Symmetry in integrals Use symmetry to evaluate the following integrals.
∫²₋₂ [(x³ ― 4x) / (x² + 1)] dx 

Verified step by step guidance

1. Step 1: Recognize the symmetry in the integrand. The integral is over the interval [-2, 2], which suggests checking for even or odd symmetry in the function f(x) = (x³ - 4x) / (x² + 1).
2. Step 2: Test for odd symmetry. A function is odd if f(-x) = -f(x). Substitute -x into f(x): f(-x) = ((-x)³ - 4(-x)) / ((-x)² + 1) = (-x³ + 4x) / (x² + 1). Simplify to see if f(-x) = -f(x).
3. Step 3: Confirm that f(x) is odd. Since f(-x) = -f(x), the function is odd. For integrals of odd functions over symmetric intervals [-a, a], the result is always 0.
4. Step 4: Conclude that the integral ∫²₋₂ [(x³ - 4x) / (x² + 1)] dx evaluates to 0 due to the odd symmetry of the integrand.
5. Step 5: Use symmetry properties to save time and avoid direct computation, as the integral of an odd function over a symmetric interval is zero.

Key Concepts

Symmetry in Functions

Symmetry in functions refers to the property where a function exhibits even or odd characteristics. An even function, f(x), satisfies f(x) = f(-x), while an odd function satisfies f(x) = -f(-x). Recognizing these properties can simplify the evaluation of integrals, particularly over symmetric intervals, as the integral of an odd function over a symmetric interval is zero.

Definite Integrals

A definite integral calculates the net area under a curve between two specified limits. It is represented as ∫[a, b] f(x) dx, where 'a' and 'b' are the bounds of integration. Understanding how to evaluate definite integrals, especially using properties like symmetry, is crucial for solving problems involving area and accumulation.

Properties of Integrals

Properties of integrals include various rules that can simplify the evaluation process. For instance, the integral of a sum can be split into the sum of integrals, and the integral of a constant multiplied by a function can be factored out. These properties, along with symmetry, allow for more efficient calculations and can lead to quicker solutions in integral evaluation.