Question

The area of the region enclosed by the parabola $y = 4x - x^2$ and $3y = (x - 4)^2$ is equal to

• A. $\frac{32}{9}$
• B. $4$
• C. $6$
• D. $\frac{14}{3}$

Answer 1

Answer: (C)

$6$

Answer 2

Set up the equations. We have two curves:

$y = 4x - x^2$

and

$3y = (x - 4)^2.$

Rewrite the second equation for $y$:

$y = \frac{(x - 4)^2}{3}.$

Find the points of intersection. To find the points of intersection, set the two expressions for $y$ equal to each other:

$4x - x^2 = \frac{(x - 4)^2}{3}.$

Multiply through by 3 to eliminate the fraction:

$3(4x - x^2) = (x - 4)^2$ $12x - 3x^2 = x^2 - 8x + 16$

Bring all terms to one side of the equation:

$-4x^2 + 20x - 16 = 0.$

Divide by $-4$:

$x^2 - 5x + 4 = 0.$

Factor the quadratic equation:

$(x - 4)(x - 1) = 0.$

Thus, the points of intersection are $x = 1$ and $x = 4$.

Set up the integral for the area. The area enclosed by the curves from $x = 1$ to $x = 4$ is given by the integral of the difference between the upper and lower functions:

$\text{Area} = \int_{1}^{4} \left( (4x - x^2) - \frac{(x - 4)^2}{3} \right) dx.$

Simplify the integrand. Expand $\frac{(x - 4)^2}{3}$:

$\frac{(x - 4)^2}{3} = \frac{x^2 - 8x + 16}{3} = \frac{x^2}{3} - \frac{8x}{3} + \frac{16}{3}.$

Now rewrite the integrand:

$\left( 4x - x^2 - \frac{x^2}{3} + \frac{8x}{3} - \frac{16}{3} \right).$

Combine like terms:

$= \int_{1}^{4} \left( -\frac{4x^2}{3} + \frac{20x}{3} - \frac{16}{3} \right) dx.$

Integrate term by term. Now integrate each term separately:

$\int_{1}^{4} -\frac{4x^2}{3} dx = -\frac{4}{3} \left[ \frac{x^3}{3} \right]_{1}^{4} = -\frac{4}{9} (64 - 1) = -\frac{4 \times 63}{9} = -28.$

$\int_{1}^{4} \frac{20x}{3} dx = \frac{20}{3} \left[ \frac{x^2}{2} \right]_{1}^{4} = \frac{20}{3} \cdot \frac{15}{2} = 50.$

$\int_{1}^{4} -\frac{16}{3} dx = -\frac{16}{3} \cdot (4 - 1) = -\frac{16 \times 3}{3} = -16.$

Add the results:

$\text{Area} = -28 + 50 - 16 = 6.$

Thus, the answer is:

6