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Chapter 5: Problem 51

Suppose \(f\) is an even function and $$\int_{-8}^{8} f(x) d x=18$$. a. Evaluate \(\int_{0}^{8} f(x) d x\) b. Evaluate \(\int_{-8}^{8} x f(x) d x\)

Short Answer

Expert verified
Answer: The integral of the given even function \(f(x)\) from \(0\) to \(8\) is 9, and the integral of the product of the given function and \(x\) from \(-8\) to \(8\) is 0.

Step by step solution

01

Review properties of even functions

Recall that for an even function \(f(x)\), we have \(f(-x) = f(x)\). This property results in the integral of an even function over a symmetric interval having a special property. More specifically, \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\).
02

Evaluate \(\int_{0}^{8} f(x) d x\)

We know that \(\int_{-8}^{8} f(x) d x=18\). Since \(f(x)\) is an even function, we can use the property mentioned in Step 1 to rewrite this as \(2 \int_{0}^{8} f(x) dx = 18\). Now, we can solve for \(\int_{0}^{8} f(x) d x\) by dividing both sides of the equation by 2:$$\int_{0}^{8} f(x) d x = \frac{18}{2} = 9$$. Thus, \(\int_{0}^{8} f(x) d x = 9\).
03

Review properties of odd functions and even functions products

Recall that if we multiply an even function by an odd function, we get an odd function. Moreover, the integral of an odd function over a symmetric interval always equals zero. That is, \(\int_{-a}^{a} h(x) dx = 0\) for any odd function \(h(x)\).
04

Evaluate \(\int_{-8}^{8} x f(x) d x\)

In this part, we want to evaluate \(\int_{-8}^{8} x f(x) d x\). Here, we can recognize that the function \(xf(x)\) is a product of an odd function (\(x\)) and an even function (\(f(x)\)). As mentioned in Step 3, this product results in an odd function. Therefore, the integral of \(x f(x)\) over the symmetric interval \([-8, 8]\) is zero:$$\int_{-8}^{8} x f(x) d x = 0$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Properties
The integral properties of functions reveal several interesting facets about how functions behave over particular intervals. One foundational element is the understanding of integrals themselves, which are essentially a measure of total accumulation, such as an area under a curve. Here are some key properties:
  • Linearity: Integration respects addition and scalar multiplication. For functions \(f(x)\) and \(g(x)\) and scalars \(a\) and \(b\), \(\int (a f(x) + b g(x)) dx = a \int f(x) dx + b \int g(x) dx\).
  • Additivity over intervals: The integral over a large interval can be broken down into smaller intervals: \(\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx\).
Understanding these properties is crucial when working with specific types of functions such as even and odd functions over symmetric intervals.
Symmetric Intervals
When we integrate functions over symmetric intervals, from \(-a\) to \(a\), peculiar properties can arise. A key concept is how this symmetry impacts even and odd functions:
  • If a function \(f(x)\) is even, meaning \(f(-x) = f(x)\), the integral over a symmetric interval can be simplified. This is because the areas on both sides of the y-axis are symmetrical. Thus, the integral can be expressed as double the integral from 0 to \(a\): \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\).
  • Symmetric intervals simplify calculations, especially for even functions, as in the given exercise where \(\int_{-8}^{8} f(x) dx = 18\) became \(2 \int_{0}^{8} f(x) dx = 18\).
These properties greatly simplify evaluating certain integrals and understanding the comprehensive behavior of the function over symmetric ranges.
Odd Function Integrals
The concept of odd functions adds another layer of intrigue when dealing with integrals. An odd function satisfies the condition \(f(-x) = -f(x)\). This property leads to some unique results:
  • When integrating an odd function over a symmetric interval, the result is always zero. This is because the contributions from \(-a\) to 0 and from 0 to \(a\) cancel each other out completely: \(\int_{-a}^{a} f(x) dx = 0\).
  • In the exercise context, the function \(x f(x)\) was noted to become odd since \(x\) itself is an odd function, and the product of an even function and an odd function is always odd. Consequently, \(\int_{-8}^{8} x f(x) dx = 0\), reflecting this integral property.
Recognizing whether a function is odd or not is essential in quickly determining the outcome of integrals over symmetric intervals and can streamline the problem-solving process significantly.

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Most popular questions from this chapter

Riemann sums for larger values of \(n\) Complete the following steps for the given function \(f\) and interval. a. For the given value of \(n\), use sigma notation to write the left, right, and midpoint Riemann sums. Then evaluate each sum using a calculator. b. Based on the approximations found in part (a), estimate the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the interval. $$f(x)=\cos 2 x \text { on }[0, \pi / 4] ; n=60$$.

What is the geometric meaning of a definite integral if the integrand changes sign on the interval of integration?

Net area and definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result. $$\int_{0}^{4}(8-2 x) d x$$

Approximating displacement The velocity in \(\mathrm{ft} / \mathrm{s}\) of an object moving along a line is given by \(v=3 t^{2}+1\) on the interval \(0 \leq t \leq 4\).a. Divide the interval [0,4] into \(n=4\) subintervals, [0,1] \([1,2],[2,3],\) and \([3,4] .\) On each subinterval, assume the object moves at a constant velocity equal to \(v\) evaluated at the midpoint of the subinterval and use these approximations to estimate the displacement of the object on [0,4] (see part (a) of the figure). b. Repeat part (a) for \(n=8\) subintervals (see part (b) of the figure).

Use symmetry to evaluate the following integrals. $$\int_{-10}^{10} \frac{x}{\sqrt{200-x^{2}}} d x$$
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