Area bounded by the curve `y=x^(2)+x-6` and the X-axis is

Options

A. `(127)/(6)`
B. `(225)/(6)`
C. `(125)/(6)`
D. None of these

Answer

To find the area bounded by the curve \( y = x^2 + x - 6 \) and the x-axis, we will follow these steps:

### Step 1: Find the points where the curve intersects the x-axis.
To find the points of intersection, we set \( y = 0 \):

\[ x^2 + x - 6 = 0 \]

### Step 2: Factor the quadratic equation.
We can factor the quadratic as follows:

\[ x^2 + 3x - 2x - 6 = 0 \]

Grouping the terms gives us:

\[ x(x + 3) - 2(x + 3) = 0 \]

Factoring out \( (x + 3) \):

\[ (x + 3)(x - 2) = 0 \]

### Step 3: Solve for \( x \).
Setting each factor to zero gives us:

\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

Thus, the curve intersects the x-axis at \( x = -3 \) and \( x = 2 \).

### Step 4: Set up the integral for the area.
The area \( A \) bounded by the curve and the x-axis from \( x = -3 \) to \( x = 2 \) is given by the integral:

\[ A = \int_{-3}^{2} (x^2 + x - 6) \, dx \]

### Step 5: Compute the integral.
Now we calculate the integral:

\[ A = \int (x^2 + x - 6) \, dx = \frac{x^3}{3} + \frac{x^2}{2} - 6x \]

### Step 6: Evaluate the definite integral.
We evaluate this from \( -3 \) to \( 2 \):

\[ A = \left[ \frac{2^3}{3} + \frac{2^2}{2} - 6 \cdot 2 \right] - \left[ \frac{(-3)^3}{3} + \frac{(-3)^2}{2} - 6 \cdot (-3) \right] \]

Calculating the upper limit:

\[ = \left[ \frac{8}{3} + 2 - 12 \right] = \left[ \frac{8}{3} + \frac{6}{3} - \frac{36}{3} \right] = \frac{8 + 6 - 36}{3} = \frac{-22}{3} \]

Calculating the lower limit:

\[ = \left[ \frac{-27}{3} + \frac{9}{2} + 18 \right] = \left[ -9 + \frac{9}{2} + 18 \right] = \left[ 9 - 9 + \frac{9}{2} \right] = \frac{9}{2} \]

### Step 7: Combine the results.
Now we combine the results:

\[ A = \left( \frac{-22}{3} \right) - \left( \frac{9}{2} \right) \]

Finding a common denominator (which is 6):

\[ A = \left( \frac{-44}{6} \right) - \left( \frac{27}{6} \right) = \frac{-44 - 27}{6} = \frac{-71}{6} \]

Since area cannot be negative, we take the absolute value:

\[ A = \frac{71}{6} \]

### Step 8: Final Answer
Thus, the area bounded by the curve and the x-axis is:

\[ \text{Area} = \frac{71}{6} \text{ square units} \]