Let $f$ be a continuous function such that $f(x)+f(1-x)=2$ for all $x\in [0,1]$. The value of the integral ${\int }_0^{1}f(x)dx$ is equal to:

Solution

1. Let $I={\int }_0^{1}f(x)dx$.
2. Using the property ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$, we have $I={\int }_0^{1}f(1-x)dx$.
3. Adding the two expressions, $$\begin{array}{rl}2I& ={\int }_0^1(f(x)+f(1-x))dx\\ & ={\int }_0^{1}2dx\\ & =2\end{array}$$.
4. Hence the answer is (B).