برای محاسبه انتگرال نامعین
$$\int |x| \mathrm{d}x$$
از روش جزءبهجزء استفاده میکنیم:
$$\begin{align}
\left\{\begin{array}{r@{\mskip\thickmuskip}l}
|x|=u\\
\mathrm{d}x=\mathrm{d}v
\end{array} \right.
\quad \implies \quad
\left\{\begin{array}{r@{\mskip\thickmuskip}l}
\frac{|x|}{x}\:\mathrm{d}x=\mathrm{d}u\\
x=v
\end{array}\right.
\end{align}$$
بنابراین داریم:
$$\begin{align}
\int |x|\mathrm{d}x =\int u \mathrm{d}v&=uv - \int v\ \mathrm{d}u\\
&=x|x| - \int x\: \frac{|x|}{x} \ \mathrm{d}x=x|x|-\int |x| \ \mathrm{d}x\\
&\Longrightarrow \int |x|\ \mathrm{d}x=\frac{x|x|}{2}
\end{align}$$
در نهایت برای محاسبه انتگرال معین $\int_a^b |x|\ \mathrm{d}x$ داریم:
$$
\int_a^b |x|\ \mathrm{d}x=\frac{x|x|}{2} |_a^b=\frac{b|b|}{2}-\frac{a|a|}{2}
$$
برای محاسبه انتگرال معین $\int_a^b \lfloor x \rfloor \mathrm{d}x$
از خاصیت جزء صحیح که برای هر عدد حقیقی $t$ داریم:
$$\lfloor t \rfloor \leq t < \lfloor t \rfloor +1$$
بنابراین:
$\begin{align*}
\left\{\begin{array}{ll}
a < \lfloor a \rfloor+1 \\
\lfloor b\rfloor \leq b
\end{array} \right.
\end{align*}$
لذا داریم:
$\begin{align*}
\int_a^b \lfloor x \rfloor \mathrm{d}x&=\int_a^{\lfloor a \rfloor+1} \lfloor x \rfloor \mathrm{d}x+\int_{\lfloor a \rfloor+1}^{\lfloor a \rfloor+2} \lfloor x \rfloor \mathrm{d}x+\dots\\
&+\int_{\lfloor b \rfloor-1}^{\lfloor b \rfloor} \lfloor x \rfloor \mathrm{d}x+\int_{\lfloor b \rfloor}^{b} \lfloor x \rfloor \mathrm{d}x\\
&=(\lfloor a \rfloor-a+1)\lfloor a \rfloor+(\lfloor a \rfloor+1)+\ldots\\
&+(\lfloor b \rfloor-1)+(b-\lfloor b \rfloor)\lfloor b \rfloor\\
&=(\lfloor a \rfloor-a+1)\lfloor a \rfloor+(\lfloor a \rfloor+1)+\ldots\\
&+(\lfloor a \rfloor+(\lfloor b \rfloor-\lfloor a \rfloor-1))+(b-\lfloor b \rfloor)\lfloor b \rfloor\\
&=(\lfloor a \rfloor-a+1)\lfloor a \rfloor+(\lfloor b \rfloor-\lfloor a \rfloor-1)(\lfloor a \rfloor)\\
&+[1+2+3+\ldots+(\lfloor b \rfloor-\lfloor a \rfloor-1)]+(b-\lfloor b \rfloor)\lfloor b \rfloor\\
&=(\lfloor a \rfloor-a+1)\lfloor a \rfloor+(\lfloor b \rfloor-\lfloor a \rfloor-1)(\lfloor a \rfloor)\\
&+\frac{(\lfloor b \rfloor-\lfloor a \rfloor-1)(\lfloor b \rfloor-\lfloor a \rfloor)}{2}+(b-\lfloor b \rfloor)\lfloor b \rfloor\\
&=(\lfloor a \rfloor-a+1)\lfloor a \rfloor+\frac{(\lfloor b \rfloor-\lfloor a \rfloor-1)(\lfloor b \rfloor+\lfloor a \rfloor)}{2}+(b-\lfloor b \rfloor)\lfloor b \rfloor\\
\end{align*}$
