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

The Correct Option is C

To find the area of the region enclosed by the two curves \( y = 4x - x^2 \) and \( 3y = (x - 4)^2 \), we need to follow these steps:

1. First, we need to convert the second equation \( 3y = (x - 4)^2 \) into \( y \) terms:

\[3y = (x - 4)^2 \quad \Rightarrow \quad y = \frac{(x - 4)^2}{3}\]

1. Equate both equations to find the points of intersection:

\[4x - x^2 = \frac{(x - 4)^2}{3}\]

Multiply throughout by 3 to clear the fraction:

\[12x - 3x^2 = (x - 4)^2\]

Expand the right side and rearrange terms:

\[12x - 3x^2 = x^2 - 8x + 16\]

Translate into standard form:

\[0 = 4x^2 - 20x + 16 \quad \Rightarrow \quad 2x^2 - 10x + 8 = 0\]

1. Solve the quadratic equation:

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -10, c = 8 \):

\[x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 8}}{2 \times 2}\]

 

\[x = \frac{10 \pm \sqrt{100 - 64}}{4} \quad \Rightarrow \quad x = \frac{10 \pm \sqrt{36}}{4} \quad \Rightarrow \quad x = \frac{10 \pm 6}{4}\]

This results in \( x = \frac{16}{4} = 4 \) and \( x = \frac{4}{4} = 1 \).

1. Compute the area between the curves over the interval \([1, 4]\):

Calculate the definite integral of the difference between these functions:

\[\int_{1}^{4} \left[(4x - x^2) - \frac{(x - 4)^2}{3}\right] \, dx\]

Simplify the integrand:

\[\int_{1}^{4} \left[4x - x^2 - \left(\frac{x^2 - 8x + 16}{3}\right)\right] \, dx\]

 

\[= \int_{1}^{4} \left[\frac{12x - 3x^2 - x^2 + 8x - 16}{3}\right] \, dx\]

 

\[= \int_{1}^{4} \left[\frac{-4x^2 + 20x - 16}{3}\right] \, dx\]

1. Split the integral and compute each term:

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

1. Calculate each integral separately:

\[\int_{1}^{4} (-4x^2) \, dx = \left[-\frac{4}{3}x^3 \right]_{1}^{4} = -\frac{4}{3}(64) + \frac{4}{3}(1) = -\frac{256}{3} + \frac{4}{3} = -\frac{252}{3} = -84\]

 

\[\int_{1}^{4} (20x) \, dx = \left[10x^2 \right]_{1}^{4} = 10(16) - 10(1) = 160 - 10 = 150\]

 

\[\int_{1}^{4} (16) \, dx = \left[16x \right]_{1}^{4} = 16(4) - 16(1) = 64 - 16 = 48\]

1. Combine the results:

\[\frac{1}{3}(-84 + 150 - 48) = \frac{1}{3}(18) = 6\]

The area of the region enclosed by the curves is \(6\).

Hence, the correct answer is \( 6 \).

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