First slide

Area of bounded Regions

Question

$If \int_{0}^{1} \frac{e^{- x} d x}{1 + e^{x}} = \log_{e} (1 + e) + K, then find the value of k is$

Moderate

A

$2 (\frac{1}{e} + \log 2)$
B

$(\frac{1}{e} - \log 2)$
C

$- (\frac{1}{e} + \log 2)$
D

$(\frac{1}{e} + \log 2)$

Solution

$I = \int_{0}^{1} \frac{e^{- x} d x}{1 + e^{x}} = \int_{0}^{1} \frac{d x}{e^{x} (1 + e^{x})}$
$Put e^{x} = z or e^{x} d x = d z or d x = \frac{d z}{e^{x}} = \frac{d z}{z}$
$\begin{array}{l} ∴ I = \int_{1}^{e} \frac{d z}{z^{2} (1 + z)} \\ = \int_{1}^{e} (\frac{1}{1 + z} - \frac{z - 1}{z^{2}}) d z \\ = {|\log (1 + z) - \log z - \frac{1}{z}|}_{1}^{e} \\ = (\log (1 + e) - \log e - \frac{1}{e}) - (\log 2 - \log 1 - 1) \\ = \log (1 + e) - \frac{1}{e} - \log 2 \\ ∴ K = - (\frac{1}{e} + \log 2) \end{array}$

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