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Chapter 4: Problem 105

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

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

Step by step solution

01

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.
02

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 \(\int_{-4}^{4} -3 d x\).
03

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 \(\int_{-4}^{4} -3 d x = -3 \times [\frac{1}{1}x]_{0}^{4} = -3[(4)-(0)] = -24\).

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Most popular questions from this chapter

Find the derivative of the function. \(y=\sinh ^{-1}(\tan x)\)

The area \(A\) between the graph of the function \(g(t)=4-4 / t^{2}\) and the \(t\) -axis over the interval \([1, x]\) is \(A(x)=\int_{1}^{x}\left(4-\frac{4}{t^{2}}\right) d t\) (a) Find the horizontal asymptote of the graph of \(g\). (b) Integrate to find \(A\) as a function of \(x\). Does the graph of \(A\) have a horizontal asymptote? Explain.

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