Problem 45 - Calculus Early Transcendentals

Question

Use symmetry to evaluate the following integrals. $$\int_{-2}^{2} \frac{x^{3}-4 x}{x^{2}+1} d x$$

Step-by-Step Solution

Answer

Short Answer: The given function, \(f(x) = \frac{x^{3}-4 x}{x^{2}+1}\), is an odd function because \(f(-x) = -f(x)\). Therefore, since the integral is over a symmetric interval \([-2, 2]\), the result of the integral is 0: $$\int_{-2}^{2} \frac{x^{3}-4 x}{x^{2}+1} d x = 0$$.

Step 1: Determine if the function is even or odd

To check if the function is even or odd, we will first find \(f(-x)\). This is achieved by replacing \(x\) with \(-x\) in our function \(f(x)\). So, $$f(-x) = \frac{(-x)^{3}-4 (-x)}{(-x)^{2}+1}$$ $$f(-x) = \frac{-x^{3}+4 x}{x^{2}+1}$$ We can then compare \(f(x)\) and \(f(-x)\). In our case, we see that \(f(-x) = -f(x)\). This means that our function is an odd function.

Step 2: Use symmetry to evaluate the integral of an odd function

Since our function is odd (\(f(-x) = -f(x)\)), we can use the property of odd function integrals: $$\int_{-a}^{a} f(x) dx = 0$$ Given that our integral is: $$\int_{-2}^{2} \frac{x^{3}-4 x}{x^{2}+1} d x$$ We can now conclude that this integral is equal to 0, since our function is odd and the integral is taken over a symmetric interval of \([-2, 2]\): $$\int_{-2}^{2} \frac{x^{3}-4 x}{x^{2}+1} d x = 0$$