int_{1 / 2}^2left|log _{10} xright| d x= | vedclass

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(B) We evaluate the integral $I = \int_{1/2}^2 |\log_{10} x| dx$. Since $\log_{10} x < 0$ for $x \in [1/2, 1)$ and $\log_{10} x \ge 0$ for $x \in [1,

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We evaluate the integral $I = \int_{1/2}^2 |\log_{10} x| dx$. Since $\log_{10} x $I = -\int_{1/2}^1 \log_{10} x dx + \int_1^2 \log_{10} x dx$. Using the change of base formula $\log_{10} x = \frac{\ln x}{\ln 10}$,we get: $I = \frac{1}{\ln 10} \left[ -\int_{1/2}^1 \ln x dx + \int_1^2 \ln x dx \right]$. Using the integral $\int \ln x dx = x \ln x - x$: $I = \frac{1}{\ln 10} \left[ -(x \ln x - x)|_{1/2}^1 + (x \ln x - x)|_1^2 \right]$. Evaluating the limits: $I = \frac{1}{\ln 10} \left[ -((1 \ln 1 - 1) - (\frac{1}{2} \ln \frac{1}{2} - \frac{1}{2})) + ((2 \ln 2 - 2) - (1 \ln 1 - 1)) \right]$. $I = \frac{1}{\ln 10} \left[ -(-1 - \frac{1}{2} \ln \frac{1}{2} + \frac{1}{2}) + (2 \ln 2 - 2 + 1) \right]$. $I = \frac{1}{\ln 10} \left[ (\frac{1}{2} + \frac{1}{2} \ln \frac{1}{2}) + (2 \ln 2 - 1) \right] = \frac{1}{\ln 10} [\frac{1}{2} - \frac{1}{2} \ln 2 + 2 \ln 2 - 1] = \frac{1}{\ln 10} [\frac{3}{2} \ln 2 - \frac{1}{2}] = \frac{1}{2 \ln 10} [3 \ln 2 - 1] = \frac{1}{2} \log_{10} (\frac{2^3}{e}) = \frac{1}{2} \log_{10} (\frac{8}{e})$.

$\int_{1 / 2}^2\left|\log _{10} x\right| d x=$

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