Question

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

Answer: The value of $\int_{0}^{8} f(x) dx$ is 9 and the value of $\int_{-8}^{8} xf(x) dx$ is 0.

Step-by-step solution

Property of even function

An even function has the property \(f(-x) = f(x)\). This results in symmetry about the y-axis.

Odd property of definite integrals

Since the function \(f(x)\) is even, its integral from -a to a can be written as twice the integral from 0 to a, that is: $$\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$$

Evaluate the integral from 0 to 8

Using the odd property of definite integrals with \(a=8\) and given that the integral from -8 to 8 is equal to 18, we can calculate the integral from 0 to 8 as: $$\int_{0}^{8} f(x) dx = \frac{1}{2} \int_{-8}^{8} f(x) dx = \frac{1}{2} \cdot 18 = 9$$ For b:

Product of even and odd functions

The product of an even function and an odd function results in an odd function. In our case, \(xf(x)\) is an odd function since the product of an even and an odd function (\(x\)) is odd.

Integral of odd function

The property of definite integrals for odd functions states that: $$\int_{-a}^{a} g(x) dx = 0$$ where \(g(x)\) is an odd function.

Evaluate the integral from -8 to 8

Since \(xf(x)\) is odd, the integral of xf(x) from -8 to 8 should equal 0: $$\int_{-8}^{8} xf(x) dx = 0$$ In summary: a. The value of \(\int_{0}^{8} f(x) dx = 9\) b. The value of \(\int_{-8}^{8} xf(x) dx = 0\)

Key concepts

Even Function Properties

Even functions possess unique symmetry properties that are both visually satisfying and mathematically significant. By definition, an even function satisfies the condition that for every point \(x\) in the domain, the corresponding point \( -x\) results in the same function value; that is, \(f(-x) = f(x)\). This equality leads to an even function displaying mirror symmetry about the y-axis.

An important consequence of this symmetry is observed when we compute definite integrals over symmetric intervals. Since the area under the curve from \( -a\) to \(0\) is exactly the same as the area from \(0\) to \(a\), we have a useful property: the integral of an even function over \( [-a, a]\) is twice the integral over \( [0, a]\). This makes calculating definite integrals of even functions considerably easier when dealing with symmetric limits of integration.

Odd Function Properties

Odd functions are characterized by another type of symmetry called rotational symmetry about the origin. This means that if you rotate the graph of an odd function \(180^\circ\) around the origin, you will get the same graph. The formal property that defines an odd function is \(f(-x) = -f(x)\) for every \(x\) in the domain.

When integrating an odd function over a symmetric interval, such as \( [-a, a]\), the areas above and below the x-axis cancel out exactly, resulting in a total integral value of zero. This property is extremely useful as it allows for rapid evaluation of integrals of odd functions over symmetric limits without performing any calculation—provided the function's oddness can be established.

Symmetry in Integrals

Symmetry plays a pivotal role simplifying the process of integration. When dealing with an even function, its symmetry about the y-axis allows us to compute the area under one half of the curve and simply double it to find the total area. This is not just a shortcut but also an application of the underlying geometric and algebraic properties of even functions. For odd functions, their rotational symmetry about the origin means that for every positive area on one side of the y-axis, there's an equal but negative area on the other side—resulting in a net area (and hence integral) of zero over symmetric limits.

The exercise provided illustrates these principles perfectly: The integral of the even function \(f(x)\) from \(0\) to \(8\) is half of the integral from \( -8\) to \(8\), and the integral of the product of \(x\) and \(f(x)\)—where \(x\) makes the product an odd function—over \( [-8, 8]\) is zero. Recognizing symmetry can dramatically reduce the complexity of many integral problems.

Integral of Product of Even and Odd Functions

When we multiply an even function by an odd function, the resulting function is odd. To understand why, recall that an even function \(f(x)\) satisfies \(f(-x) = f(x)\), while an odd function \(g(x)\) satisfies \(g(-x) = -g(x)\). Thus, if we take an even function \(f(x)\) and multiply it by \(x\) (which is an odd function), we get \(xf(x)\). For any negative \(x\), \(xf(-x) = -x \cdot f(-x) = -x \cdot f(x) = -(xf(x))\), which shows that \(xf(x)\) is indeed odd.

From this arises an invaluable integral property: the integral of the product of an even and odd function over a symmetric interval is zero. This is because the areas on one side of the y-axis are perfectly negated by the areas on the other side, leading to a net integral value of zero. In the given exercise, the second part leverages this property by asserting that since \(xf(x)\) is odd, the integral of \(xf(x)\) from \( -8\) to \(8\) is zero without any need for further calculation.