What is the value of the definite integral \[\int\limits_0^{2a} {\dfrac{{f\left( x \right)}}{{f\left( x \right) + f\left( {2a - x} \right)}}} dx\]?
A. \[a\]
B. \[\dfrac{a}{2}\]
C. \[2a\]
D. 0

Hint: Here, a definite integral is given. First, simplify the integral by using the integration rule \[\int\limits_0^a {f\left( x \right)dx = } \int\limits_0^a {f\left( {a - x} \right)dx} \]. Then, add both integrals and simplify the equation. After that, solve the integrals using the integration formula. In the end, apply the upper and lower limits and solve the equation to get the required answer.



Formula Used:\[\int\limits_0^a {f\left( x \right)dx = } \int\limits_0^a {f\left( {a - x} \right)dx} \]
\[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_0^{2a} {\dfrac{{f\left( x \right)}}{{f\left( x \right) + f\left( {2a - x} \right)}}} dx\].
Let consider,
\[I = \int\limits_0^{2a} {\dfrac{{f\left( x \right)}}{{f\left( x \right) + f\left( {2a - x} \right)}}} dx\] \[.....\left( 1 \right)\]
Apply the integration rule \[\int\limits_0^a {f\left( x \right)dx = } \int\limits_0^a {f\left( {a - x} \right)dx} \].
\[I = \int\limits_0^{2a} {\dfrac{{f\left( {2a - x} \right)}}{{f\left( {2a - x} \right) + f\left( {2a - \left( {2a - x} \right)} \right)}}} dx\]
\[I = \int\limits_0^{2a} {\dfrac{{f\left( {2a - x} \right)}}{{f\left( {2a - x} \right) + f\left( x \right)}}} dx\] \[.....\left( 2 \right)\]

Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\]
\[I + I = \int\limits_0^{2a} {\dfrac{{f\left( x \right)}}{{f\left( x \right) + f\left( {2a - x} \right)}}} dx + \int\limits_0^{2a} {\dfrac{{f\left( {2a - x} \right)}}{{f\left( {2a - x} \right) + f\left( x \right)}}} dx\]
\[ \Rightarrow 2I = \int\limits_0^{2a} {\left[ {\dfrac{{f\left( x \right)}}{{f\left( x \right) + f\left( {2a - x} \right)}} + \dfrac{{f\left( {2a - x} \right)}}{{f\left( {2a - x} \right) + f\left( x \right)}}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_0^{2a} {\left[ {\dfrac{{f\left( x \right) + f\left( {2a - x} \right)}}{{f\left( x \right) + f\left( {2a - x} \right)}}} \right]} dx\]
\[ \Rightarrow 2I = \int\limits_0^{2a} 1 dx\]
Solve the integral by using the integration formula \[\int\limits_a^b {ndx = \left[ {nx} \right]} _a^b = n\left( {b - a} \right)\].
\[ \Rightarrow 2I = \left[ x \right]_0^{2a}\]
Apply the limits.
\[ \Rightarrow 2I = 2a - 0\]
\[ \Rightarrow 2I = 2a\]
\[ \Rightarrow I = a\]
Thus, \[\int\limits_0^{2a} {\dfrac{{f\left( x \right)}}{{f\left( x \right) + f\left( {2a - x} \right)}}} dx = a\].



Option ‘A’ is correct



Note: The integration rule \[\int\limits_0^a {f\left( x \right)dx = } \int\limits_0^a {f\left( {a - x} \right)dx} \] is obtained from another integration rule \[\int\limits_a^b {f\left( x \right)dx = } \int\limits_a^b {f\left( {a + b - x} \right)dx} \]. So, we can also apply this rule to solve the given integral.