Question Number 8471 by PradipGos. last updated on 12/Oct/16
$$\underset{−\infty} {\overset{\infty} {\int}}{e}^{−\mathrm{2}\mid{x}\mid{d}\underset{} {{x}}} \:\:\:\:\:\:?\:\: \\ $$
Commented by FilupSmith last updated on 12/Oct/16
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}: \\ $$$$\int_{−\infty} ^{\:\infty} {e}^{−\mathrm{2}\mid{x}\mid} {dx} \\ $$$$??? \\ $$
Commented by PradipGos. last updated on 12/Oct/16
$${yes} \\ $$
Answered by FilupSmith last updated on 12/Oct/16
![S=∫_(−∞) ^( ∞) e^(−2∣x∣) dx e^(−2∣x∣) =e^(−2∣−x∣) ∴S=2∫_0 ^( ∞) e^(−2∣x∣) dx for x≥0, ∣x∣=x ∴S=2∫_0 ^( ∞) e^(−2x) dx S=2[(1/(−2))e^(−2x) ]_0 ^∞ S=−(e^(−∞) −e^0 ) S=−(0−1) S=1](Q8480.png)
$${S}=\int_{−\infty} ^{\:\infty} {e}^{−\mathrm{2}\mid{x}\mid} {dx} \\ $$$${e}^{−\mathrm{2}\mid{x}\mid} ={e}^{−\mathrm{2}\mid−{x}\mid} \\ $$$$\therefore{S}=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} {e}^{−\mathrm{2}\mid{x}\mid} {dx} \\ $$$$\mathrm{for}\:\:{x}\geqslant\mathrm{0},\:\:\mid{x}\mid={x} \\ $$$$\therefore{S}=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} {e}^{−\mathrm{2}{x}} {dx} \\ $$$${S}=\mathrm{2}\left[\frac{\mathrm{1}}{−\mathrm{2}}{e}^{−\mathrm{2}{x}} \right]_{\mathrm{0}} ^{\infty} \\ $$$${S}=−\left({e}^{−\infty} −{e}^{\mathrm{0}} \right) \\ $$$${S}=−\left(\mathrm{0}−\mathrm{1}\right) \\ $$$${S}=\mathrm{1} \\ $$
Commented by PradipGos. last updated on 12/Oct/16
$${thank}\:{you} \\ $$