Let $$f\left( x \right)$$ be a continuous function such that $$f\left( {a - x} \right) + f\left( x \right) = 0$$ for all $$x\, \in \,\left[ {0,\,a} \right].$$ Then $$\int_0^a {\frac{{dx}}{{1 + {e^{f\left( x \right)}}}}} $$ is equal to :
A.
$$a$$
B.
$$\frac{a}{2}$$
C.
$$f\left( a \right)$$
D.
$$\frac{1}{2}f\left( a \right)$$
Answer :
$$\frac{a}{2}$$
Solution :
$$\eqalign{
& I = \int_0^a {\frac{{dx}}{{1 + {e^{f\left( x \right)}}}}} = \int_0^a {\frac{{dx}}{{1 + {e^{f\left( {a - x} \right)}}}}} = \int_0^a {\frac{{dx}}{{1 + {e^{ - f\left( x \right)}}}}} \left( {{\text{from the question}}} \right) \cr
& \,\,\,\,\, = \int_0^a {\frac{{{e^{f\left( x \right)}}dx}}{{{e^{f\left( x \right)}} + 1}}} \cr
& \therefore I + I = \int_0^a {\frac{{dx}}{{1 + {e^{f\left( x \right)}}}}} + \int_0^a {\frac{{{e^{f\left( x \right)}}}}{{1 + {e^{f\left( x \right)}}}}} dx \cr
& \therefore 2I = \int_0^a {dx} = a \cr} $$
Releted MCQ Question on Calculus >> Application of Integration
Releted Question 1
The area bounded by the curves $$y = f\left( x \right),$$ the $$x$$-axis and the ordinates $$x = 1$$ and $$x = b$$ is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$ Then $$f\left( x \right)$$ is-