Let \(f(x)\) be a continuous function from \(\mathbb{R}\) to \(\mathbb{R}\) such that
\[f(x) = 1 - f(2 - x)\]
Which one of the following options is the CORRECT value of \(\int_{0}^{2} f(x)\,dx\) ?
Answer
(B)
Explanation
Let \(I = \int_{0}^{2} f(x)\,dx\). The substitution \(x \to 2 - x\) maps \([0, 2]\) onto itself, so also \(I = \int_{0}^{2} f(2 - x)\,dx\). Adding the two forms and using \(f(x) + f(2 - x) = 1\):
\[2I = \int_{0}^{2} \big[f(x) + f(2 - x)\big]\,dx = \int_{0}^{2} 1\,dx = 2,\]
so \(I = 1\). Geometrically, the graph is symmetric about the point \((1, \frac{1}{2})\), so the parts above and below the line \(y = \frac{1}{2}\) cancel and the integral is \(2 \times \frac{1}{2} = 1\). For example, \(f(x) = \frac{x}{2}\) satisfies the condition, and its integral from 0 to 2 is 1.
Answer: B