Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{-100}^{100}\left(v+\sin v+v \cos v+\sin ^{3} v\right)^{5} d v\)

Step-by-Step Solution

Answer

The integral evaluates to 0 because the integrand is an odd function.

Step 1: Analyze the Integrand

First, notice the integrand is composed of the function \( f(v) = v + \sin v + v \cos v + \sin^3 v \). This function does not seem to have any obvious symmetry like being odd or simple periodicity that would immediately simplify the integration bounds.

Step 2: Consider Symmetry or Simplification

One might consider checking if \( f(v) \) is an odd function. A function \( f(v) \) is odd if \( f(-v) = -f(v) \) for all \( v \). Calculate \( f(-v) \) to see if there is symmetry:- \( f(-v) = -v + \sin(-v) - v \cos(-v) + \sin^3(-v) \)- Substitute the identities: \( \sin(-v) = -\sin(v) \) and \( \cos(-v) = \cos(v) \)- \( f(-v) = -v - \sin v - v \cos v - \sin^3 v \)- \( f(-v) = -(v + \sin v + v \cos v + \sin^3 v) = -f(v) \)Thus, \( f(v) \) is an odd function.

Step 3: Use the Properties of Definite Integrals

Since \( f(v) \) is an odd function and the integration limits are symmetric around zero (−100 to 100), the integral of an odd function over symmetric limits is zero. Therefore, \( \int_{-100}^{100} (v + \sin v + v \cos v + \sin^3 v)^5 dv = 0 \).

Key Concepts

Odd Functions

An odd function is a function that satisfies the condition \( f(-x) = -f(x) \) for all \( x \) in its domain. This means that the graph of an odd function is symmetric with respect to the origin. Odd functions have a unique property that makes them interesting in calculus, especially when dealing with definite integrals.
When checking if a function is odd, we substitute \( -x \) into the function and see if we end up with the negative of the original function.

• Formulaically, odd functions have this: \( f(-x) = -f(x) \)
• Graphically, the graph will mirror across the origin. If you rotate the graph 180 degrees about the origin, it should look the same.

This property is particularly useful when evaluating integrals because it allows us to simplify calculations, especially over symmetric intervals.

Symmetry in Calculus

Symmetry is a powerful tool in calculus that helps simplify problems and computations. When integrands exhibit symmetry, it can significantly reduce the complexity of evaluating definite integrals.
There are generally two types of symmetry considered: symmetry about the y-axis (even functions) and symmetry about the origin (odd functions).

• Even functions: Have the property \( f(x) = f(-x) \). Example functions are cosine functions and parabolas.
• Odd functions: Exhibit the property \( f(-x) = -f(x) \), like sine functions and cubics.

When calculating a definite integral over a symmetric interval, the symmetry can often simplify the process. Particularly, for odd functions integrated across symmetric limits about zero, the integral evaluates to zero, which can save a lot of time in solving problems.
This makes symmetry an invaluable concept in solving integral problems efficiently.

Properties of Integrals

Integrals have several important properties that can be employed to simplify calculations or derive solutions to problems. Understanding these can make solving definite integrals much easier. In the context of odd functions and symmetry, a very useful property is:

• If \( f(x) \) is an odd function, then \( \int_{-a}^{a} f(x) \, dx = 0 \).
• The integration limits need to be symmetric around zero for this property to hold.

This property exists because the positive area from \( 0 \) to \( a \) cancels with the negative area from \( -a \) to \( 0 \).
Other properties of integrals that could be found useful are:

• Linearity: \( \int (cf(x) + g(x)) \, dx = c\int f(x) \, dx + \int g(x) \, dx \).
• Reversal of limits: \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).
• Splitting the interval: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \).

Utilizing these properties can greatly simplify the computation of integrals especially when combined with insights from symmetry.