Find the area of the region enclosed by the curves \( y = x^2 \) and \( y = 4x - x^2 \).
["# Find the Area of the Region Enclosed by the Curves ( y = x^2 ) and ( y = 4x - x^2 )", "Understanding how to compute the area enclosed by two intersecting curves is a fundamental concept in calculus and geometric analysis. In this article, we explore the precise steps to find the area bounded by the parabola ( y = x^2 ) and the quadratic curve ( y = 4x - x^2 ). This problem highlights key techniques such as finding intersection points, determining function dominance, and applying definite integration.", "---", "## Step 1: Find the Points of Intersection", "The region enclosed by the curves is bounded where they intersect. To find these points, set the two equations equal:", "[\nx^2 = 4x - x^2\n]", "Solve for ( x ):", "[\nx^2 + x^2 - 4x = 0 \implies 2x^2 - 4x = 0 \implies 2x(x - 2) = 0\n]", "Thus, the solutions are:", "[\nx = 0 \quad \ ext{and} \quad x = 2\n]", "The curves intersect at ( x = 0 ) and ( x = 2 ). These are the left and right boundaries of the enclosed region.", "---", "## Step 2: Determine Which Curve Rises Above the Other", "To compute the area between the curves from ( x = 0 ) to ( x = 2 ), determine which function lies above the other in this interval.", "Evaluate both functions at a test point, such as ( x = 1 ):", "- ( y = x^2 = 1^2 = 1 )\n- ( y = 4x - x^2 = 4(1) - (1)^2 = 3 )", "Since ( 3 > 1 ), the curve ( y = 4x - x^2 ) lies above ( y = x^2 ) on the interval ([0, 2]).", "---", "## Step 3: Set Up the Integral Expression", "The area ( A ) between the curves from ( x = 0 ) to ( x = 2 ) is given by the integral of the top function minus the bottom function:", "[\nA = \int_{0}^{2} \left[(4x - x^2) - x^2\right] , dx\n]", "Simplify the integrand:", "[\nA = \int_{0}^{2} (4x - 2x^2) , dx\n]", "---", "## Step 4: Compute the Integral", "Integrate term by term:", "[\n\int (4x - 2x^2) , dx = 4 \cdot \frac{x^2}{2} - 2 \cdot \frac{x^3}{3} = 2x^2 - \frac{2}{3}x^3\n]", "Now evaluate from 0 to 2:", "[\nA = \left[2x^2 - \frac{2}{3}x^3\right]_0^2 = \left(2(2)^2 - \frac{2}{3}(2)^3\right) - 0\n]", "Calculate:", "[\n= \left(2 \cdot 4 - \frac{2}{3} \cdot 8\right) = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}\n]", "---", "## Final Result", "The area of the region enclosed by the curves ( y = x^2 ) and ( y = 4x - x^2 ) is:", "[\n\boxed{\frac{8}{3}} \ ext{ square units}\n]", "---", "### Why This Matters", "This problem demonstrates a powerful application of algebra and integration to quantify geometric spaces bounded by non-linearity—essential in engineering, physics, and data visualization. Understanding how to derive intersection points, analyze function dominance, and perform definite integrals builds the foundation for advanced analysis of curves in real-world modeling.", "For further exploration, try finding the area between other pairs of curves or investigating regions bounded by one curve and a straight line. The principles remain consistent, and practice deepens mastery in integral calculus."]
