Question

The area enclosed between the curves y = log_e (x + e), x = log_e \[\left( \frac{1}{y} \right)\] and the x-axis is _______ .

विकल्प

• 2

• 1

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Answer 1

2

 

The point of intersection of the curves \[y = \log_e \left( x + e \right)\text{ and }x = \log_e \left( \frac{1}{y} \right)\] is (0, 1)
\[y = \log_e \left( x + e \right)\]
\[ \Rightarrow x + e = e^y \]
\[ \Rightarrow x = e^y - e\]
\[\text{ Here taking, }x_1 = e^y - e\]
\[\text{ and }x_2 = \log_e \left( \frac{1}{y} \right)\]
Therefore, area of the required region,
\[A = \int_0^1 \left( x_2 - x_1 \right) dy ..........\left[\text{Where, }x_1 = e^y - e \text{ and }x_2 = \log_e \left( \frac{1}{y} \right) \right]\]
\[A = \int_0^1 \log_e \left( \frac{1}{y} \right) d y - \int_0^1 \left( e^y - e \right) d y\]
\[A = \int_0^1 \log_e \left( \frac{1}{y} \right) d y - \left[ e^y - ey \right]_0^1 . . . . . \left( 1 \right)\]
\[\text{ Let }I = \int \log_e \left( \frac{1}{y} \right) d y\]
\[\text{ Putting }\frac{1}{y} = t\]
\[ \Rightarrow - \frac{1}{y^2}dy = dt\]
\[ \Rightarrow dy = - y^2 dt\]
\[ \Rightarrow dy = - \frac{1}{t^2} dt\]
Therefore, integral becomes
\[I = \int - \frac{1}{t^2} \log_e t dt\]
\[ = - \log_e t \int\frac{1}{t^2}dt - \int\frac{1}{t} \times \frac{1}{t}dt\]
\[ = \frac{1}{t} \log_e t + \frac{1}{t}\]
\[ = y \log_e \frac{1}{y} + y\]
\[\text{ Now, }\left( 1 \right)\text{ becomes }\]
\[A = \left[ y \log_e \frac{1}{y} + y \right]_0^1 - \left[ e^y - ey \right]_0^1 \]
\[ = \left[ y \log_e \left( \frac{1}{y} \right) + y - e^y + ey \right]_0^1 \]
\[ = \left[ \log_e \left( 1 \right) + 1 - e^1 + e\left( 1 \right) \right] - \left[ 0 + 0 - e^\left( 0 \right) + e\left( 0 \right) \right]\]
\[ = 2\]