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
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)