Question Number 138428 by tugu last updated on 13/Apr/21
$${what}\:{the}\:{area}\:{of}\:\:{area}\:{bounded}\:{by}\:{line} \\ $$$${y}=\:\mid{ln}\:{x}\mid\:{and}\:{y}=\:\mathrm{2}\: \\ $$
Answered by Ñï= last updated on 13/Apr/21
$$\mid{lnx}\mid=\mathrm{2} \\ $$$$\Rightarrow{x}={e}^{\mathrm{2}} ,{e}^{−\mathrm{2}} \\ $$$${S}'=\int_{\mathrm{1}} ^{{e}^{−\mathrm{2}} } {lnxdx}+\int_{\mathrm{1}} ^{{e}^{\mathrm{2}} } {lnxdx} \\ $$$$=\left\{{xlnx}−{x}\right\}_{\mathrm{1}} ^{{e}^{−\mathrm{2}} } +\left\{{xlnx}−{x}\right\}_{\mathrm{1}} ^{{e}^{\mathrm{2}} } \\ $$$$=−\mathrm{3}{e}^{−\mathrm{2}} +{e}^{\mathrm{2}} +\mathrm{2} \\ $$$${S}=\mathrm{2}\left({e}^{\mathrm{2}} −{e}^{−\mathrm{2}} \right)−\left({e}^{\mathrm{2}} −\mathrm{3}{e}^{−\mathrm{2}} +\mathrm{2}\right) \\ $$$$={e}^{\mathrm{2}} +{e}^{−\mathrm{2}} −\mathrm{2} \\ $$
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](Q138441.png)
$${y}=\mathrm{ln}\:{x}\:\Rightarrow{x}={e}^{{y}} \\ $$$${y}=−\mathrm{ln}\:{x}\:\Rightarrow{x}={e}^{−{y}} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{2}} \left({e}^{{y}} −{e}^{−{y}} \right){dy}=\left[{e}^{{y}} +{e}^{−{y}} \right]_{\mathrm{0}} ^{\mathrm{2}} ={e}^{\mathrm{2}} +{e}^{−\mathrm{2}} −\mathrm{2} \\ $$