إيجاد المساحة بين القطع المكافئة

Problem

Find the area of the region enclosed between the parabolas $y = x^2$ and $y = 2x - x^2$.

Step 1: Find the intersection points

Set the two curves equal to each other to find where they meet:

$$x^2 = 2x - x^2$$

Move everything to one side:

$$2x^2 - 2x = 0$$

Factor:

$$2x(x - 1) = 0$$

So the intersection points occur at

$$x = 0 \quad \text{and} \quad x = 1$$

These correspond to the origin and the point where the curves cross again.

Step 2: Set up the area integral

The area between two curves is found by integrating top minus bottom. Here, the top curve is $2x - x^2$ and the bottom curve is $x^2$.

So the area is

$$\int_0^1 \bigl[(2x - x^2) - x^2\bigr]\,dx$$

which simplifies to

$$\int_0^1 (2x - 2x^2)\,dx$$

The parabola $2x - x^2$ stays above $x^2$ on the interval from $0$ to $1$.

Step 3: Evaluate the integral

Use the power rule and integrate term by term:

$$\int (2x - 2x^2)\,dx = x^2 - \frac{2}{3}x^3$$

Now evaluate from $0$ to $1$:

$$\left(x^2 - \frac{2}{3}x^3\right)\Bigg|_0^1 = \left(1 - \frac{2}{3}\right) - 0 = \frac{1}{3}$$

Answer

The exact area is $\dfrac{1}{3}$ square units.