Let S be the area of the region enclosed by y = e^{- \textrm{ } x^{2}}, y = 0 and x = 1. Then

Engineering

Mathematics

estimation of definite integral

Question

Let S be the area of the region enclosed by y = $e^{- x^{2}}$ , y = 0 and x = 1. Then

$S \geq 1 - \frac{1}{e}$

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

$S \leq \frac{1}{4} (1 + \frac{1}{\sqrt{e}})$

$S \geq \frac{1}{e}$

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$S = \int_{0}^{1} e^{- x^{2}} dx$

For 0 $\leq x \leq$ 1

$x^{2} \leq x \Rightarrow - x^{2} \geq - x \Rightarrow e^{- x^{2}} \geq e^{- x}$

$\Rightarrow \int_{0}^{1} e^{- x^{2}} dx > \int_{0}^{1} e^{- x} dx or S > {[- e^{- x}]}_{0}^{1} \Rightarrow S > 1 - \frac{1}{e}$

$\Rightarrow S > \frac{1}{e} (As 1 - \frac{1}{e} > \frac{1}{e})$

$Again S = \int_{0}^{1 / \sqrt{2}} e^{- x^{2}} dx + \int_{1 / \sqrt{2}}^{1} e^{- x^{2}} dx < \int_{0}^{1 / \sqrt{2}} 1 dx + \int_{1 / \sqrt{2}}^{1} e^{- 1 / 2} dx$

$\Rightarrow S < \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}} (1 - \frac{1}{\sqrt{2}})$

$Also, (1 - \frac{1}{e}) - \frac{1}{4} (1 + \frac{1}{\sqrt{e}}) = \frac{3}{4} - \frac{1}{e} - \frac{1}{4 \sqrt{e}} = \frac{3 e - 4 - \sqrt{e}}{4 e}$

$\Rightarrow S > \frac{1}{4} (1 + \frac{1}{\sqrt{e}})$