=0}^{11} \int_k^{k+1} \lfloor x \rfloor^2 \,dx &= \sum_{k=0}^{11} \int_k^{k+1} k^2 \,dx \\ &= \sum_{k=0}^{11} \left. k^2 x \right|_k^{k+1} \\ &= \sum_{k=0}^{11} \left( k^2(k+1) - k^2(k) \right) \\ &= \sum_{k=0}^{11} k^2 \\ &= \frac{11 \cdot 12 \cdot 23}{6} \\ &= 506 \end{aligned}$

Adding the two terms together gives our result: $4 + 506 = \boxed{510}$ .