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 determine the area bounded by the curves \( y = 4x - x^2 \) and \( 3y = (x - 4)^2 \), the following procedure is employed:

1. The second equation, \( 3y = (x - 4)^2 \), is rewritten in terms of \( y \):

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

1. The points of intersection are found by equating the expressions for \( y \):

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

Multiplying by 3 to eliminate the denominator yields:

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

Expanding the right side and rearranging terms gives:

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

This simplifies to the standard quadratic form:

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

1. The quadratic equation is solved for \( x \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2, b = -10, c = 8 \):

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

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

The solutions are \( x = \frac{16}{4} = 4 \) and \( x = \frac{4}{4} = 1 \).

1. The area between the curves is computed over the interval \([1, 4]\) by integrating the difference between the functions:

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

The integrand is simplified as follows:

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

 

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

 

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

1. The integral is separated into individual terms for computation:

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

1. Each integral is evaluated:

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

 

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

 

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

1. The results are combined:

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

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

Therefore, the area is \( 6 \).