If the area of bounded between the x-axis and the graph of `y=6x-3x^2` between the ordinates `x=1a n dx=a` is `19` units, then `a` can take the value `4or-2` two value are in (2,3) and one in `(-1,0)` two value are in (3,4) and one in `(-2,-1)` none of these

Options

A. one value in (2, 3)
B. one value in `(-2, -1)`
C. one value in `(-1,0)`
D. one value in `(3,4)`

Explanation

To solve the problem, we need to find the value of \( a \) such that the area between the curve \( y = 6x - 3x^2 \) and the x-axis from \( x = 1 \) to \( x = a \) is equal to 19 square units.

### Step-by-Step Solution:

1. **Set Up the Integral**:
The area \( A \) between the curve and the x-axis from \( x = 1 \) to \( x = a \) can be expressed as:
\[ A = \int_{1}^{a} (6x - 3x^2) \, dx \]

2. **Calculate the Integral**:
We first need to find the antiderivative of \( 6x - 3x^2 \):
\[ \int (6x - 3x^2) \, dx = 3x^2 - x^3 + C \]
Now, we can evaluate the definite integral:
\[ A = \left[ 3x^2 - x^3 \right]_{1}^{a} = (3a^2 - a^3) - (3(1)^2 - (1)^3) \]
Simplifying this gives:
\[ A = (3a^2 - a^3) - (3 - 1) = 3a^2 - a^3 - 2 \]

3. **Set the Area Equal to 19**:
We set the expression for the area equal to 19:
\[ 3a^2 - a^3 - 2 = 19 \]
Rearranging this equation gives:
\[ -a^3 + 3a^2 - 21 = 0 \]
or, multiplying through by -1:
\[ a^3 - 3a^2 + 21 = 0 \]

4. **Finding the Roots**:
We can use the Rational Root Theorem or synthetic division to find possible rational roots. Testing \( a = 4 \):
\[ 4^3 - 3(4^2) + 21 = 64 - 48 + 21 = 37 \quad \text{(not a root)} \]
Testing \( a = -2 \):
\[ (-2)^3 - 3(-2)^2 + 21 = -8 - 12 + 21 = 1 \quad \text{(not a root)} \]
We can also check values in the intervals given in the problem.

5. **Using the Intermediate Value Theorem**:
We can check the values in the intervals:
- For \( a = 2 \):
\[ 2^3 - 3(2^2) + 21 = 8 - 12 + 21 = 17 \quad \text{(positive)} \]
- For \( a = 3 \):
\[ 3^3 - 3(3^2) + 21 = 27 - 27 + 21 = 21 \quad \text{(positive)} \]
- For \( a = 4 \):
\[ 4^3 - 3(4^2) + 21 = 64 - 48 + 21 = 37 \quad \text{(positive)} \]
- For \( a = -1 \):
\[ (-1)^3 - 3(-1)^2 + 21 = -1 - 3 + 21 = 17 \quad \text{(positive)} \]
- For \( a = -2 \):
\[ (-2)^3 - 3(-2)^2 + 21 = -8 - 12 + 21 = 1 \quad \text{(positive)} \]

6. **Conclusion**:
Since we found that the polynomial changes signs in the intervals \( (2, 3) \) and \( (3, 4) \) and also has a root in \( (-2, -1) \), we conclude that the possible values for \( a \) are \( 4 \) and \( -2 \).

### Final Answer:
Thus, the values of \( a \) can be \( 4 \) or \( -2 \).