Math, asked by Anonymous, 1 month ago

Evaluate the definite integral:

\displaystyle \int \limits_0^ \infty \dfrac{\sqrt{x}}{ {e}^{x}} dx

Answers

Answered by kinzal
14

We can write this equation it like  \sf \displaystyle \int \limits_{ 0}^{  \infty }  \sqrt{x} e ^{ - x} dx \\

Let u equal square root of x

u = √x

x = u²

Then Differentiate both sides we get

dx = 2udu _____(1)

So, Now   \sf \displaystyle \int \limits_{x = 0}^{x =  \infty }  \sqrt{x} e ^{ - x} dx \\

Now if we put x = 0 in u = √x we get, u = 0

And if we put x = ∞ and we get u =

Hence, and Also we compare tha given equation to the eq ___(1) then we get,

 \sf \displaystyle   \int \limits_{u = 0}^{u =  \infty } ue ^{ - u ^{2}}2 udu   \\

  \sf \displaystyle   \int \limits_{u = 0}^{u =  \infty } ue ^{ - u ^{2}}2 udu   \\

 \sf \displaystyle   2 \int \limits_{ 0}^{  \infty } ue ^{ - u ^{2}}du  \\

Now,

D(Differentiation) : + u → - 1

I (Integer) :  \sf ue^{-u^{2} }  \sf -  \frac{1}{2}  {ue}^{ { - u}^{2} } \\

So,

 \displaystyle 2 \bigg[ \bigg( -\frac{-1}{2} ue^{-u^{2}} \bigg)_0^∞ + \int \limits_0^∞ \frac{1}{2} e^{-u^{2} } du \bigg] \\

Now, focus on  \sf \bigg( -\frac{1}{2} ue^{-u^{2}} \bigg)_0^∞ \\

Now put  \sf Lim_{ \: \: u → ∞ } \bigg( \frac{-u}{2e^{-u^2}} \bigg) \\

Now, differentiate  \sf Lim_{ \: \: u → ∞ } \bigg( \frac{\frac{d}{dv}(-u)}{\frac{d}{du}(2e^{-u^2})} \bigg) \\

 \sf Lim_{ \: \: u → ∞ } \bigg( \frac{-1}{4ue^{u^2}} \bigg) \\

 \sf \frac{-1}{∞} \\

 \sf = 0 \\

Now, come to complete our sum

 \sf 2 \bigg[ (0-0) + \displaystyle \int \limits_{∞}^{0} \frac{1}{2} e^{-u^{2}} du \bigg]  \\

 \sf 2 \bigg[ (0-0) + \frac{1}{2} \displaystyle \int \limits_{∞}^{0}  e^{-u^{2}} du \bigg]  \\

Now, focus on  \sf 2 \bigg[ (0-0) + \frac{1}{2}\displaystyle \int \limits_{∞}^{0}  e^{-u^{2}} du \bigg]  \\

we already know,

 \displaystyle \int \limits_{∞}^{-∞} e^{-u^{2}} du = \sqrt{π} \\

If You go from negative infinity to positive infinity, this right here you get  \sqrt{π}\\

(Because of Gaussian Integral)

 \displaystyle \int \limits_{∞}^{0} e^{-u^{2}} du = \frac{\sqrt{π}}{2} \\

Now , finally We have,

 \sf \cancel{2}\bigg( \frac{1}{\cancel{2}} × \frac{\sqrt{π}}{2} \bigg) \\

 \sf\boxed{ \frac{\sqrt{π}}{2} }\\

I hope it helps you ❤️✔️

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