To find the area of the region enclosed by the curves \( y = x^2 \) and \( y = 4x - x^2 \), we first determine the points of intersection by setting the equations equal:

["Title: How to Find the Area Enclosed by the Curves ( y = x^2 ) and ( y = 4x - x^2 ) – Step-by-Step Guide", "---", "Introduction\nUnderstanding how to compute the area enclosed by two curves is a fundamental skill in calculus, with applications across physics, engineering, and economics. In this article, we’ll walk through the process of finding the area bounded by the parabolas ( y = x^2 ) and ( y = 4x - x^2 ). We’ll start by identifying their points of intersection, then set up the integral to compute the enclosed area—perfect for students and learners seeking clarity on this essential topic.", "---", "### Step 1: Find the Points of Intersection\nTo determine the region enclosed by two curves, we first locate where they intersect. This requires solving:\n[\nx^2 = 4x - x^2\n]\nBring all terms to one side:\n[\nx^2 - (4x - x^2) = 0 \quad \Rightarrow \quad 2x^2 - 4x = 0\n]\nFactor the expression:\n[\n2x(x - 2) = 0\n]\nThus, the solutions are:\n[\nx = 0 \quad \ ext{and} \quad x = 2\n]\nThese x-values correspond to the endpoints of the region. Substituting back into either equation (say, ( y = x^2 )) gives the y-coordinates:\n- At ( x = 0 ): ( y = 0^2 = 0 )\n- At ( x = 2 ): ( y = 2^2 = 4 )", "So the curves intersect at points ( (0, 0) ) and ( (2, 4) ).", "---", "### Step 2: Identify Which Curve Is Above the Other\nTo find the area between the curves, we need to know which function lies above the other over the interval ([0, 2]).\nLet’s sample a point in the middle, say ( x = 1 ):\n- ( y = x^2 = 1^2 = 1 )\n- ( y = 4x - x^2 = 4(1) - 1^2 = 3 )", "Since ( 3 > 1 ), ( 4x - x^2 ) is above ( x^2 ) on ([0, 2]).", "---", "### Step 3: Set Up the Integral for Enclosed Area\nThe area ( A ) between two curves from ( x = a ) to ( x = b ), where ( f(x) \geq g(x) ), is given by:\n[\nA = \int_a^b \left[ f(x) - g(x) \right],dx\n]\nHere, ( f(x) = 4x - x^2 ) and ( g(x) = x^2 ), so:\n[\nA = \int_0^2 \left[(4x - x^2) - x^2\right],dx = \int_0^2 (4x - 2x^2),dx\n]", "---", "### Step 4: Evaluate the Integral\nNow compute the definite integral:\n[\nA = \int_0^2 (4x - 2x^2),dx = \left[ 2x^2 - \frac{2}{3}x^3 \right]_0^2\n]\nEvaluate at the upper limit:\n[\n2(2)^2 - \frac{2}{3}(2)^3 = 2(4) - \frac{2}{3}(8) = 8 - \frac{16}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}\n]\nAt the lower limit (0), the expression is 0.", "Thus, the enclosed area is:\n[\nA = \frac{8}{3} \ ext{ square units}\n]", "---", "### Conclusion\nFinding the area between two curves involves Locating intersection points, determining which function bounds the region above, setting up the proper integral, and evaluating it. For the curves ( y = x^2 ) and ( y = 4x - x^2 ), the enclosed area between ( x = 0 ) and ( x = 2 ) is exactly ( \frac{8}{3} ) square units. Mastering this process strengthens your calculus toolkit and prepares you for complex real-world problems involving area, volume, and motion.", "Keywords:\narea between curves, definite integral, calculus practice, find enclosed area, ( y = x^2 ), ( y = 4x - x^2 ), integration steps", "---", "Note: This step-by-step approach ensures clarity and reproducibility—ideal for students, teachers, and self-learners seeking to understand area between curves in calculus. For visualization, plot the two functions in Desmos or a graphing tool to observe the enclosed region visually."]