Question

\( \int_{-1}^{1} x(1-x)(1+x)\, dx = \)

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

Hint:

Always check symmetry before integrating—odd functions over symmetric limits directly give zero.

Answer 1

Answer: (E)

Concept: For definite integrals over symmetric limits \([-a,a]\):
• If \(f(x)\) is an odd function, then \(\int_{-a}^{a} f(x)\,dx = 0\)
• If \(f(x)\) is an even function, symmetry can simplify evaluation

Step 1: Simplify the integrand
\[ x(1-x)(1+x) = x(1 - x^2) \] \[ = x - x^3 \]

Step 2: Identify nature of function
\[ f(x) = x - x^3 \] Check: \[ f(-x) = -x + x^3 = -(x - x^3) = -f(x) \] Thus, function is odd

Step 3: Apply symmetry property
\[ \int_{-1}^{1} (x - x^3)\,dx = 0 \]

Step 4: Verification by direct integration
\[ \int (x - x^3)dx = \frac{x^2}{2} - \frac{x^4}{4} \] Evaluate: \[ \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{-1}^{1} \] At \(x=1\): \( \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \) At \(x=-1\): \( \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \) Difference: \[ \frac{1}{4} - \frac{1}{4} = 0 \] Final Answer: \[ \boxed{0} \]