Definite Integration Question 122
Question: $ \int_1^{2}{\frac{1}{x^{2}}{e^{\frac{-1}{x}}}dx=} $
[DCE 2001]
Options:
A) $ \sqrt{e}+1 $
B) $ \sqrt{e}-1 $
C) $ \frac{\sqrt{e}+1}{e} $
D) $ \frac{\sqrt{e}-1}{e} $
Show Answer
Answer:
Correct Answer: D
Solution:
Put $ t=-\frac{1}{x}\Rightarrow dt=\frac{1}{x^{2}}dx $ , then it reduces to $ \int _{-1}^{-1/2}{e^{t}dt=[e^{t}] _{-1}^{-1/2}={e^{-1/2}}-{e^{-1}}}=\frac{\sqrt{e}-1}{e} $ .
BETA
