$\int_{1 / 2}^2\left|\log _{10} x\right| d x$
$\begin{aligned} & =\int_{1 / 2}^1\left|\log _{10} x\right| d x+\int_1^2\left|\log _{10} x\right| d x \\ & =-\int_{1 / 2}^1 \log _{10} x d x+\int_1^2 \log _{10} x d x \\ & =\frac{1}{\log _e 10}\left[\int_1^2 \log _e x d x-\int_{1 / 2}^1 \log _e x d x\right] \\ & =\frac{1}{\log _e 10}\left(\left[x \log _e x-x\right]_1^2-\left[x \log _e x-x\right]_{1 / 2}^1\right) \\ & =\frac{1}{\log _e 10}\left[\left(2 \log _e 2-2-1 \log _e 1+1\right)\right. \\ & =\frac{1}{\log _e 10}\left[2 \log _e 2-1+1-\frac{1}{2}+\frac{1}{2} \log _e \frac{1}{2}\right] \\ & \left.=\frac{1}{2 \log _e 10} \log _e\left(\frac{8}{e}\right)=\frac{1}{2} \log _{10}\left(\frac{8}{e}\right) \cdot\left(1 \log _e 1-1-\frac{1}{2} \log _e \frac{1}{2}+\frac{1}{2}\right)\right]\end{aligned}$