Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f\) be an odd function and \(g\) be an even function, and suppose that \(\int_{0}^{1}|f(x)| d x=\int_{0}^{1} g(x) d x=3 .\) Use geometric reasoning to calculate each of the following: (a) \(\int_{-1}^{1} f(x) d x\) (b) \(\int_{-1}^{1} g(x) d x\) (c) \(\int_{-1}^{1}|f(x)| d x\) (d) \(\int_{-1}^{1}[-g(x)] d x\) (e) \(\int_{-1}^{1} x g(x) d x\) (f) \(\int_{-1}^{1} f^{3}(x) g(x) d x\)

Step-by-Step Solution

Answer

(a) 0, (b) 6, (c) 6, (d) -6, (e) 0, (f) 0.

Step 1: Understanding Odd and Even Functions

An odd function, such as \(f(x)\), satisfies \(f(-x) = -f(x)\). An even function, like \(g(x)\), satisfies \(g(-x) = g(x)\). These properties are key to solving the integrals.

Step 2: Solving Part (a): Integral of an Odd Function

For the integral \(\int_{-1}^{1} f(x) \, dx\), since \(f(x)\) is odd, any integral of an odd function over a symmetric interval \([-a, a]\) is zero. Therefore, \(\int_{-1}^{1} f(x) \, dx = 0\).

Step 3: Solving Part (b): Integral of an Even Function

For \(\int_{-1}^{1} g(x) \, dx\), since \(g(x)\) is even, \(\int_{-a}^{a} g(x) \, dx = 2 \int_{0}^{a} g(x) \, dx\). Therefore, \(\int_{-1}^{1} g(x) \, dx = 2 \times 3 = 6\).

Step 4: Solving Part (c): Integral of Absolute Value of an Odd Function

For \(\int_{-1}^{1} |f(x)| \, dx\), both halves are symmetric, and due to the absolute value, it becomes \(2 \times \int_{0}^{1} |f(x)| \, dx\). Thus, \(\int_{-1}^{1} |f(x)| \, dx = 2 \times 3 = 6\).

Step 5: Solving Part (d): Integral of Negated Even Function

For \(\int_{-1}^{1} [-g(x)] \, dx\), we can use linearity: \(\int_{-1}^{1} [-g(x)] \, dx = -\int_{-1}^{1} g(x) \, dx = -6\).

Step 6: Solving Part (e): Integral of \(x\) Times an Even Function

For \(\int_{-1}^{1} x g(x) \, dx\), due \(x\) is odd and \(g(x)\) is even, the product \(xg(x)\) is odd. Hence, \(\int_{-1}^{1} x g(x) \, dx = 0\).

Step 7: Solving Part (f): Integral Involving Odd Function to an Odd Power Times an Even Function

For \(\int_{-1}^{1} f^3(x) g(x) \, dx\), since \(f^3(x)\) is odd and \(g(x)\) is even, \(f^3(x)g(x)\) is odd. Thus, \(\int_{-1}^{1} f^3(x) g(x) \, dx = 0\).

Key Concepts

Definite Integrals

When solving problems involving definite integrals, one calculates the net area under a curve between two points on the x-axis. Definite integrals are written in the form \( \int_{a}^{b} f(x) \, dx \). This notation tells us that we are summing up all the tiny values of the function \(f(x)\) starting from point \(a\) and ending at point \(b\). This is not just about counting the positive areas (above the x-axis) but also subtracting the negative areas (below the x-axis).

Definite integrals have several important properties that can make calculations easier. For instance, integrating over a symmetric interval, like \([-a, a]\), can simplify if the function has particular properties, such as being odd or even. Such simplifications are crucial for quickly finding solutions and understanding the geometric implications of functions.

Symmetric Intervals

Symmetric intervals, like \([-a, a]\), are intervals that mirror around the y-axis. They are particularly handy when dealing with symmetric functions, like odd and even functions. When evaluating integrals over these intervals, the properties of the function can substantially simplify calculations.

For odd functions, which satisfy \(f(-x) = -f(x)\), their integral over symmetric intervals is zero. This is because the areas below and above the x-axis cancel each other out. For even functions, which satisfy \(g(-x) = g(x)\), the integral over a symmetric interval is twice the integral from 0 to \(a\). This is because the function produces identical areas on both sides of the y-axis.

• For odd functions: \( \int_{-a}^{a} f(x) \, dx = 0 \)
• For even functions: \( \int_{-a}^{a} g(x) \, dx = 2 \int_{0}^{a} g(x) \, dx \)

These properties can solve or simplify problems quickly when the function type is known.

Absolute Value Function

The absolute value function affects the integral in a distinctive way, as it treats all negative values as positive. When working with an absolute value of a function within an integral, like \(\int |f(x)| \, dx\), the result reflects the total length of the function, regardless of direction.

If the original function \(f(x)\) is odd, \(|f(x)|\) becomes even over symmetric intervals due to the absolute value conversion. This transformation means that you can calculate the integral of the absolute value over a symmetric interval by doubling the integral from 0 to \(a\). In our case, for an odd function \(f\) with given properties, we found:

• \(\int_{-1}^{1} |f(x)| \, dx = 2 \times \int_{0}^{1} |f(x)| \, dx = 6\).

Understanding the absolute value function helps in visualizing how the negative parts mirror into the positive domain, emphasizing magnitude without considering signs.

Geometric Reasoning

Geometric reasoning allows us to visualize and solve integrals comprehensively by considering the shapes and properties involved. It relies on understanding the graphical representations and symmetries of functions.

For example, when dealing with even and odd functions, geometric reasoning informs that:

• The integral of an odd function over a symmetric interval results in zero, since the areas on either side of the y-axis cancel each other out.
• The integral of an even function is just twice the integral from 0 to its symmetric endpoint, simplifying calculations significantly.

This approach is not only intuitive but also efficient. It drives home the fact that integrals are not just algebraic calculations but can be insightful geometric problems. By visualizing the shapes and areas functions cover, geometric reasoning aids in understanding and predicting the behavior of integrals across different scenarios. Thus, knowledge of graph behaviors complements mathematical techniques, providing a holistic view of problem-solving in calculus.