The total area between `y=4x-x^(2) and y=x` from x = 0 and x = 4 is

Options

A. `17//3`
B. `37//6`
C. `19/3`
D. 4

Answer

To find the total area between the curves \(y = 4x - x^2\) and \(y = x\) from \(x = 0\) to \(x = 4\), we will follow these steps:

### Step 1: Find the points of intersection
To find the points where the two curves intersect, we set \(4x - x^2 = x\).

\[ 4x - x^2 = x \]

Rearranging gives:

\[ -x^2 + 4x - x = 0 \implies -x^2 + 3x = 0 \]

Factoring out \(x\):

\[ x(-x + 3) = 0 \]

This gives us the solutions:

\[ x = 0 \quad \text{and} \quad x = 3 \]

### Step 2: Set up the area integrals
The area between the curves from \(x = 0\) to \(x = 3\) can be found by integrating the difference of the upper curve and the lower curve. Here, \(y = 4x - x^2\) is above \(y = x\) in this interval.

The area \(A_1\) from \(x = 0\) to \(x = 3\) is given by:

\[ A_1 = \int_0^3 ((4x - x^2) - x) \, dx = \int_0^3 (4x - x^2 - x) \, dx = \int_0^3 (3x - x^2) \, dx \]

### Step 3: Calculate \(A_1\)
Now we calculate the integral:

\[ A_1 = \int_0^3 (3x - x^2) \, dx \]

Calculating the integral:

\[ = \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_0^3 \]

Evaluating at the limits:

\[ = \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) - \left( \frac{3(0)^2}{2} - \frac{(0)^3}{3} \right) \]

\[ = \left( \frac{27}{2} - 9 \right) = \frac{27}{2} - \frac{18}{2} = \frac{9}{2} \]

### Step 4: Calculate the area from \(x = 3\) to \(x = 4\)
For the interval from \(x = 3\) to \(x = 4\), the upper curve is \(y = x\) and the lower curve is \(y = 4x - x^2\).

The area \(A_2\) from \(x = 3\) to \(x = 4\) is given by:

\[ A_2 = \int_3^4 (x - (4x - x^2)) \, dx = \int_3^4 (x - 4x + x^2) \, dx = \int_3^4 (x^2 - 3x) \, dx \]

### Step 5: Calculate \(A_2\)
Now we calculate the integral:

\[ A_2 = \int_3^4 (x^2 - 3x) \, dx \]

Calculating the integral:

\[ = \left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_3^4 \]

Evaluating at the limits:

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

Calculating each term:

\[ = \left( \frac{64}{3} - \frac{48}{2} \right) - \left( \frac{27}{3} - \frac{27}{2} \right) \]

\[ = \left( \frac{64}{3} - 24 \right) - \left( 9 - \frac{27}{2} \right) \]

Finding a common denominator for each part:

\[ = \left( \frac{64}{3} - \frac{72}{3} \right) - \left( \frac{18}{2} - \frac{27}{2} \right) \]

\[ = \left( \frac{-8}{3} \right) - \left( \frac{-9}{2} \right) \]

Finding a common denominator for the subtraction:

\[ = \frac{-8 \cdot 2 + 9 \cdot 3}{6} = \frac{-16 + 27}{6} = \frac{11}{6} \]

### Step 6: Total Area
Now, we add the two areas together:

\[ \text{Total Area} = A_1 + A_2 = \frac{9}{2} + \frac{11}{6} \]

Finding a common denominator (which is 6):

\[ = \frac{27}{6} + \frac{11}{6} = \frac{38}{6} = \frac{19}{3} \]

### Final Answer
The total area between the curves from \(x = 0\) to \(x = 4\) is:

\[ \boxed{\frac{19}{3}} \]