Question

\( \int_{-1}^{1} \max\{x, x^3\} \, dx \) is equal to:

• A. \( \frac{3}{4} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( 1 \)
• E. \( 0 \)

Hint:

Always sketch or test values for "max/min" functions. The intersection points (\( x^3 = x \implies x = 0, \pm 1 \)) are the natural boundaries for splitting your definite integral.

Answer 1

The Correct Option is B

Concept: The function \( f(x) = \max\{x, x^3\} \) selects the larger of the two values at any given point \( x \). To integrate this over the interval \( [-1, 1] \), we must divide the integral into sub-intervals based on which function is greater in those specific regions.

Step 1: Compare \( x \) and \( x^3 \) over the interval \( [-1, 1] \).
Analyze the behavior:
• For \( x \in [-1, 0] \): \( x^3 \geq x \) (e.g., at \( x = -0.5 \), \( -0.125 > -0.5 \)). So, \( \max\{x, x^3\} = x^3 \).
• For \( x \in [0, 1] \): \( x \geq x^3 \) (e.g., at \( x = 0.5 \), \( 0.5 > 0.125 \)). So, \( \max\{x, x^3\} = x \).

Step 2: Split the integral and evaluate.
\[ \int_{-1}^{1} \max\{x, x^3\} \, dx = \int_{-1}^{0} x^3 \, dx + \int_{0}^{1} x \, dx \] Evaluate the first part: \[ \left[ \frac{x^4}{4} \right]_{-1}^{0} = \frac{0^4}{4} - \frac{(-1)^4}{4} = -\frac{1}{4} \] Evaluate the second part: \[ \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

Step 3: Sum the results.
\[ \text{Total Integral} = -\frac{1}{4} + \frac{1}{2} = -\frac{1}{4} + \frac{2}{4} = \frac{1}{4} \]