Step 1: Understanding the Concept:
Calculus provides two different results for a function like \(x^3\) over a symmetric interval \([-1, 1]\). The definite integral \(\int_{-1}^{1} x^3 dx\) will result in \(0\) because \(x^3\) is an odd function, meaning the area below the x-axis (from \(-1\) to \(0\)) is treated as negative and cancels out the area above the x-axis (from \(0\) to \(1\)).
However, the "Total Area" in a geometric sense is the sum of magnitudes. Geometric area is always non-negative. To find the total physical area bounded between a curve and the x-axis, we must take the absolute value of the function before integrating, effectively reflecting the negative portion above the axis.
Key Formula or Approach:
The total area \(A\) is given by:
\[ \text{Area} = \int_{a}^{b} |f(x)| \, dx \]
We must split the integral at the points where the function crosses the x-axis (the roots). For \(x^3\), the crossing occurs at \(x=0\).
Step 2: Detailed Explanation:
The interval is \([-1, 1]\). We split this into \([-1, 0]\) and \([0, 1]\).
For interval \([-1, 0]\): The function \(y = x^3\) is negative.
\[ \text{Area}_1 = \left| \int_{-1}^{0} x^3 \, dx \right| = \left| \left[ \frac{x^4}{4} \right]_{-1}^{0} \right| = \left| 0 - \frac{(-1)^4}{4} \right| = \left| -\frac{1}{4} \right| = \frac{1}{4} \]
For interval \([0, 1]\): The function \(y = x^3\) is positive.
\[ \text{Area}_2 = \int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1^4}{4} - 0 = \frac{1}{4} \]
Total Area:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]
Step 3: Final Answer:
While the integral is zero, the geometric area is the sum of the absolute values of the areas of the two loops, resulting in \(1/2\) square units.