The area bounded by the curve `y=(x-1)(x-2)(x-3)` and `x`-axis lying between the ordinates `x = 0 and x = 4` is

Options

A. a) `(7)/(4)`
B. b) `(9)/(4)`
C. c) `0`
D. d) `(11)/(4)`

Answer

To find the area bounded by the curve \( y = (x-1)(x-2)(x-3) \) and the x-axis between the ordinates \( x = 0 \) and \( x = 4 \), we will follow these steps:

### Step 1: Identify the Roots of the Curve
The roots of the curve \( y = (x-1)(x-2)(x-3) \) are \( x = 1, 2, 3 \). This means the curve intersects the x-axis at these points.

### Step 2: Determine the Area Segments
We need to calculate the area in segments because the curve is below the x-axis between \( x = 0 \) and \( x = 1 \), above the x-axis between \( x = 1 \) and \( x = 2 \), below the x-axis between \( x = 2 \) and \( x = 3 \), and above the x-axis between \( x = 3 \) and \( x = 4 \).

### Step 3: Set Up the Integral for Each Segment
We will calculate the area for each segment using definite integrals:

1. **Area \( A_1 \) from \( x = 0 \) to \( x = 1 \)**:
\[ A_1 = -\int_0^1 (x-1)(x-2)(x-3) \, dx \]

2. **Area \( A_2 \) from \( x = 1 \) to \( x = 2 \)**:
\[ A_2 = \int_1^2 (x-1)(x-2)(x-3) \, dx \]

3. **Area \( A_3 \) from \( x = 2 \) to \( x = 3 \)**:
\[ A_3 = -\int_2^3 (x-1)(x-2)(x-3) \, dx \]

4. **Area \( A_4 \) from \( x = 3 \) to \( x = 4 \)**:
\[ A_4 = \int_3^4 (x-1)(x-2)(x-3) \, dx \]

### Step 4: Evaluate Each Integral
We first expand \( (x-1)(x-2)(x-3) \):
\[ y = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6 \]

Now we can evaluate each integral:

1. **For \( A_1 \)**:
\[ A_1 = -\int_0^1 (x^3 - 6x^2 + 11x - 6) \, dx \]
Evaluating this integral:
\[ = -\left[ \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x \right]_0^1 \]
\[ = -\left( \frac{1}{4} - 2 + \frac{11}{2} - 6 \right) = -\left( \frac{1}{4} - 8 + \frac{11}{2} \right) = -\left( \frac{1}{4} + \frac{22}{4} - \frac{32}{4} \right) = -\left( -\frac{9}{4} \right) = \frac{9}{4} \]

2. **For \( A_2 \)**:
\[ A_2 = \int_1^2 (x^3 - 6x^2 + 11x - 6) \, dx \]
Evaluating this integral:
\[ = \left[ \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x \right]_1^2 \]
\[ = \left( \frac{16}{4} - 16 + 22 - 12 \right) - \left( \frac{1}{4} - 2 + \frac{11}{2} - 6 \right) \]
\[ = (4 - 16 + 22 - 12) - (-\frac{9}{4}) = -2 + \frac{9}{4} = \frac{1}{4} \]

3. **For \( A_3 \)**:
\[ A_3 = -\int_2^3 (x^3 - 6x^2 + 11x - 6) \, dx \]
Evaluating this integral:
\[ = -\left[ \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x \right]_2^3 \]
\[ = -\left( \frac{81}{4} - 54 + \frac{99}{2} - 18 - \left( \frac{16}{4} - 16 + 22 - 12 \right) \right) \]
\[ = -\left( \frac{81}{4} - 54 + \frac{198}{4} - 18 + 2 \right) = -\left( \frac{279}{4} - 72 \right) = -\left( \frac{279 - 288}{4} \right) = \frac{9}{4} \]

4. **For \( A_4 \)**:
\[ A_4 = \int_3^4 (x^3 - 6x^2 + 11x - 6) \, dx \]
Evaluating this integral:
\[ = \left[ \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x \right]_3^4 \]
\[ = \left( 64 - 128 + 88 - 24 \right) - \left( \frac{81}{4} - 54 + \frac{99}{2} - 18 \right) \]
\[ = (64 - 128 + 88 - 24) - \left( \frac{279}{4} - 72 \right) = 0 + \frac{9}{4} = \frac{9}{4} \]

### Step 5: Calculate Total Area
Now we sum up all the areas:
\[ \text{Total Area} = A_1 + A_2 + A_3 + A_4 = \frac{9}{4} + \frac{1}{4} + \frac{9}{4} + \frac{9}{4} = \frac{28}{4} = 7 \]

### Final Answer
The area bounded by the curve and the x-axis between \( x = 0 \) and \( x = 4 \) is \( 7 \).