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Prove that \( \int_{0}^{\pi / 2} e^{-x^{2}} d x=\frac{\sqrt{\pi}}{2} \) |
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Answer» Let x2 = y ⇒ 2x dx = dy ⇒ dx = dy/dx = dy/2√y \(\therefore\int\limits_0^{\infty}e^{-x^2}dx\) = \(\int\limits_0^{\infty}\frac1{2\sqrt y}e^{-y}dy\)\(=\frac12\int\limits_0^{\infty}y^{\frac{-1}2}e^{-y}dy\) \(=\frac12\int\limits_0^{\infty}y^{\frac12-1}e^{-y}dy = \frac{\Gamma(\frac12)}2\) \((\because\int\limits_0^{\infty}x^{n-1}e^{-x}dx=\Gamma(x))\) \(=\frac{\sqrt\pi}2(\because\Gamma(\frac12)=\sqrt\pi)^2\) |
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