Question:

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

Updated On: Nov 19, 2024
  • \( \frac{32}{9} \)
  • \( 4 \)
  • \( 6 \)
  • \( \frac{14}{3} \)
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The Correct Option is C

Solution and Explanation

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

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