The area (in sq. units) of the region bounded by the X-axis and the curve `y=1-x-6x^(2)` is

Options

A. `125/216`
B. `125/512`
C. `25/216`
D. `25/512`

Answer

To find the area of the region bounded by the X-axis and the curve \( y = 1 - x - 6x^2 \), we will follow these steps:

### Step 1: Find the Points of Intersection
We need to determine where the curve intersects the X-axis. This occurs when \( y = 0 \):

\[ 1 - x - 6x^2 = 0 \]

Rearranging gives us:

\[ 6x^2 + x - 1 = 0 \]

### Step 2: Solve the Quadratic Equation
We can solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6, b = 1, c = -1 \).

Calculating the discriminant:

\[ b^2 - 4ac = 1^2 - 4 \cdot 6 \cdot (-1) = 1 + 24 = 25 \]

Now applying the quadratic formula:

\[ x = \frac{-1 \pm \sqrt{25}}{2 \cdot 6} = \frac{-1 \pm 5}{12} \]

This gives us two solutions:

\[ x_1 = \frac{4}{12} = \frac{1}{3}, \quad x_2 = \frac{-6}{12} = -\frac{1}{2} \]

### Step 3: Set Up the Integral
The area \( A \) between the curve and the X-axis from \( x = -\frac{1}{2} \) to \( x = \frac{1}{3} \) is given by:

\[ A = \int_{-\frac{1}{2}}^{\frac{1}{3}} (1 - x - 6x^2) \, dx \]

### Step 4: Compute the Integral
Now we compute the integral:

\[ A = \int_{-\frac{1}{2}}^{\frac{1}{3}} (1 - x - 6x^2) \, dx \]

Calculating the integral:

\[ = \left[ x - \frac{x^2}{2} - 2x^3 \right]_{-\frac{1}{2}}^{\frac{1}{3}} \]

### Step 5: Evaluate the Integral at the Limits
First, we evaluate at \( x = \frac{1}{3} \):

\[ = \frac{1}{3} - \frac{(\frac{1}{3})^2}{2} - 2\left(\frac{1}{3}\right)^3 \]
\[ = \frac{1}{3} - \frac{1/9}{2} - 2 \cdot \frac{1}{27} \]
\[ = \frac{1}{3} - \frac{1}{18} - \frac{2}{27} \]

Finding a common denominator (which is 54):

\[ = \frac{18}{54} - \frac{3}{54} - \frac{4}{54} = \frac{18 - 3 - 4}{54} = \frac{11}{54} \]

Now evaluate at \( x = -\frac{1}{2} \):

\[ = -\frac{1}{2} - \frac{(-\frac{1}{2})^2}{2} - 2\left(-\frac{1}{2}\right)^3 \]
\[ = -\frac{1}{2} - \frac{1/4}{2} + \frac{1}{4} \]
\[ = -\frac{1}{2} - \frac{1}{8} + \frac{1}{4} \]

Finding a common denominator (which is 8):

\[ = -\frac{4}{8} - \frac{1}{8} + \frac{2}{8} = -\frac{4 + 1 - 2}{8} = -\frac{3}{8} \]

### Step 6: Combine the Results
Now we combine the results from both limits:

\[ A = \left(\frac{11}{54}\right) - \left(-\frac{3}{8}\right) \]

Finding a common denominator (which is 216):

\[ = \frac{11 \cdot 4}{216} + \frac{3 \cdot 27}{216} = \frac{44 + 81}{216} = \frac{125}{216} \]

### Final Answer
Thus, the area of the region bounded by the X-axis and the curve is:

\[ \boxed{\frac{125}{216}} \text{ square units.} \]