Problem 52 - Calculus Early Transcendentals

Question

Suppose \(f\) is an odd function, \(\int_{0}^{4} f(x) d x=3\), and \(\int_{0}^{8} f(x) d x=9\). a. Evaluate \(\int_{-4}^{8} f(x) d x\) b. Evaluate \(\int_{-8}^{4} f(x) d x\)

Step-by-Step Solution

Answer

Question: Given an odd function, \(f(x)\), and the following definite integrals: \(\int_{0}^{4} f(x) dx = 3\) and \(\int_{0}^{8} f(x) dx = 9\), determine the value of the following integrals: (a) \(\int_{-4}^{8} f(x) dx\) and (b) \(\int_{-8}^{4} f(x) dx\). Answer: (a) \(\int_{-4}^{8} f(x) dx = 6\) and (b) \(\int_{-8}^{4} f(x) dx = -6\).

Step 1: An odd function is a function that satisfies the following condition: \(f(-x) = -f(x)\). This property implies that the graph of an odd function is symmetric about the origin. #Step 2: Define the property of the definite integral of an odd function#

For an odd function over a symmetric interval, the definite integral is equal to zero. Mathematically, for an odd function \(f(x)\), \(\int_{-a}^{a} f(x) dx = 0\). #Step 3: Evaluate the integral \(\int_{-4}^{8} f(x) d x\)#

Step 2: We know that \(\int_{0}^{4} f(x) dx = 3\) and \(\int_{0}^{8} f(x) dx = 9\). To evaluate the integral \(\int_{-4}^{8} f(x) dx\), we can split the integral into \(\int_{-4}^{0} f(x) dx + \int_{0}^{8} f(x) dx\). Since \(f(x)\) is an odd function, we have \(\int_{-4}^{0} f(x) dx = -\int_{0}^{4} f(x) dx\). Thus, this becomes: \(-\int_{0}^{4} f(x) dx + \int_{0}^{8} f(x) dx = -3 + 9 = 6\). So, \(\int_{-4}^{8} f(x) d x = \boxed{6}\). #Step 4: Evaluate the integral \(\int_{-8}^{4} f(x) d x\)#

Similarly, we can split the integral into \(\int_{-8}^{0} f(x) dx + \int_{0}^{4} f(x) dx\). Since \(f(x)\) is an odd function, we have \(\int_{-8}^{0} f(x) dx = -\int_{0}^{8} f(x) dx\). This becomes: \(-\int_{0}^{8} f(x) dx + \int_{0}^{4} f(x) dx = -9 + 3 = -6\). So, \(\int_{-8}^{4} f(x) d x = \boxed{-6}\).

Key Concepts

Odd Functions

Odd functions hold a special property, which is expressed as \( f(-x) = -f(x) \). This means that for any point \( x \) on the graph of the function, its corresponding \( -x \) point will have a value equal in magnitude but opposite in sign.
This unique characteristic makes odd functions symmetrical around the origin. When you graph an odd function, you’ll notice that if you rotate it 180 degrees around the origin, it looks the same.
This property becomes particularly useful when dealing with integrals of odd functions over symmetric intervals, as it can often simplify calculations significantly.

Symmetric Intervals

A symmetric interval is one that is equally spaced around a central point, often zero. In mathematical terms, this can be denoted as \([-a, a]\). The integral of an odd function over a symmetric interval simplifies elegantly due to the function’s symmetry.
For instance, given an odd function \( f(x) \), the definite integral over a symmetric interval is always zero: \[ \int_{-a}^{a} f(x) dx = 0 \] This is because the areas under the curve on opposite sides of the origin cancel each other out.
Symmetric intervals can greatly ease calculations by allowing these areas to be quickly evaluated under certain conditions.

Properties of Integration

Integration is a fundamental concept that allows us to calculate the area under a curve. For odd functions, their inherent properties can make the integration over certain intervals much simpler.
When integrating an odd function across a non-symmetric interval, one can separate it into known integral components.

• For example, to evaluate \( \int_{-4}^{8} f(x) dx \), it is split into \( \int_{-4}^{0} f(x) dx + \int_{0}^{8} f(x) dx \).
• Due to the odd nature of \( f(x) \), we know \( \int_{-4}^{0} f(x) dx = -\int_{0}^{4} f(x) dx \).

This property not only simplifies calculations but also reduces the chance of errors when working out integrals involving odd functions.
Understanding these properties deeply helps you quickly solve integrals of odd functions over any given intervals.