Area between the parabolas y = x² and y = 2x − x²

This problem asks to find the area of a plane region bounded by two parabolas, using a double integral.

Problem

Find the area of the region D bounded by the curves:

y = x^{2}, y = 2 x - x^{2}

Solution

Step 1

Find the intersection points of the parabolas by equating their equations:

Why find the intersection points?

The intersection points determine the limits of integration in x. Without them, it is impossible to set the bounds of the region D correctly.

x^{2} = 2 x - x^{2} \Rightarrow 2 x^{2} - 2 x = 0 \Rightarrow 2 x (x - 1) = 0

x_{1} = 0, x_{2} = 1

Step 2

Find the corresponding y values by substituting x into either equation:

O (0; 0), A (1; 1)

Step 3

Determine the region D. On the segment [0; 1] the parabola y = 2x − x² lies above the parabola y = x², so the region is given by:

D : 0 \leq x \leq 1, x^{2} \leq y \leq 2 x - x^{2}

Why is y = 2x − x² above?

The vertex of y = 2x − x² = 1 − (x−1)² is at (1; 1), branches pointing downward. The vertex of y = x² is at (0; 0), branches pointing upward. Between the intersection points the first lies above the second — easily checked by substituting x = 0.5.

Step 4

The area of the region via a double integral:

S = \iint_{D} d x d y = \int_{0}^{1} d x \int_{x^{2}}^{2 x - x^{2}} d y

Step 5

Evaluate the inner integral over y:

\int_{x^{2}}^{2 x - x^{2}} d y = (2 x - x^{2}) - x^{2} = 2 x - 2 x^{2}

Step 6

After substituting the limits, we obtain a definite integral over x and evaluate it using the Newton–Leibniz formula:

S = \int_{0}^{1} (2 x - 2 x^{2}) d x = {[x^{2} - \frac{2 x^{3}}{3}]}_{0}^{1} = 1 - \frac{2}{3} = \frac{1}{3}

Answer

S = 1/3.

Frequently asked questions

How do I find the area of a region using a double integral?

The area of a region D is computed as S = ∬_D dx dy. The double integral is reduced to an iterated one by setting the limits along the boundaries: first over y (from the lower to the upper boundary), then over x.

How do I determine the order of traversing the region?

First find the intersection points of the boundary curves — they set the limits in x. Then, for each x, determine which curve is lower and which is upper — these are the limits in y. In our case 0 ≤ x ≤ 1 and x² ≤ y ≤ 2x − x².

What if the region is bounded by more than two curves?

In that case, split the region into several parts, each bounded by exactly two curves (upper and lower). Then compute the area of each part and sum the results.

Can this area be computed without a double integral?

Yes, the area between two curves on a segment [a; b] can be found as S = ∫_a^b (y_top(x) − y_bottom(x)) dx. This is essentially a collapsed iterated integral. But the double-integral method is more universal and works for more complex regions.