If the area bounded by the curves `{(x, y)|x^(2)-y+1 ge 0} and {(x, y)|x+y-3 ge0}` is k square units, then the value of 3k is equal to

Options

A. 27
B. 9
C. `(9)/(2)`
D. `(27)/(2)`

Answer

To solve the problem, we need to find the area bounded by the curves defined by the equations \(x^2 - y + 1 \geq 0\) and \(x + y - 3 \geq 0\).

### Step-by-Step Solution:

1. **Rewrite the Inequalities**:
- The first inequality \(x^2 - y + 1 \geq 0\) can be rewritten as \(y \leq x^2 + 1\).
- The second inequality \(x + y - 3 \geq 0\) can be rewritten as \(y \geq 3 - x\).

2. **Find Points of Intersection**:
- To find the area bounded by these curves, we first need to find their points of intersection. Set the equations equal to each other:
\[ x^2 + 1 = 3 - x \]
- Rearranging gives:
\[ x^2 + x - 2 = 0 \]
- Factor the quadratic:
\[ (x - 1)(x + 2) = 0 \]
- Thus, the solutions are \(x = 1\) and \(x = -2\).

3. **Find Corresponding y-values**:
- For \(x = 1\):
\[ y = 1^2 + 1 = 2 \]
- For \(x = -2\):
\[ y = (-2)^2 + 1 = 5 \]
- The points of intersection are \((-2, 5)\) and \((1, 2)\).

4. **Set Up the Integral for Area**:
- The area \(A\) between the curves from \(x = -2\) to \(x = 1\) is given by:
\[ A = \int_{-2}^{1} \left[(3 - x) - (x^2 + 1)\right] \, dx \]
- Simplifying the integrand:
\[ A = \int_{-2}^{1} (3 - x - x^2 - 1) \, dx = \int_{-2}^{1} (2 - x - x^2) \, dx \]

5. **Evaluate the Integral**:
- Now, we compute the integral:
\[ A = \int_{-2}^{1} (2 - x - x^2) \, dx \]
- The antiderivative is:
\[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \]
- Evaluating from \(-2\) to \(1\):
\[ A = \left[2(1) - \frac{1^2}{2} - \frac{1^3}{3}\right] - \left[2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3}\right] \]
- Calculate the upper limit:
\[ = 2 - \frac{1}{2} - \frac{1}{3} = \frac{12}{6} - \frac{3}{6} - \frac{2}{6} = \frac{7}{6} \]
- Calculate the lower limit:
\[ = -4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = -\frac{18}{3} + \frac{8}{3} = -\frac{10}{3} \]
- Thus:
\[ A = \frac{7}{6} - \left(-\frac{10}{3}\right) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2} \]

6. **Find \(3k\)**:
- Given that \(k = \frac{9}{2}\), we find:
\[ 3k = 3 \times \frac{9}{2} = \frac{27}{2} \]

### Final Answer:
The value of \(3k\) is \(\frac{27}{2}\).