Question

$\displaystyle\int_{1/2}^{2}|\log_{10}\,x|dx=$

• A. $\log_{10}\left(\frac{8}{e}\right)$
• B. $\frac{1}{2}\log_{10}\left(\frac{8}{e}\right)$
• C. $log_{10}\left(\frac{2}{e}\right)$
• D. none of these.

Answer 1

Answer: (B)

$\frac{1}{2}\log_{10}\left(\frac{8}{e}\right)$

Answer 2

$\int log \, x \, dx=x \,log x-x$ [By Using integration by parts] $log x is -ve. for x, < 1$ and $+ve$ for $x > 1$. Also $log_{10} \,x= \frac{log\,x}{log\,10}$ $\therefore$ given integral $=\frac{1}{log\,10}\left[\int\limits_{1 /2}^{1}-log\,x\,dx+\int\limits_{1}^{2} log\,x\,dx\right]$ $=-\frac{1}{log\,10} \left[x\,log\,x-x\right]_{1 /2}^{1}+\frac{1}{log\,10} \left[x\,log\,x-x\right]_{1}^{2}$ $=-\frac{1}{log\,10}\left[-1-\frac{1}{2}log\frac{1}{2}+\frac{1}{2}\right]+\frac{1}{log\,10} \left[2\,log\,2-2+1\right]$ $=\frac{1}{log\,10}\left[\frac{1}{2}-\frac{1}{2}log\,2\right]+\frac{2\,log\,2-1}{log\,10}$ $=\frac{4\,log\,2-2+1-log\,2}{2\,log\,10}=\frac{3\,log\,2-1}{2\,log\,10}$ $=\frac{1}{2} \frac{log\,8-log\,e}{log\,10}=\frac{1}{2} \frac{log\left(\frac{8}{e}\right)}{log\,10}$ $=\frac{1}{2}log_{10} \left(\frac{8}{e}\right)$