Question

A function f(x) is defined on the interval with values in R. It satisfies \( \int_0^2 f(x)[x-f(x)]dx = \frac{2}{3} \). Find the value of f(1).}

Hint:

When an integral equation involves quadratic terms of an unknown function, try to arrange it into the integral of a perfect square set to zero. This is a common trick that leads to a direct solution for the function.

Answer 1

Step 1: Identify a matching integral.
Note that $\int_0^2 \frac{x^2}{4} dx = \left[ \frac{x^3}{12} \right]_0^2 = \frac{8}{12} = \frac{2}{3}$.
Step 2: Restructure the given integral.
$\int_0^2 (xf(x) - f(x)^2) dx = 2/3$. Subtracting this from the integral in Step 1: $\int_0^2 (\frac{x^2}{4} - xf(x) + f(x)^2) dx = \frac{2}{3} - \frac{2}{3} = 0$.
Step 3: Solve for f(x).
$\int_0^2 (\frac{x}{2} - f(x))^2 dx = 0$. The integral of a squared non-negative function is zero only if the function is zero: $x/2 - f(x) = 0 \implies f(x) = x/2$. Thus, $f(1) = 1/2 = 0.5$.