Question

Let $A$ be the area of the region
$\left\{(x, y): y \geq x^2, y \geq(1-x)^2, y \leq 2 x(1-x)\right\}$. 
Then $540 A$ is equal to ______.

Answer 1

Correct Answer: 25

$A=2\underset{\frac{1}{3}}{\overset{\frac{1}{2}}{\int }}\left(2x-2x^2-(1-x{)}^2\right)dx$
$=2{\left[2x^2-x^3-x\right]}_{1/3}^{1/2}$
$\therefore A=\frac{5}{108}\Rightarrow 540A=\frac{5}{108}\times 540=25$
So , the correct answer is 25.

Answer 2

We are given the region defined by:

{ (x, y) : y ≥ x², y ≥ (1 − x)², y ≤ 2x(1 − x) }

Here, the region is bounded from below by the curves y = x² and y = (1 − x)² and from above by y = 2x(1 − x). Since the conditions require y to be greater than or equal to both quadratic curves, the lower boundary of the region is the upper envelope of these curves:

Lower boundary = max{ x², (1 − x)² }

Step 1. Find Intersection Points

a) Intersection of y = x² and y = 2x(1 − x):

x² = 2x(1 − x) ⟹ x² = 2x − 2x² ⟹ 3x² − 2x = 0 ⟹ x(3x − 2) = 0

Thus, x = 0 or x = 2/3.

b) Intersection of y = (1 − x)² and y = 2x(1 − x):

(1 − x)² = 2x(1 − x)

If (1 − x) ≠ 0, divide by (1 − x):

1 − x = 2x ⟹ x = 1/3.

Also, note that when 1 − x = 0, x = 1. Thus, intersections occur at x = 1/3 and x = 1.

Step 2. Determine the x-Interval Where the Region Exists

The region is defined by:
y ≥ x² and y ≥ (1 − x)² with y ≤ 2x(1 − x). For the region to be nonempty, we need:

max{x², (1 − x)²} ≤ 2x(1 − x)

Analysis shows that the valid x-interval is from x = 1/3 to x = 2/3.
Moreover, the identity of the lower boundary changes:

• For x in [1/3, 1/2]: max{x², (1 − x)²} = (1 − x)².
• For x in [1/2, 2/3]: max{x², (1 − x)²} = x².

Step 3. Set Up the Area Integrals

The total area A is given by:

A = ∫[x=1/3]^[1/2] [2x(1 − x) − (1 − x)²] dx + ∫[x=1/2]^[2/3] [2x(1 − x) − x²] dx

Step 4. Evaluate the Integrals

Integral 1 (x in [1/3, 1/2]):

Simplify the integrand:

2x(1 − x) − (1 − x)² = (2x − 2x²) − (1 − 2x + x²) = 4x − 3x² − 1

Thus, we compute:

A₁ = ∫[1/3]^[1/2] (4x − 3x² − 1) dx

The antiderivative is:

F(x) = 2x² − x³ − x

Evaluate at x = 1/2:

F(1/2) = 2(1/4) − (1/8) − (1/2) = 1/2 − 1/8 − 1/2 = -1/8

Evaluate at x = 1/3:

F(1/3) = 2(1/9) − (1/27) − (1/3) = 2/9 − 1/27 − 1/3 = (6 − 1 − 9)/27 = -4/27

Thus,

A₁ = F(1/2) − F(1/3) = (-1/8) − (-4/27) = -1/8 + 4/27

Convert to a common denominator:

-1/8 = -27/216,  4/27 = 32/216  ⟹ A₁ = (32 − 27)/216 = 5/216

Integral 2 (x in [1/2, 2/3]):

Simplify the integrand:

2x(1 − x) − x² = 2x − 2x² − x² = 2x − 3x²

Thus, we compute:

A₂ = ∫[1/2]^[2/3] (2x − 3x²) dx

The antiderivative is:

G(x) = x² − x³

Evaluate at x = 2/3:

G(2/3) = (4/9) − (8/27) = (12 − 8)/27 = 4/27

Evaluate at x = 1/2:

G(1/2) = (1/4) − (1/8) = 1/8

Thus,

A₂ = G(2/3) − G(1/2) = 4/27 − 1/8

Converting to a common denominator:

4/27 = 32/216,  1/8 = 27/216  ⟹ A₂ = (32 − 27)/216 = 5/216

Step 5. Total Area and Final Answer

The total area is:

A = A₁ + A₂ = 5/216 + 5/216 = 10/216 = 5/108

We are asked to find 540A:

540A = 540 × (5/108) = (540/108) × 5 = 5 × 5 = 25

Hence, 540A = 25.