Problem 32 - Calculus

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 has symmetry about the origin, meaning that for every point \((x, f(x))\), the point \((-x, -f(x))\) is also on the graph. An even function is symmetric about the y-axis, indicating that for every point \((x, g(x))\), the point \((-x, g(x))\) is also on the graph.

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

For an odd function, the integral over a symmetric interval around zero is zero because the positive and negative parts cancel each other out. Thus, we have:\[\int_{-1}^{1} f(x) \, dx = 0.\]

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

The integral of an even function over a symmetric interval around zero is twice the integral from 0 to 1 because the areas are mirrored:\[\int_{-1}^{1} g(x) \, dx = 2 \int_{0}^{1} g(x) \, dx = 2 \times 3 = 6.\]

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

Because \(|f(x)|\) is non-negative and identical on both sides of the y-axis, the integral from -1 to 1 will be twice that from 0 to 1:\[\int_{-1}^{1} |f(x)| \, dx = 2 \int_{0}^{1} |f(x)| \, dx = 2 \times 3 = 6.\]

Step 5: Solving Part (d): Integral of Negative of an Even Function Over Symmetric Limits

Since multiplying by -1 scales an even function evenly across the y-axis, the integral becomes:\[\int_{-1}^{1} [-g(x)] \, dx = -\int_{-1}^{1} g(x) \, dx = -6.\]

Step 6: Solving Part (e): Integral Involving a Factor of x with an Even Function

When multiplying an even function by an odd function like \(x\), the resultant is odd. Hence, the integral over symmetric limits is zero:\[\int_{-1}^{1} x g(x) \, dx = 0.\]

Step 7: Solving Part (f): Integral of a Product Involving f to an Odd Power and an Even Function

Since \(f(x)\) is odd and raising it to an odd power maintains the oddness, the product \(f^3(x)g(x)\) remains odd. Integrating an odd function over symmetric limits gives:\[\int_{-1}^{1} f^3(x)g(x) \, dx = 0.\]

Key Concepts

Odd Function

An odd function is a special type of function where its graph has rotational symmetry about the origin. This means that if you rotate the graph 180 degrees around the origin, it fits exactly onto itself.
For any point \((x, f(x))\), there is a corresponding point \((-x, -f(x))\). This property can be expressed mathematically as \(f(-x) = -f(x)\) for all \(x\).
When we talk about integrating an odd function over a symmetric interval, such as \([-a, a]\), it’s noteworthy that the positive area on one side of the y-axis cancels out the negative area on the other side. Thus, the integral of an odd function over such an interval is zero:

• Symmetry makes calculations easier.
• The areas above and below the x-axis cancel each other out.

For example, if \(f(x)\) is odd, then \(\int_{-1}^{1} f(x) \, dx = 0\) because both halves of the interval contribute equal but opposite areas.

Even Function

An even function is characterized by its symmetry about the y-axis. This means the graph looks the same to the left and right of the y-axis.
A common feature of even functions is that for every point \((x, g(x))\), there is also a point \((-x, g(x))\). This can be written as \(g(-x) = g(x)\) for all \(x\).
One important property of even functions when it comes to integral calculus is how they behave over symmetric intervals. If you integrate an even function over the interval \([-a, a]\), the result is essentially twice the integral from \(0\) to \(a\):

• Symmetry about the y-axis simplifies integration.
• You only need to calculate one side and double it.

For instance, if \(g(x)\) is even and \(\int_{0}^{1} g(x) \, dx = 3\), then \(\int_{-1}^{1} g(x) \, dx = 2 \times 3 = 6\). This symmetry lets us handle calculations more efficiently.

Symmetric Interval

A symmetric interval refers to an interval centered at zero, usually described as \([-a, a]\). This interval is particularly important when dealing with odd and even functions.
No matter whether you're integrating or analyzing the behavior of functions, symmetry plays a key role in simplifying problems.
For odd functions like \(f(x)\), integration over a symmetric interval naturally results in zero because of the way these functions "cancel out" across the origin. Conversely, for even functions like \(g(x)\), integrating over symmetric intervals provides twice the result of integrating over half of it. Why is this?

• For odd functions, their mirrored negative counterparts cancel out.
• For even functions, mirrored positive areas simply mirror, doubling the contribution.

Understanding symmetric intervals helps us apply these properties efficiently in problem solving. It means once you recognize the type of function and that the interval is symmetric, many calculations become straightforward.