Discussion
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
The integral equals 1.
Let \(I = \int_{0}^{2} f(x)\,dx\). Substituting \(x \to 2 - x\) maps [0, 2] to itself, so \(\int_{0}^{2} f(2 - x)\,dx = I\) as well. Integrating the given relation over [0, 2]:
\[ \begin{aligned} I &= \int_{0}^{2} [1 - f(2 - x)]\,dx \\ &= 2 - I \\ I &= 1 \end{aligned} \]
The relation says f takes values symmetric about 1/2 around x = 1 (for example \(f(x) = x/2\) works), so the average of f over [0, 2] is 1/2 and the integral is 1. 0, 2 and \(-1\) all fail the equation \(I = 2 - I\).
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