Let f(x) = x * sin(2x). Find the area beneath the graph of y = f(x), bounded by the x-axis, the y-axis and the line x = π/2.

After short consideration, we see that we must integrate f(x) with respect to x, between 0 and π/2, using integration by parts. Taking the formula for integration by parts, we set u = x and dv/dx = sin(2x). This implies that du/dx = 1 and v = -cos(2x)/2. Substituting this into the formula, we see that the integral is equal to evaluating (-x * cos(2x) / 2) between π/2 and 0, minus the integral of (-cos(2x)/2) between π/2 and 0.

This is equivalent to evaluating the expression ((-x * cos(2x) / 2) + (sin(2x) / 4)) between π/2 and 0. This is equal to (π/4 + 0) - (0 + 0) = π/4. Hence the area specified in the question is equal to π/4 units squared.