Question

int_(0)^( infinity )e^(-a^(2)x^(2)dx)`​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that,

By definition of gamma function, we have

$\boxed{\displaystyle\int_0^ \infty \tt \: {e}^{ - x} {x}^{n - 1} = \Gamma n}$

and

$\boxed{ \bf \Gamma \: \dfrac{1}{2} \: = \sqrt{\pi} }$

Consider,

$\rm :\longmapsto\:Let \: I \: = \displaystyle\int _{0}^\infty \sf \: {e}^{ - {a}^{2} {x}^{2}}dx$

To solve this integral, we use method of Substitution

$\red{\rm :\longmapsto\: {a}^{2} {x}^{2} = y } \\ \red{\rm :\longmapsto\:2 {a}^{2}x = \dfrac{dy}{dx} } \\ \red{\rm :\longmapsto\:dx =\dfrac{dy}{2 {a}^{2} x}} \\ \red{\rm :\longmapsto\:dx = \dfrac{dy}{2a \sqrt{y} } }$

So, Above integral can be rewritten as

$\rm :\longmapsto\:\: I \: = \displaystyle\int _{0}^\infty \sf \: {e}^{ -y}\dfrac{1}{2a \sqrt{y}}dy$

$\rm :\longmapsto\:\: I \: =\dfrac{1}{2a} \displaystyle\int _{0}^\infty \sf \: {e}^{ -y}\dfrac{1}{\sqrt{y}}dy$

$\rm :\longmapsto\:\: I \: =\dfrac{1}{2a} \displaystyle\int _{0}^\infty \sf \: {e}^{ -y} \: {y}^{ - \frac{1}{2} } dy$

can be rewritten as

$\rm :\longmapsto\:\: I \: =\dfrac{1}{2a} \displaystyle\int _{0}^\infty \sf \: {e}^{ -y} \: {y}^{\frac{1}{2} - 1} dy$

$\bf\implies \:I \: = \dfrac{1}{2a} \Gamma \: \bigg(\dfrac{1}{2} \bigg)$

$\bf\implies \:I \: = \dfrac{1}{2a} \sqrt{\pi}$

Hence,

$\bf :\longmapsto\: \displaystyle\int _{0}^\infty \bf \: {e}^{ - {a}^{2} {x}^{2}}dx = \dfrac{ \sqrt{\pi} }{2a}$

Additional Information :-

$\green{\boxed{ \tt \Gamma \: (n + 1) = n \: \Gamma \:(n)}}$

$\green{\boxed{ \tt \:\Gamma \:1 = 1 }}$

$\green{\boxed{ \tt \:\Gamma \:(n + 1) = n! \: if \: n \in \: natural \: number }}$

$\green{\boxed{ \tt \:B(m, n) = \dfrac{\Gamma \:m \: \Gamma \:n}{\Gamma \:(m + n)} \: where \: m \: and \: n \: > 0} \:}$

Answer 2

Step-by-step explanation:

Answer

To find integral using limit of sum method :-

$\displaystyle\boxed{\sf \int_{a}^{b} f( {x})dx = (b - a)}$

$\lim_{n \to \infty}(\frac{1}{n} )[f(a)+f(a+n) + f(a + \{n - 1\}h)]$

where

$\sf h = \frac{b-a}{n}$

For given question :-

$\rm a = 0a=0$

$\rm b = 2b=2$

$\rm f(x) = ( x^2 + 1 )f(x)=(x)$

$\displaystyle\sf \int_{0}^{2} ( {x}^{2} + 1)dx = 2 \lim_{n \to \infty}( \frac{1}{n} )[f(0) + f(\frac{2}{n}) + f(\frac{4}{n}) + .. + f(\frac{2 \{{n - 1} \}}{n} )]$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[1 + \{ \frac{2^2}{n^2} + 1\} + \{ \frac{4^2}{n^2} + 1 \} + .. + \{ \frac{(2n-2)^2}{n^2} + 1 \}]=2$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )([ 1 + 1 + 1 .. 1(n - times) ] + \frac{1}{n^2} )$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[ n + \frac{2^2}{n^2}(1^2 + 2^2 ..+ (n-1)^2]=2$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[ n+\frac{4}{n^2}\{(n-1)n(\frac{2n-1}{6}\}]=2$

$\displaystyle\sf =2 \lim_{n \to \infty}( \frac{1}{n} )[n + \frac{2}{3} \{\frac{(n-1)(2n-1)}{n} \}]=2$

$\displaystyle\sf = 2 \lim_{n \to \infty}( \frac{1}{n} )[n+\frac{2}{3}(1-\frac{1}{n} )(2-\frac{1}{n})]=2$

$\sf n \to \infty , \frac{1}{n} \to 0$

$\displaystyle\sf \int_{0}^{2} ( {x}^{2} + 1)dx = 2 (1+ \frac{4}{3}$

$\displaystyle\boxed{\sf\red{ \int_{0}^{2} ( {x}^{2} + 1)dx = \frac{14}{3}}}$