Step 1: Understanding the Concept:
To find the area bounded by a curve and the X-axis, we first need to determine where the curve intersects the axis.
The intersections occur where \( y = 0 \). These points define the intervals of integration.
Because "area" is a positive physical quantity, we must be careful with regions where the curve lies below the X-axis (\( y<0 \)).
In such regions, the definite integral will be negative. We must take the absolute value of the integral for those parts to ensure we are adding positive areas.
The given function is a cubic polynomial, which typically changes sign at its roots.
Step 2: Key Formula or Approach:
The total area \( A \) is given by:
\[ A = \int_{a}^{b} |f(x)| \, dx \]
We first expand the expression:
\[ y = x(x^2 - x - 2) = x^3 - x^2 - 2x \]
Find the roots by setting \( y = 0 \):
\[ x(x - 2)(x + 1) = 0 \implies x = -1, 0, 2 \]
The intervals are \( [-1, 0] \) and \( [0, 2] \).
Step 3: Detailed Explanation:
First, check the sign of the function in each interval to decide the integration setup.
For \( x \in (-1, 0) \), let \( x = -0.5 \): \( y = (-0.5)(-2.5)(0.5) = 0.625>0 \). (Above X-axis)
For \( x \in (0, 2) \), let \( x = 1 \): \( y = (1)(-1)(2) = -2<0 \). (Below X-axis)
The total area is:
\[ \text{Area} = \int_{-1}^{0} (x^3 - x^2 - 2x) \, dx + \left| \int_{0}^{2} (x^3 - x^2 - 2x) \, dx \right| \]
Let's compute the indefinite integral first:
\[ \int (x^3 - x^2 - 2x) \, dx = \frac{x^4}{4} - \frac{x^3}{3} - x^2 \]
Now, apply limits for the first part \( I_1 \) (Interval \([-1, 0]\)):
\[ I_1 = \left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 \right]_{-1}^{0} = (0) - \left( \frac{1}{4} + \frac{1}{3} - 1 \right) = - \left( \frac{3+4-12}{12} \right) = \frac{5}{12} \]
Apply limits for the second part \( I_2 \) (Interval \([0, 2]\)):
\[ I_2 = \left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 \right]_{0}^{2} = \left( \frac{16}{4} - \frac{8}{3} - 4 \right) - (0) = 4 - \frac{8}{3} - 4 = -\frac{8}{3} \]
The area from the second part is \( |I_2| = \frac{8}{3} = \frac{32}{12} \).
Total Area:
\[ \text{Area} = \frac{5}{12} + \frac{32}{12} = \frac{37}{12} \]
Step 4: Final Answer:
The total area is \( \frac{37}{12} \) square units.
This matches option (A).