উচ্চতর গণিত ১ম পত্র প্রশ্ন

AI-তৈরি নমুনা উত্তর

দেওয়া আছে,
$I = \int_{10}^{100} \log_{10}{x} \, dx$

আমরা জানি, $\log_{10}{x} = \frac{\ln x}{\ln 10}$

$\therefore I = \frac{1}{\ln 10} \int_{10}^{100} \ln x \, dx$

এখন, $\int \ln x \, dx = x \ln x - x$

I &= \frac{1}{\ln 10} \Big[ x \ln x - x \Big]_{10}^{100} \\ &= \frac{1}{\ln 10} \Big[ (100 \ln 100 - 100) - (10 \ln 10 - 10) \Big] \\ &= \frac{1}{\ln 10} \Big[ 100 \ln(10^2) - 100 - 10 \ln 10 + 10 \Big] \\ &= \frac{1}{\ln 10} \Big[ 200 \ln 10 - 100 - 10 \ln 10 + 10 \Big] \\ &= \frac{1}{\ln 10} \Big[ 190 \ln 10 - 90 \Big] \\ &= 190 - \frac{90}{\ln 10} \quad \text{বা,} \quad 190 - 90\log_{10}e \approx 150.913 \end{aligned}$$ **উত্তর:** $190 - \frac{90}{\ln 10}$ বা $190 - 90\log_{10}e$