Menu Close

A-1-0-e-x-1-x-amp-aA-e-b-0-1-e-x-1-x-than-a-b-a-b-Z-




Question Number 148026 by mnjuly1970 last updated on 25/Jul/21
  A :=∫_(−1) ^( 0) e^( x +(1/x))   & aA+e^b = ∫_0 ^( ∞) ((1/e))^( x+(1/x))     than :  a+ b =?     a , b ∈ Z
$$ \\ $$$$\mathrm{A}\::=\int_{−\mathrm{1}} ^{\:\mathrm{0}} {e}^{\:{x}\:+\frac{\mathrm{1}}{{x}}} \:\:\&\:{a}\mathrm{A}+{e}^{{b}} =\:\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{{e}}\right)^{\:{x}+\frac{\mathrm{1}}{{x}}} \\ $$$$\:\:{than}\::\:\:{a}+\:{b}\:=?\:\:\:\:\:{a}\:,\:{b}\:\in\:\mathbb{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by qaz last updated on 25/Jul/21
A=∫_(−1) ^0 e^(x+(1/x)) dx=−∫_1 ^0 e^(−x−(1/x)) dx=∫_0 ^1 e^(−x−(1/x)) dx  aA+e^b   =∫_0 ^∞ ((1/e))^(x+(1/x)) dx  =∫_0 ^1 e^(−x−(1/x)) dx+∫_1 ^∞ e^(−x−(1/x)) dx  =A+∫_1 ^0 e^(−x−(1/x)) d((1/x))  =A+∫_0 ^1 e^(−x) d(e^(−(1/x)) )  =A+e^(−x−(1/x)) ∣_0 ^1 +∫_0 ^1 e^(−(1/x)) e^(−x) dx  =2A+e^(−2)   ⇒a+b=0
$$\mathrm{A}=\int_{−\mathrm{1}} ^{\mathrm{0}} \mathrm{e}^{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{dx}=−\int_{\mathrm{1}} ^{\mathrm{0}} \mathrm{e}^{−\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{dx} \\ $$$$\mathrm{aA}+\mathrm{e}^{\mathrm{b}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{e}}\right)^{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{dx}+\int_{\mathrm{1}} ^{\infty} \mathrm{e}^{−\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{dx} \\ $$$$=\mathrm{A}+\int_{\mathrm{1}} ^{\mathrm{0}} \mathrm{e}^{−\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{d}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$=\mathrm{A}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{x}} \mathrm{d}\left(\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \right) \\ $$$$=\mathrm{A}+\mathrm{e}^{−\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}} \mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{e}^{−\mathrm{x}} \mathrm{dx} \\ $$$$=\mathrm{2A}+\mathrm{e}^{−\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{b}=\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 25/Jul/21
thanks alot m qaz...
$${thanks}\:{alot}\:{m}\:{qaz}… \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *