Question

$\sf \: Evaluate \: the \: definite \: integral: \\ \boxed{ \ \: \: \: \: \: \: \: \: \: \: \displaystyle \int \limits_0^ \infty \dfrac{\sqrt{x}}{ {e}^{x}} dx \: \: \: \: \: \: \: \: \: \: \: \: }$
Try to solve it !!​

Answer 1

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

Can be written as:

$\displaystyle{ \bold \red{\int_{0}^{\infty}\left( \sqrt{x} e^{- x} \right)dx}}$

First, calculate the corresponding indefinite integral:

$\displaystyle{ \bold \red{\int{\sqrt{x} e^{- x} d x}=- \sqrt{x} e^{- x} - \frac{\sqrt{\pi} \operatorname{erfc}{\left(\sqrt{x} \right)}}{2}}}$

Since there is infinity in the upper bound, this is improper integral of type 1.

To evaluate an integral on an interval, we use the Fundamental Theorem of Calculus. However, we need to use the limit, if an endpoint of the interval is special (infinite).

${ \displaystyle{ \bold \red{\int_{0}^{\infty}\left( \sqrt{x} e^{- x} \right)dx=\lim_{x \to \infty}\left(- \sqrt{x} e^{- x} - \frac{\sqrt{\pi} \operatorname{erfc}{\left(\sqrt{x} \right)}}{2}\right)-\left(- \sqrt{x} e^{- x} - \frac{\sqrt{\pi} \operatorname{erfc}{\left(\sqrt{x} \right)}}{2}\right)|_{\left(x=0\right)}=\frac{\sqrt{\pi}}{2}}}}$

Answer:

${\displaystyle{ \bold \red{ \int_{0}^{\infty}\left( \sqrt{x} e^{- x} \right)dx=\frac{\sqrt{\pi}}{2}\approx 0.886226925452758}}}$

Answer 2

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