To determine the area of the region defined by the given inequalities and expression, we proceed step-by-step.
Region Definition:
The inequality \(y^2 \leq 4x\) defines the region below the parabola \(y^2 = 4x\), which opens to the right with its vertex at the origin.
The condition \(x<4\) restricts the region to the left of the vertical line \(x = 4\).
The expression \(\frac{xy(x - 1)(x - 2)}{(x - 3)(x - 4)}>0\) requires that the numerator and denominator are either both positive or both negative.
Identifying Valid x-intervals:
The critical points for the numerator are \(x = 0, 1, 2\).
The critical points for the denominator are \(x = 3, 4\).
Since \(x eq 3\) to avoid division by zero, the x-axis is divided into the intervals: \((-\infty, 0), (0, 1), (1, 2), (2, 3), (3, 4)\).
Evaluating Expression Sign:
The sign of the expression is evaluated in each interval:
Interval \((0, 1):\): The expression is positive.
Interval \((1, 2):\): The expression is positive.
Interval \((2, 3):\): The expression is positive.
Interval \((3, 4):\): The expression is negative.
Area Calculation:
The valid x-intervals where the inequality holds are \((0, 1), (1, 2), (2, 3)\).
The inequality \(y^2 \leq 4x\) implies \(-\sqrt{4x} \leq y \leq \sqrt{4x}\). For a given x, the length of the vertical segment is \(2\sqrt{4x}\).
The area is calculated by integrating this length over the valid x-intervals. Considering the intervals \((0,1), (1,2), (2,3)\), the integration is from \(x=0\) to \(x=3\).
The integral of \(2\sqrt{4x}\) from \(x=0\) to \(x=3\) is calculated as follows:
The computed area from \(x=0\) to \(x=3\) is:
The total area is obtained by doubling this result due to symmetry across the x-axis (or by summing the areas from the individual positive intervals).
