Question

Find the total area of the region bounded between the curve \( y = x^3 \), the x-axis, and the vertical lines \( x = -1 \) and \( x = 1 \).

• A. \( 0 \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( 1 \)

Hint:

Always look at the symmetry of the function. Because \( y = x^3 \) is an odd function, integrating it directly from \(-1\) to \(1\) gives 0. For geometric area, utilize its structural balance by integrating the positive slice and doubling it: \( 2 \times \int_{0}^{1} x^3 \, dx = \frac{1}{2} \).

Answer 1

The Correct Option is C

Concept: When calculating the geometric area under a curve that dips below the x-axis within the integration interval, you cannot integrate continuously across the entire range. Doing so allows negative values below the axis to cancel out positive values above it. Instead, you must split the integral at the origin crossing point and sum the absolute values of the separate region areas: \[ \text{Total Area} = \int_{a}^{c} |f(x)|\,dx + \int_{c}^{b} f(x)\,dx \]

Step 1: Identify the behavior of the cubic function across the boundaries.
The function \( y = x^3 \) crosses the x-axis at \( x = 0 \).

• Across the left interval section \( [-1, 0] \), the function values are negative (\( y \le 0 \)).
• Across the right interval section \( [0, 1] \), the function values are positive (\( y \ge 0 \)).

Step 2: Set up the split definite integration equations.
Apply absolute value adjustments to the negative area section by introducing an explicit negative sign: \[ \text{Area} = \int{-1}^{0} (-x^3) \, dx + \int_{0}^{1} x^3 \, dx \] Apply the power integration rule to evaluate each section: \[ = \left[ -\frac{x^4}{4} \right]_{-1}^{0} + \left[ \frac{x^4}{4} \right]_{0}^{1} \]

Step 3: Evaluate the limits to find the final area.
Substitute the boundary numbers into the integrated terms: \[ \text{Area} = \left( 0 - \left( -\frac{(-1)^4}{4} \right) \right) + \left( \frac{(1)^4}{4} - 0 \right) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \text{ square units} \]