Question

The value of the integral \( I = \int_{-3}^3 (x^3 - x) dx \) is

• A. 0
• B. 3/2
• C. 1/2
• D. x

Hint:

Always check for odd/even symmetry when integrating over symmetric intervals \([-a, a]\). This often simplifies the problem significantly. The integral of an odd function over a symmetric interval is always zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to evaluate a definite integral over a symmetric interval (from -3 to 3).

Step 2: Key Formula or Approach:
For a definite integral over a symmetric interval \([-a, a]\):
1. If \( f(x) \) is an odd function (\( f(-x) = -f(x) \)), then \( \int_{-a}^a f(x) dx = 0 \).
2. If \( f(x) \) is an even function (\( f(-x) = f(x) \)), then \( \int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx \).

Step 3: Detailed Explanation:
Given integral: \( I = \int_{-3}^3 (x^3 - x) dx \)
Let \( f(x) = x^3 - x \).
First, check if \( f(x) \) is an odd or even function:
Substitute \(-x\) for \(x\):
\[ f(-x) = (-x)^3 - (-x) \]
\[ f(-x) = -x^3 + x \]
Factor out -1:
\[ f(-x) = -(x^3 - x) \]
Since \( f(-x) = -f(x) \), the function \( f(x) = x^3 - x \) is an odd function.
Because the integral is over a symmetric interval \([-3, 3]\) and the integrand is an odd function, the value of the integral is 0.
Alternatively, directly integrate:
\[ \int (x^3 - x) dx = \frac{x^4}{4} - \frac{x^2}{2} + C \]
Evaluate the definite integral:
\[ \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-3}^3 = \left( \frac{3^4}{4} - \frac{3^2}{2} \right) - \left( \frac{(-3)^4}{4} - \frac{(-3)^2}{2} \right) \]
\[ = \left( \frac{81}{4} - \frac{9}{2} \right) - \left( \frac{81}{4} - \frac{9}{2} \right) \]
\[ = 0 \]

Step 4: Final Answer:
The value of the integral is 0.