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The value of \(\mathop \int \limits_{ - 1}^1 \frac{{{e^x}}}{{1\; + \;{e^x}}}dx\)
1. e – 1
2. e
3. 1
4. 0
5. None of these

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Correct Answer - Option 3 : 1

Concept:

Definite Integral properties:

  • \(\mathop \int \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \int \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

 

Calculation:

Let \({\rm{I}} = \mathop \int \limits_{ - 1}^1 \frac{{{{\rm{e}}^{\rm{x}}}}}{{1{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\)     ---(1)

Here, f(a + b – x) = f(1 – 1 – x) = f(-x)

So, replacing x with -x

\({\rm{I}} = \mathop \smallint \limits_{ - 1}^1 \frac{{{{\rm{e}}^{ - {\rm{x}}}}}}{{1{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{ - {\rm{x}}}}}}{\rm{\;}}\)

\( \Rightarrow {\rm{I}} = \mathop \smallint \limits_{ - 1}^1 \frac{1}{{1{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{\rm{x}}}}}\)     ---(2)

Adding (1) and (2), we get

\(2{\rm{I}} = \mathop \smallint \limits_{ - 1}^1 \left\{ {\frac{{{{\rm{e}}^{\rm{x}}}}}{{1{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{\rm{x}}}}} + \frac{1}{{1{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{\rm{x}}}}}} \right\}{\rm{dx}}\)

\(\Rightarrow 2{\rm{I}} = \mathop \smallint \limits_{ - 1}^1 1{\rm{\;dx}}\)

\(\Rightarrow 2{\rm{I}} = \left[ {\rm{x}} \right]_{ - 1}^1\)

⇒ 2I = 2

⇒ I = 1

Alternative method:

\({\rm{I}} = \mathop \smallint \limits_{ - 1}^1 \frac{{{{\rm{e}}^{\rm{x}}}}}{{1{\rm{\;}} + {\rm{\;}}{{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\)

Let 1 + ex = t

Differentiating both sides

⇒ ex dx = dt

x

-1

1

t

1 + e-1

1 + e

 

Now, \({\rm{I}} = \mathop \smallint \limits_{1 + {{\rm{e}}^{ - 1}}}^{1 + {\rm{e}}} \left\{ {\frac{{{\rm{dt}}}}{{\rm{t}}}} \right\}\) 

\(= \left[ {\ln {\rm{t}}} \right]_{1 + {{\rm{e}}^{ - 1}}}^{1 + e}\)

\(= \ln \left( {1 + e} \right) - \ln \left( {1 + \frac{1}{e}} \right)\)

\(= \ln \left( {1 + e} \right) - \ln \left( {\frac{{e + 1}}{e}} \right)\)

= ln (1 + e) – ln (e + 1) + ln e

= 1 (∵ ln e = 1)

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