Question

Let \(f(x)\) be a continuous function defined in \([0,2] \to \mathbb{R}\) and satisfying the equation
\[ \int_0^2 f(x)[x - f(x)]\,dx = \frac{2}{3} \]
The value of \(f(1)\) is

• A. 1
• B. 2
• C. \(\frac{1}{2}\)
• D. 0

Hint:

Complete the square in \(f(x)\) inside the integrand and compare the leftover term with the value of \(\int_0^2 \frac{x^2}{4}\,dx\).

Answer 1

The Correct Option is C

Step 1: Rewrite the integrand.
The given condition is \(\int_0^2 f(x)[x - f(x)]\,dx = \frac{2}{3}\).
Expand the integrand as \(xf(x) - f(x)^2\), so the equation becomes \(\int_0^2 \left[xf(x) - f(x)^2\right] dx = \frac{2}{3}\).

Step 2: Complete the square.
Treat \(xf(x) - f(x)^2\) as a quadratic expression in \(f(x)\). It can be written as
\[ xf(x) - f(x)^2 = \frac{x^2}{4} - \left(f(x) - \frac{x}{2}\right)^2 \]
Check this identity by expanding the right side: \(\frac{x^2}{4} - f(x)^2 + xf(x) - \frac{x^2}{4} = xf(x) - f(x)^2\), which matches the left side, so the identity holds.

Step 3: Split the integral and evaluate the known part.
Using the identity, the given equation becomes
\[ \int_0^2 \frac{x^2}{4}\,dx - \int_0^2 \left(f(x) - \frac{x}{2}\right)^2 dx = \frac{2}{3} \]
Compute the first integral directly:
\[ \int_0^2 \frac{x^2}{4}\,dx = \frac{1}{4}\left[\frac{x^3}{3}\right]_0^2 = \frac{1}{4}\cdot\frac{8}{3} = \frac{2}{3} \]
So the equation reduces to \(\frac{2}{3} - \int_0^2 \left(f(x) - \frac{x}{2}\right)^2 dx = \frac{2}{3}\).

Step 4: Use the fact that a nonnegative integral equal to zero forces the integrand to be zero.
From Step 3, \(\int_0^2 \left(f(x) - \frac{x}{2}\right)^2 dx = 0\).
The integrand \(\left(f(x)-\frac{x}{2}\right)^2\) is a square, so it can never be negative, and \(f\) is given to be continuous. A continuous function that is never negative and whose integral is zero must be zero at every single point in the interval.
So \(f(x) - \frac{x}{2} = 0\) for every \(x\) in \([0,2]\), which gives \(f(x) = \frac{x}{2}\).

Step 5: Check the result and rule out the other options.
Check: with \(f(x)=\frac{x}{2}\), \(\int_0^2 \frac{x}{2}\left(x - \frac{x}{2}\right) dx = \int_0^2 \frac{x^2}{4}\,dx = \frac{2}{3}\), which matches the given condition, so \(f(x)=x/2\) is indeed the correct (and only) function satisfying the equation.
A constant function \(f(x)=c\) does not work either: it gives \(\int_0^2 c(x-c)\,dx = 2c - 2c^2\), and setting \(2c-2c^2=\frac{2}{3}\) has no real solution for \(c\), so the answer cannot be a fixed constant like 1, 2 or 0.
Since \(f(x) = \frac{x}{2}\), \(f(1) = \frac{1}{2}\), which rules out options (A) 1, (B) 2 and (D) 0.

Final Answer:
\(f(1) = \frac{1}{2}\), option (C). \[ \boxed{f(1) = \frac{1}{2}} \]