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Question Number 138428 by tugu last updated on 13/Apr/21

what the area of  area bounded by line  y= ∣ln x∣ and y= 2

$${what}\:{the}\:{area}\:{of}\:\:{area}\:{bounded}\:{by}\:{line} \\ $$$${y}=\:\mid{ln}\:{x}\mid\:{and}\:{y}=\:\mathrm{2}\: \\ $$

Answered by Ñï= last updated on 13/Apr/21

∣lnx∣=2  ⇒x=e^2 ,e^(−2)   S′=∫_1 ^e^(−2)  lnxdx+∫_1 ^e^2  lnxdx  ={xlnx−x}_1 ^e^(−2)  +{xlnx−x}_1 ^e^2    =−3e^(−2) +e^2 +2  S=2(e^2 −e^(−2) )−(e^2 −3e^(−2) +2)  =e^2 +e^(−2) −2

$$\mid{lnx}\mid=\mathrm{2} \\ $$$$\Rightarrow{x}={e}^{\mathrm{2}} ,{e}^{−\mathrm{2}} \\ $$$${S}'=\int_{\mathrm{1}} ^{{e}^{−\mathrm{2}} } {lnxdx}+\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } {lnxdx} \\ $$$$=\left\{{xlnx}−{x}\right\}_{\mathrm{1}} ^{{e}^{−\mathrm{2}} } +\left\{{xlnx}−{x}\right\}_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \\ $$$$=−\mathrm{3}{e}^{−\mathrm{2}} +{e}^{\mathrm{2}} +\mathrm{2} \\ $$$${S}=\mathrm{2}\left({e}^{\mathrm{2}} −{e}^{−\mathrm{2}} \right)−\left({e}^{\mathrm{2}} −\mathrm{3}{e}^{−\mathrm{2}} +\mathrm{2}\right) \\ $$$$={e}^{\mathrm{2}} +{e}^{−\mathrm{2}} −\mathrm{2} \\ $$

Answered by mr W last updated on 13/Apr/21

y=ln x ⇒x=e^y   y=−ln x ⇒x=e^(−y)   A=∫_0 ^2 (e^y −e^(−y) )dy=[e^y +e^(−y) ]_0 ^2 =e^2 +e^(−2) −2

$${y}=\mathrm{ln}\:{x}\:\Rightarrow{x}={e}^{{y}} \\ $$$${y}=−\mathrm{ln}\:{x}\:\Rightarrow{x}={e}^{−{y}} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{2}} \left({e}^{{y}} −{e}^{−{y}} \right){dy}=\left[{e}^{{y}} +{e}^{−{y}} \right]_{\mathrm{0}} ^{\mathrm{2}} ={e}^{\mathrm{2}} +{e}^{−\mathrm{2}} −\mathrm{2} \\ $$

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