Definite Integration Question 174
Question: Area bounded by the curves $ y=e^{x},y={e^{-x}} $ and the straight line $ x=1 $ is (in sq. units)
Options:
A) $ e+\frac{1}{e} $
B) $ e+\frac{1}{e}+2 $
C) $ e+\frac{1}{e}-2 $
D) $ e-\frac{1}{e}+2 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given curves are $ y=e^{x} $ and $ y={e^{-x}} $ Now, $ e^{x}={e^{-x}}\Rightarrow x=0 $
$ \therefore $ Area $ =A=\int\limits_0^{1}{( e^{x}-{e^{-x}} )dx=( e^{x}+{e^{-x}} )_0^{1}} $
$ =[ ( e+{e^{-1}} )-( e^{0}+{e^{-0}} ) ]=e+\frac{1}{e}-2 $ .
BETA
