The area bounded by the curves `3y = (x- 4)^2` and `y = 4x - x^2` is

Options

A. 10
B. 6
C. 14
D. 27

Answer

To find the area bounded by the curves \(3y = (x - 4)^2\) and \(y = 4x - x^2\), we will follow these steps:

### Step 1: Rewrite the equations
First, we rewrite the equations in terms of \(y\):
1. From \(3y = (x - 4)^2\), we get:
\[ y = \frac{(x - 4)^2}{3} \]
2. The second equation is already in the required form:
\[ y = 4x - x^2 \]

### Step 2: Find points of intersection
To find the points of intersection, we set the two equations equal to each other:
\[ \frac{(x - 4)^2}{3} = 4x - x^2 \]
Multiplying both sides by 3 to eliminate the fraction:
\[ (x - 4)^2 = 12x - 3x^2 \]
Expanding the left side:
\[ x^2 - 8x + 16 = 12x - 3x^2 \]
Rearranging all terms to one side gives:
\[ 4x^2 - 20x + 16 = 0 \]
Dividing the entire equation by 4:
\[ x^2 - 5x + 4 = 0 \]
Factoring the quadratic:
\[ (x - 1)(x - 4) = 0 \]
Thus, the points of intersection are:
\[ x = 1 \quad \text{and} \quad x = 4 \]

### Step 3: Set up the integral for the area
The area \(A\) between the curves from \(x = 1\) to \(x = 4\) is given by:
\[ A = \int_{1}^{4} \left( (4x - x^2) - \frac{(x - 4)^2}{3} \right) dx \]

### Step 4: Simplify the integrand
Now we simplify the integrand:
\[ A = \int_{1}^{4} \left( 4x - x^2 - \frac{(x^2 - 8x + 16)}{3} \right) dx \]
Distributing the \(\frac{1}{3}\):
\[ A = \int_{1}^{4} \left( 4x - x^2 - \frac{x^2}{3} + \frac{8x}{3} - \frac{16}{3} \right) dx \]
Combining like terms:
\[ A = \int_{1}^{4} \left( \left(4 - \frac{8}{3}\right)x - \left(1 + \frac{1}{3}\right)x^2 + \frac{16}{3} \right) dx \]
This simplifies to:
\[ A = \int_{1}^{4} \left( \frac{12 - 8}{3}x - \frac{4}{3}x^2 + \frac{16}{3} \right) dx \]
\[ A = \int_{1}^{4} \left( \frac{4}{3}x - \frac{4}{3}x^2 + \frac{16}{3} \right) dx \]

### Step 5: Integrate
Now we can integrate:
\[ A = \frac{1}{3} \int_{1}^{4} \left( 4x - 4x^2 + 16 \right) dx \]
Calculating the integral:
\[ A = \frac{1}{3} \left[ 2x^2 - \frac{4}{3}x^3 + 16x \right]_{1}^{4} \]
Evaluating at the bounds:
\[ = \frac{1}{3} \left[ \left(2(4^2) - \frac{4}{3}(4^3) + 16(4)\right) - \left(2(1^2) - \frac{4}{3}(1^3) + 16(1)\right) \right] \]
Calculating each term:
\[ = \frac{1}{3} \left[ \left(32 - \frac{64}{3} + 64\right) - \left(2 - \frac{4}{3} + 16\right) \right] \]
\[ = \frac{1}{3} \left[ \left(32 + 64 - \frac{64}{3}\right) - \left(18 - \frac{4}{3}\right) \right] \]
Combining and simplifying gives:
\[ = \frac{1}{3} \left[ 96 - \frac{64}{3} - 18 + \frac{4}{3} \right] \]
\[ = \frac{1}{3} \left[ 78 - \frac{60}{3} \right] \]
\[ = \frac{1}{3} \left[ 78 - 20 \right] = \frac{1}{3} \times 58 = \frac{58}{3} \]

### Final Area Calculation
After evaluating the integral, we find the area bounded by the curves is:
\[ A = 6 \]