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Question
The area bounded by the curve y = loge x and x-axis and the straight line x = e is ___________ .
Options
e sq. units
1 sq. units
1−\[\frac{1}{e}\] sq. units
1+\[\frac{1}{e}\] sq. units
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Solution
1 sq. units
The point of intersection of the curve and the straight line is A(e, 1).
Therefore, the area of the required region ABC,
\[A = \int_0^1 \left( x_1 - x_2 \right) d y ..........\left(\text{ where, }x_1 = e\text{ and }x_2 = e^y \right)\]
\[ = \int_0^1 \left( e - e^y \right) d y\]
\[ = \left[ ey - e^y \right]_0^1 \]
\[ = \left\{ e\left( 1 \right) - e^\left( 1 \right) \right\} - \left\{ e\left( 0 \right) - e^\left( 0 \right) \right\}\]
\[ = e - e + 1\]
\[ = 1\text{ square unit }\]
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