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Higher Order Derivatives

Area bounded ...

Question

Area bounded by y $= e^{x}$ and lines $x =$ 0 and $y = e$ is given by:

e - \int_{0}^{1} e^{x}

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e - \int_{1}^{e} ln y d y

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e - 1

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1

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Solution

The correct options are
A $e - \int_{0}^{1} e^{x}$ dx
D $1$
From figure
Required area is, $A = \int_{1}^{e} ln y d y = e - \int_{0}^{1} e^{x} d x$
$= e - {(e^{x})}_{0}^{x}$
$= e - (e - 1) = 1$

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= e^{2 x}

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y = e^{x}, y = e^{- x}

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