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 \(\int_{-8}^{4} f(x) d x\)

Short answer

Answer: The value of the integral $\int_{-4}^{8} f(x) dx$ is 6, while the value of the integral $\int_{-8}^{4} f(x) dx$ is -6.

Step-by-step solution

Part a: Split the integral

We start by splitting the given integral into two parts: $$ \int_{-4}^{8} f(x) dx = \int_{-4}^{0} f(x) dx + \int_{0}^{8} f(x) dx $$

Part a: Use given integrals and odd function property

We know that \(\int_{0}^{8} f(x) dx = 9\) and \(\int_{0}^{4} f(x) dx = 3\). Additionally, knowing that \(f\) is an odd function, we have \(\int_{-4}^{0} f(x)d x = -\int_{0}^{4} f(x) d x\). Therefore, we can determine the value of the first part of the integral: $$ \int_{-4}^{0} f(x) dx = -\int_{0}^{4} f(x) d x = -3 $$

Part a: Get the final answer

Now, we substitute the results we computed back into the original expression: $$ \int_{-4}^{8} f(x) dx = \int_{-4}^{0} f(x) dx + \int_{0}^{8} f(x)dx = -3 + 9 = 6 $$ So, the value of the integral \(\int_{-4}^{8} f(x) dx\) is equal to \(6\).

Part b: Split the integral

Similar to part a, we start by splitting the given integral into two parts: $$ \int_{-8}^{4} f(x) dx = \int_{-8}^{0} f(x) dx + \int_{0}^{4} f(x)dx $$

Part b: Use given integrals and odd function property

We have \(\int_{0}^{4} f(x) dx = 3\) and \(\int_{0}^{8} f(x) dx = 9\). Knowing that \(f\) is an odd function, we can express the integral from \(-8\) to \(0\) based on the integral from \(0\) to \(8\) as follows: $$ \int_{-8}^{0} f(x) dx = -\int_{0}^{8} f(x) d x = -\int_{0}^{4} f(x) d x -\int_{4}^{8} f(x) d x = -3 - 6 $$

Part b: Get the final answer

Now, we substitute the expression we determined back into the split integral: $$ \int_{-8}^{4} f(x) dx = \int_{-8}^{0} f(x) dx + \int_{0}^{4} f(x)dx = (-3 - 6) + 3 = -6 $$ So, the value of the integral \(\int_{-8}^{4} f(x) dx\) is equal to \(-6\).

Key concepts

Odd Functions

Understanding odd functions is essential in solving integrals when dealing with symmetric intervals around zero. A function is termed as "odd" if it satisfies the condition \(f(-x) = -f(x)\) for all values of \(x\) within its domain. This symmetrical property implies that the graph of an odd function is symmetric about the origin.
One important aspect of odd functions is that, when integrating over symmetrical intervals, such as \([-a, a]\), the integral of an odd function becomes zero, because the areas above and below the x-axis cancel each other out. For example, \(\int_{-a}^{a} f(x) \, dx = 0\).

• Odd functions exhibit symmetry about the origin.
• They fulfill the condition: \(f(-x) = -f(x)\).
• The integral over a symmetric interval around zero equals zero, making computations more straightforward.

Therefore, recognizing that a function is odd provides shortcuts when performing integrations, especially when balancing the calculations between negative and positive intervals.

Integral Properties

Integral properties are essential concepts that simplify the process of solving definite integrals, especially when combined with the properties of odd functions. These properties include linearity, addition, and the zero-boundary integral over symmetric intervals.
The linearity property implies that you can factor out constants from an integral, allowing for easier computation:

• \(\int cf(x) \, dx = c \int f(x) \, dx\) where \(c\) is a constant.

When dealing with odd functions, we leverage the symmetry property. For an odd function \(f(x)\), you have:

• \(\int_{a}^{b} f(x) \, dx = -\int_{-b}^{-a} f(x) \, dx\).

This property is particularly useful for simplifying calculations where the integral limits are expanded or reversed.
Applying these properties in practice allows you to break down more complex integrals into manageable parts, or sometimes even to predict their outcome, saving time and effort in computations.

Splitting Integrals

Splitting integrals is a technique often used to simplify complex integration problems by breaking them into smaller, more manageable parts. This technique is especially effective when the integral involves an even or odd function, or if there is a change in behavior of the function across the interval.
When solving problems like \(\int_{-4}^{8} f(x) \, dx\) or \(\int_{-8}^{4} f(x) \, dx\), splitting the integral helps in leveraging the properties of the function across different sub-intervals.

• For example, breaking down into \(\int_{-4}^{0} f(x) \, dx + \int_{0}^{8} f(x) \, dx\), helps apply the zero-boundary property of odd functions effectively.
• This simplification also often involves recognizing how to utilize given integral values, such as \(\int_{0}^{4} f(x) \, dx\) and \(\int_{0}^{8} f(x) \, dx\).

In essence, by splitting integrals along these strategic lines, you can methodically analyze the behavior of the function and apply known results or properties, leading to a simplified solution path.