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The area bounded by the curve $y = x e^{- x^{2}}$ ,y = 0 and the maximum ordinate is

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$\frac{1}{2}$

$\frac{1}{2 \sqrt{e}}$

$\frac{1}{2} (1 - \frac{1}{\sqrt{e}})$

$\frac{1}{2} (1 + \frac{1}{e})$

answer is C.

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Detailed Solution

$y = x e^{- x^{2}}, y^{'} (x) = (1 - 2 x^{2}) e^{- x^{2}} = 0$

$\Rightarrow x = \pm \frac{1}{\sqrt{2}}, y is max . at x = \frac{1}{\sqrt{2}}$

$∴$ Required area $= \int_{0}^{\frac{1}{\sqrt{2}}} x e^{- x^{2}} d x = - {\frac{1}{2} e^{- x^{2}}|}_{0}^{\frac{1}{\sqrt{2}}} = \frac{1}{2} (1 - e^{- \frac{1}{2}})$

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