The value of _ ,0 ^ , /2 e^ x^2 e^ x^2 + e^ ( 2 , , - , ,x ) ^2 dx is

The value of _ ,0 ^ , /2 e^ x^2 e^ x^2 + e^ ( 2 , , - , ,x ) ^2 d…

MCQ

The value of $\int_{\,0}^{\,\pi /2} {\frac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\frac{\pi }{2}\,\, - \,\,x} \right)}^2}}}}}dx} $ is

✓
$\pi /4$
B
$\pi /2$
C
${e^{{\pi ^2}/16}}$
D
${e^{{\pi ^2}/4}}$

✓

Answer

Correct option: A.

$\pi /4$

a
(a) $I = \int_0^{\pi /2} {\frac{{{e^{{x^2}}}\,\,\,\,\,\,dx}}{{{e^{{x^2}}} + {e^{\left( {\frac{\pi }{2}\,\, - x} \right)}}^2}}} $ and

$I = \int_0^{\pi /2} {\frac{{{e^{{{\left( {\frac{\pi }{2} - x} \right)}^2}}}\,\,\,\,\,\,dx}}{{{e^{{{\left( {\frac{\pi }{2}\, - x} \right)}^2}}} + {e^{{x^2}}}}}} $

$\left[ \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right]$

$ \Rightarrow 2I = \int_0^{\pi /2} {1dx = (x)_0^{\pi /2}} $

$ \Rightarrow I = \frac{\pi }{4}$.

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