Note: The provided question with \( y = x^5 \) results in an answer of \( \frac{1}{3} \), which is not among the given options. It is probable that the question intended to be \( y = x^3 \), as handwritten '3' and '5' can be mistaken for one another. The calculated answer for \( y = x^3 \) is present in the options. This solution proceeds under the assumption that the intended function was \( y = x^3 \).
Step 1: Conceptual Understanding:
The objective is to determine the area enclosed by a curve and the x-axis. As area is inherently non-negative, the integral of the absolute value of the function, \( |f(x)| \), must be computed. For the function \( y = x^3 \), the values are negative for \( x < 0 \) and positive for \( x > 0 \).
Step 2: Primary Formula/Method:
The area \( A \) is calculated via the definite integral of \( |x^3| \) from -1 to 1: \[ A = \int_{-1}^{1} |x^3| \, dx \] The integral must be divided at \( x = 0 \), the point where the function's sign changes.
Step 3: Detailed Calculation:
The integral is divided into two segments:
\[ A = \int_{-1}^{0} |x^3| \, dx + \int_{0}^{1} |x^3| \, dx \] For \( x \) in the interval \( [-1, 0) \), \( x^3 \) is negative, thus \( |x^3| = -x^3 \).
For \( x \) in the interval \( [0, 1] \), \( x^3 \) is positive, thus \( |x^3| = x^3 \).
\[ A = \int_{-1}^{0} (-x^3) \, dx + \int_{0}^{1} (x^3) \, dx \] Integration proceeds as follows: \[ A = \left[ -\frac{x^4}{4} \right]_{-1}^{0} + \left[ \frac{x^4}{4} \right]_{0}^{1} \] The first segment's evaluation: \[ \left( -\frac{0^4}{4} \right) - \left( -\frac{(-1)^4}{4} \right) = 0 - \left( -\frac{1}{4} \right) = \frac{1}{4} \] The second segment's evaluation: \[ \left( \frac{1^4}{4} \right) - \left( \frac{0^4}{4} \right) = \frac{1}{4} - 0 = \frac{1}{4} \] The total area is the sum of these segments: \[ A = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]
Step 4: Conclusive Result:
Under the assumption that the intended curve was \( y = x^3 \), the computed area is \( \frac{1}{2} \) square units.