The correct option is D 2 1/e ∫^1_0(1 + e^ x) dx nbsp;= (x e^ x)^1_0 = (1 1e) (0 1) = nbsp;(2 1e) ...

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Byju's Answer

Standard IX

Mathematics

Property 6

∫011+e-xdx=

Question

$\int_{0}^{1} (1 + e^{- x}) d x =$

-1

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1 + e^{- 1}

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2 - \frac{1}{e}

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Solution

The correct option is D $2 - \frac{1}{e}$
$\int_{0}^{1} (1 + e^{- x}) d x = {(x - e^{- x})}_{0}^{1}$
= $(1 - \frac{1}{e}) - (0 - 1)$
= $(2 - \frac{1}{e})$

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∫10(1+e−x)dx=

The correct option is D 2−1e∫10(1+e−x)dx=(x−e−x)10= (1−1e)−(0−1)= (2−1e)

$\int_{0}^{1} (1 + e^{- x}) d x =$

The correct option is D $2 - \frac{1}{e}$
$\int_{0}^{1} (1 + e^{- x}) d x = {(x - e^{- x})}_{0}^{1}$
= $(1 - \frac{1}{e}) - (0 - 1)$
= $(2 - \frac{1}{e})$