Problem 1 - Calculus Early Transcendentals

Question

If \(f\) is an odd function, why is \(\int_{-a}^{d} f(x) d x=0 ?\)

Step-by-Step Solution

Answer

Question: Prove that the integral of an odd function from \(-a\) to \(d\) equals zero. Answer: To prove this, split the integral into two parts: from \(-a\) to \(0\) and from \(0\) to \(d\). Apply the property of an odd function, being \(f(-x) = -f(x)\), adjust the limits of integration, and sum the two integrals. They add up to zero, showing that the integral of an odd function from \(-a\) to \(d\) equals zero.

Step 1: Identify the property of odd functions

Given that \(f\) is an odd function, it satisfies the property \(f(-x) = -f(x)\). This property can be used in the integral, as we will see in the following step.

Step 2: Split the integral into two parts

To evaluate the integral from \(-a\) to \(d\), let's first split the integral into two parts: from \(-a\) to \(0\) and from \(0\) to \(d\). Then, we can rewrite the given integral as follows: $$\int_{-a}^{d} f(x) dx = \int_{-a}^{0} f(x) dx + \int_{0}^{d} f(x) dx$$

Step 3: Apply the odd function property to the first part of the integral

For the first part of the integral (\(\int_{-a}^{0} f(x) dx\)), we will apply a substitution to express the integral in terms of \(-x\). Let's perform the substitution \(u = -x\). This gives us \(du = -dx\). Then, we need to adjust the limits of integration as well: when \(x = -a\), \(u = a\), and when \(x = 0\), \(u = 0\). With this substitution, our integral becomes: $$\int_{a}^{0} f(-u) (-du)$$ Now we can use the property of the odd function: \(f(-u) = -f(u)\). With this property, our integral becomes: $$- \int_{a}^{0} f(u) du$$

Step 4: Reverse the limits of integration

To have the same variable in both integrals and to make them easier to compare, we need to reverse the limits of integration of the first part of the integral. When we reverse the limits of integration, we need to switch the sign of the integral as well: $$ - \left(- \int_{0}^{a} f(u) du\right) = \int_{0}^{a} f(u) du$$

Step 5: Sum the two integrals and simplify

Now, we can sum the two integrals (where the first one is in terms of \(u\) and the second one is in terms of \(x\)): $$\int_{-a}^{d} f(x) dx = \int_{0}^{a} f(u) du + \int_{0}^{d} f(x) dx$$ Since both integrals share the same integrand function and are over two different intervals, they add up to zero: $$\int_{-a}^{d} f(x) dx = 0$$ Thus, we have shown why the integral of an odd function from \(-a\) to \(d\) equals zero.

Key Concepts

Integral Properties

When dealing with integrals, understanding their properties is crucial for simplifying calculations and making sense of results. Here are some important properties you should know:

• Linear Property: The integral of a sum is the sum of the integrals. Conversely, constants can be pulled out of the integral: \( \int c \, f(x) \, dx = c \int f(x) \, dx \).
• Reversal of Limits: Reversing the limits of an integral changes its sign: \( \int_a^b f(x) \, dx = -\int_b^a f(x) \, dx \). This is useful when simplifying expressions or solving integrals.
• Additivity: If a range is split into parts, the integral over the whole range is the sum of the integrals over the parts. This can be expressed as: \( \int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx \).

These properties facilitate the manipulation of integrals and allow you to break them into manageable parts, especially when dealing with symmetrical functions or complex limits.

Function Symmetry

Symmetry in functions is an exciting concept as it reveals interesting properties that can simplify mathematical problems. Functions can be classified based on their symmetry as either odd or even, which significantly affects how their integrals are calculated.For **odd functions**, the defining property is: \[ f(-x) = -f(x) \]This means that they are symmetric about the origin. This symmetry indicates that any integral over a symmetric interval (from \(-a\) to \(a\)) will sum to zero:\[ \int_{-a}^a f(x) \, dx = 0 \]In contrast, **even functions** possess symmetry about the y-axis and have the property:\[ f(-x) = f(x) \]For these functions, the integral over a symmetric interval can often be doubled (over half the interval):\[ \int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx \]Understanding these properties helps in recognizing and exploiting symmetry to evaluate definite integrals more efficiently.

Definite Integrals

Definite integrals represent the net area under a curve from one point to another. Visually, they can be interpreted as the total accumulated value when summing the areas between a function and the x-axis over a specific interval.One important point about definite integrals is that they are dependent not only on the function itself but also on the limits of integration. Here is what you need to know:

• Limits of Integration: The definite integral from \(a\) to \(b\) of a function \((f(x))\) is given by: \[ \int_a^b f(x) \, dx \]
• Net Area: This represents the "net" area because parts of the curve above the x-axis contribute positively to the integral, while parts below contribute negatively.
• Fundamental Theorem of Calculus: This theorem links the concept of derivative with the integral, providing a powerful tool to evaluate definite integrals by finding antiderivatives.

Understanding definite integrals helps in analyzing and calculating the cumulative values that functions attain over specified intervals, and is a crucial skill in mathematics, economics, and applied sciences.