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प्रश्न
विकल्प
1
e − 1
e + 1
0
उत्तर
1
\[\int_1^e \log x d x\]
\[ = \int_1^e \log x x^0 d x\]
\[ = \left[ x \log x \right]_1^e - \int_1^e \frac{1}{x}x d x\]
\[ = \left[ x \log x \right]_1^e - \left[ x \right]_1^e \]
\[ = \left( e - 0 \right) - \left( e - 1 \right)\]
\[ = e - e + 1\]
\[ = 1\]
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