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 Concept:
To integrate a "min" function, we must find where the two internal functions intersect and determine which is smaller on different intervals.

Step 2: Detailed Explanation:
1. Intersection Points:
$1 - x = 1 - x^{3} \implies x^{3} - x = 0 \implies x(x^{2} - 1) = 0 \implies x = -1, 0, 1$.
2. Determine "min" in intervals:
$\bullet$ For $x \in [-1, 0]$: Let $x = -0.5$. $1-x = 1.5$; $1-x^{3} = 1.125$. Min is $1-x^{3$}.
$\bullet$ For $x \in [0, 1]$: Let $x = 0.5$. $1-x = 0.5$; $1-x^{3} = 0.875$. Min is $1-x$.
$\bullet$ For $x \in [1, 2]$: Let $x = 1.5$. $1-x = -0.5$; $1-x^{3} = -2.375$. Min is $1-x^{3$}.
3. Set up the integrals:
$I = \int_{-1}^{0} (1 - x^{3}) dx + \int_{0}^{1} (1 - x) dx + \int_{1}^{2} (1 - x^{3}) dx$.
4. Calculate:
$\int_{-1}^{0} (1 - x^{3}) dx = [x - \frac{x^{4}}{4}]_{-1}^{0} = 0 - (-1 - \frac{1}{4}) = \frac{5}{4}$.
$\int_{0}^{1} (1 - x) dx = [x - \frac{x^{2}}{2}]_{0}^{1} = (1 - \frac{1}{2}) - 0 = \frac{1}{2}$.
$\int_{1}^{2} (1 - x^{3}) dx = [x - \frac{x^{4}}{4}]_{1}^{2} = (2 - \frac{16}{4}) - (1 - \frac{1}{4}) = -2 - \frac{3}{4} = -\frac{11}{4}$.
5. Sum:
$I = \frac{5}{4} + \frac{2}{4} - \frac{11}{4} = \frac{-4}{4} = -1$.

Step 3: Final Answer:
The integral value is -1.
This matches option (A).