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

w h a t t h e a r e a o f a r e a b o u n d e d b y l i n e

y = ∣ l n x ∣ a n d y = 2

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

∣ l n x ∣= 2

\Rightarrow x = e^{2}, e^{- 2}

S^{'} = \int_{1}^{e^{- 2}} l n x d x + \int_{1}^{e^{2}} l n x d x

= {x l n x - x}_{1}^{e^{- 2}} + {x l n x - x}_{1}^{e^{2}}

= - 3 e^{- 2} + e^{2} + 2

S = 2 (e^{2} - e^{- 2}) - (e^{2} - 3 e^{- 2} + 2)

= e^{2} + e^{- 2} - 2

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

y = \ln x \Rightarrow x = e^{y}

y = - \ln x \Rightarrow x = e^{- y}

A = \int_{0}^{2} (e^{y} - e^{- y}) d y = {[e^{y} + e^{- y}]}_{0}^{2} = e^{2} + e^{- 2} - 2

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