The area is \( \int_0^2 [(4x - x^2) - x^2] \, dx = \int_0^2 (4x - 2x^2) \, dx \).

["Exploring the Area Under the Curve: Solving ( \int_0^2 (4x - 2x^2) , dx )", "When it comes to calculating the area under a curve, integration is a powerful mathematical tool. One common problem in calculus involves finding the area between a function and the x-axis over a given interval. This article dives into the detailed solution of the definite integral:\n[\n\int_0^2 \left[(4x - x^2) - x^2\right] , dx = \int_0^2 (4x - 2x^2) , dx\n]", "---", "### Understanding the Integral Setup", "The expression ( (4x - x^2) - x^2 ) simplifies to ( 4x - 2x^2 ), as combining like terms yields:\n[\n4x - x^2 - x^2 = 4x - 2x^2\n]\nThis function defines a downward-opening parabola, and our goal is to find the area between the curve ( y = 4x - 2x^2 ) and the x-axis from ( x = 0 ) to ( x = 2 ).", "---", "### Step-by-Step Integration", "We compute the definite integral:\n[\n\int_0^2 (4x - 2x^2) , dx\n]", "#### 1. Compute the Antiderivative", "Find the indefinite integral:\n[\n\int (4x - 2x^2) , dx = 4 \int x , dx - 2 \int x^2 , dx = 4 \cdot \frac{x^2}{2} - 2 \cdot \frac{x^3}{3} + C = 2x^2 - \frac{2}{3}x^3 + C\n]", "#### 2. Evaluate the Definite Integral", "Now, evaluate the antiderivative at the bounds:\n[\n\left[2x^2 - \frac{2}{3}x^3\right]_0^2 = \left(2(2)^2 - \frac{2}{3}(2)^3\right) - \left(2(0)^2 - \frac{2}{3}(0)^3\right)\n]\n[\n= \left(2 \cdot 4 - \frac{2}{3} \cdot 8\right) - 0 = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}\n]", "---", "### Interpretation and Practical Insight", "The result ( \frac{8}{3} ) represents the net area between the parabola ( y = 4x - 2x^2 ) and the x-axis from ( x = 0 ) to ( x = 2 ). Since the function is non-negative over ( [0, 2] ) (as verified graphically or by testing values), the area equals the net value:\n[\n\ ext{Area} = \int_0^2 (4x - 2x^2) , dx = \frac{8}{3} \approx 2.67\n]", "This method highlights how integration quantifies area with precision, even for curved boundaries — essential in fields ranging from engineering to economics.", "---", "### Final Answer", "[\n\boxed{\int_0^2 (4x - 2x^2) , dx = \frac{8}{3}}\n]", "---", "Keywords: Definite integral, area under curve, calculus, definite integral calculation, ( \int_0^2 (4x - 2x^2) , dx ), integration technique, definite integral proof, function area, mathematical computation.", "---", "By mastering integrals like this, students and professionals unlock deeper insights into geometry, physics, and applied mathematics. This example demonstrates how algebra, limits, and continuity converge in calculus to solve real-world area problems efficiently."]