Question

What is the value of $\int_{-1}^{2} \min\{1 - x, 1 - x^3\} \, dx$?

• A. -1
• B. 0
• C. 1
• D. 2

Hint:

Whenever you have a $\min(f(x), g(x))$ or $\max(f(x), g(x))$ term in an integral, first find their intersection points by solving $f(x) = g(x)$.
These intersection points will define the limits of the piecewise intervals for integration.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are required to evaluate a definite integral from $x = -1$ to $x = 2$.
The integrand is defined as the minimum of two functions: $f(x) = 1 - x$ and $g(x) = 1 - x^3$.
To integrate, we must determine which function is smaller on different sub-intervals within $[-1, 2]$.

Step 2: Key Formula or Approach: We compare the two functions by analyzing the inequality:
\[ 1 - x \le 1 - x^3 \] This simplifies to:
\[ x^3 - x \le 0 \implies x(x^2 - 1) \le 0 \implies x(x - 1)(x + 1) \le 0 \] Using the wavy curve method, we find the intervals where $1 - x$ is smaller than or equal to $1 - x^3$, and vice versa.

Step 3: Detailed Explanation:

• Let us analyze the sign of $P(x) = x(x - 1)(x + 1)$ on $[-1, 2]$:
The critical points are $x = -1, 0, 1$.

• For $x \in [-1, 0]$:
$P(x) \ge 0 \implies x^3 - x \ge 0 \implies 1 - x \ge 1 - x^3$.
Thus, $\min\{1 - x, 1 - x^3\} = 1 - x^3$.

• For $x \in [0, 1]$:
$P(x) \le 0 \implies x^3 - x \le 0 \implies 1 - x \le 1 - x^3$.
Thus, $\min\{1 - x, 1 - x^3\} = 1 - x$.

• For $x \in [1, 2]$:
$P(x) \ge 0 \implies x^3 - x \ge 0 \implies 1 - x \ge 1 - x^3$.
Thus, $\min\{1 - x, 1 - x^3\} = 1 - x^3$.

• Now, we express the total integral as the sum of three integrals over these sub-intervals:
\[ I = \int_{-1}^{0} (1 - x^3) \, dx + \int_{0}^{1} (1 - x) \, dx + \int_{1}^{2} (1 - x^3) \, dx \]
• Let us evaluate each integral separately: 1. For the first sub-interval $[-1, 0]$:
\[ I_1 = \left[ x - \frac{x^4}{4} \right]_{-1}^{0} = 0 - \left( -1 - \frac{1}{4} \right) = \frac{5}{4} \] 2. For the second sub-interval $[0, 1]$:
\[ I_2 = \left[ x - \frac{x^2}{2} \right]_{0}^{1} = \left( 1 - \frac{1}{2} \right) - 0 = \frac{1}{2} \] 3. For the third sub-interval $[1, 2]$:
\[ I_3 = \left[ x - \frac{x^4}{4} \right]_{1}^{2} = \left( 2 - \frac{16}{4} \right) - \left( 1 - \frac{1}{4} \right) = (2 - 4) - \frac{3}{4} = -2 - \frac{3}{4} = -\frac{11}{4} \]
• Summing these values gives the final integral:
\[ I = I_1 + I_2 + I_3 = \frac{5}{4} + \frac{1}{2} - \frac{11}{4} = \frac{5 + 2 - 11}{4} = \frac{-4}{4} = -1 \]

Step 4: Final Answer:
The value of the definite integral is -1.