Question

Write the integral as the sum of the integral of an odd function and the integral of an even function. Use this simplification to evaluate the integral. $$ \int_{-4}^{4}\left(x^{3}+6 x^{2}-2 x-3\right) d x $$

Short answer

The definite integral of the given function from -4 to 4 is \(64 - 24 = 40\).

Step-by-step solution

Identification of Odd and Even Functions

From the given integral, each term in the integrand \((x^3,6x^2,-2x,-3)\) must be analyzed separately to identify if they are odd or even. An odd function satisfies the property \(f(-x)=-f(x)\) while an even function satisfies \(f(-x)=f(x)\). After analyzing, \(x^3\) and \(-2x\) are recognized as odd functions, while \(6x^2\) and \(-3\) are identified as even functions.

Splitting the Integral

Next, the given integral should be written as the sum of even and odd functions. The odd integrals are \(\int_{-4}^{4} x^{3} d x\) and \(\int_{-4}^{4} -2x d x\). The even integrals are \(\int_{-4}^{4} 6x^{2} d x\) and \(\int_{-4}^{4} -3 d x\).

Evaluating the Integrals

Because the integral of an odd function over a symmetric interval is zero, the odd integrals go to zero. The even integrals are evaluated as usual: \(\int_{-4}^{4} 6x^{2} d x = 2 \times [\frac{1}{3}x^3]_{0}^{4} = 2[\frac{1}{3}(4)^3] - 2[\frac{1}{3}(0)^3] = 64\) and \(\int_{-4}^{4} -3 d x = -3 \times [\frac{1}{1}x]_{0}^{4} = -3[(4)-(0)] = -24\).