find-1-2-e-x-2x-1-dx-

Question Number 74889 by abdomathmax last updated on 03/Dec/19

f i n d \int_{- \frac{1}{2}}^{+ \infty} e^{- x} \sqrt{2 x + 1} d x

Commented by mathmax by abdo last updated on 06/Dec/19

l e t I = \int_{- \frac{1}{2}}^{\infty} e^{- x} \sqrt{2 x + 1} d x c h a n g e m e n t \sqrt{2 x + 1} = t g i v e

2 x + 1 = t^{2} \Rightarrow 2 x = t^{2} - 1 \Rightarrow x = \frac{t^{2} - 1}{2} \Rightarrow

I = \int_{0}^{\infty} e^{- \frac{t^{2} - 1}{2}} t t d t = \sqrt{e} \int_{0}^{\infty} t^{2} e^{- \frac{t^{2}}{2}} d t

=_{\frac{t}{\sqrt{2}} = u} \sqrt{e} \int_{0}^{\infty} 2 u^{2} e^{- u^{2}} \sqrt{2} d u = 2 \sqrt{2 e} \int_{0}^{\infty} u^{2} e^{- u^{2}} d u b u t

\int_{0}^{\infty} u e^{- u^{2}} u d u =_{b y p s r t s} {[- \frac{1}{2} {u e}^{- u^{2}}]}_{0}^{+ \infty} - \int_{0}^{+ \infty} (- \frac{1}{2} e^{- u^{2}}) d u

= \frac{1}{2} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{1}{2} \times \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{4} \Rightarrow I = 2 \sqrt{2 e} \frac{\sqrt{π}}{4} = \frac{\sqrt{2} \sqrt{e} \sqrt{π}}{2}

= \sqrt{\frac{π e}{2}}

★ I = \sqrt{\frac{π e}{2}} ★

Answered by mind is power last updated on 04/Dec/19

u = \sqrt{2 x + 1} \Rightarrow x = \frac{u^{2} - 1}{2} \Rightarrow dx = udu

= \int_{0}^{+ \infty} e^{- \frac{u^{2} - 1}{2}} u^{2} du

= \sqrt{e} \int_{0}^{+ \infty} - u . (- {ue}^{- \frac{u^{2}}{2}}) du by part

\int_{0}^{+ \infty} (- u) . (- u . e^{- \frac{u^{2}}{2}}) du = [- u . e^{- \frac{u^{2}}{2}}] + \int e^{- \frac{u^{2}}{2}} du

= \int_{0}^{+ \infty} e^{- \frac{u^{2}}{2}} du = \sqrt{2} \int_{0}^{+ \infty} e^{- u^{2}} du = \frac{\sqrt{π}}{\sqrt{2}}

we get

\sqrt{e} . \sqrt{\frac{π}{2}} = \sqrt{\frac{π . e}{2}}

Commented by mathmax by abdo last updated on 04/Dec/19

t h a n k y o u s i r .