Definite Integration Question 41
Question: The value of $ \int _{0}^{\pi /2}{\frac{{e^{x^{2}}}}{{e^{x^{2}}}+{e^{{{( \frac{\pi }{2}-x )}^{2}}}}}dx} $ is
[AMU 1999]
Options:
A) $ \pi /4 $
B) $ \pi /2 $
C) $ {e^{{{\pi }^{2}}/16}} $
D) $ {e^{{{\pi }^{2}}/4}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_0^{\pi /2}{\frac{{e^{x^{2}}}dx}{{e^{x^{2}}}+{e^{( \frac{\pi }{2}-x )}}^{2}}} $ and $ I=\int_0^{\pi /2}{\frac{{e^{{{( \frac{\pi }{2}-x )}^{2}}}}dx}{{e^{{{( \frac{\pi }{2}-x )}^{2}}}}+{e^{x^{2}}}}} $
$ [ \because \int_0^{a}{f(x)dx=\int_0^{a}{f(a-x)dx}} ] $
$ \Rightarrow 2I=\int_0^{\pi /2}{1dx=(x)_0^{\pi /2}} $
$ \Rightarrow I=\frac{\pi }{4} $ .
BETA
