Definite Integration Question 144
Question: The area enclosed between the curves $ y={\log _{e}}(x+e),x={\log _{e}}( \frac{1}{y} ) $ and the x-axis is
Options:
A) 2 sq. units
B) 1 sq. units
C) 4 sq. units
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ y={\log _{e}}(x+e),x={\log _{e}}( \frac{1}{y} ) $ or $ y={e^{-x}} $ . For $ y={\log _{e}}(x+e), $ shift the graph of $ y={\log _{e}}x,e $ units to the hand side.
Required area $ =\int\limits _{1-e}^{0}{{\log _{e}}(x+e)dx+\int\limits_0^{\infty }{{e^{-x}}dx}} $
$ =| x{\log _{e}}(x+e) | _{1-e}^{0}-\int\limits _{1-e}^{0}{\frac{x}{x+e}dx-| {e^{-x}} |_0^{\infty }} $
$ =\int\limits_0^{1-e}{( 1-\frac{e}{x+e} )dx-{e^{-\infty }}+e^{0}} $
$ =| x-e\log (x+e) |_0^{1-e}-0+1 $
$ =1-e+e\log e+1=2 $ sq. units.
BETA
