Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Choose the correct answer and explain your reasoning. (a) \(\int_{-a}^{a} \frac{1}{1+x^{2}} d x=\) (i) 0 (ii) \(\int_{0}^{a} \frac{1}{1+x^{2}} d x\) (iii) \(2 \int_{0}^{a} \frac{1}{1+x^{2}} d x\) (iv) \(-\int_{0}^{a} \frac{1}{1+x^{2}} d x\) (b) \(\int_{-a}^{a} \frac{x}{1+x^{2}} d x=\) (i) 0 (ii) \(\int_{0}^{a} \frac{x}{1+x^{2}} d x\) (iii) \(2 \int_{0}^{a} \frac{x}{1+x^{2}} d x\) (iv) \(-\int_{0}^{a} \frac{x}{1+x^{2}} d x\)

Step-by-Step Solution

Answer

The answer to part (a) is \(2 \int_{0}^{a} \frac{1}{1+x^{2}} dx\) and the answer to part (b) is 0.

Step 1: Analyzing the parity of functions

An even function is a function which satisfies the property \(f(-x) = f(x)\). If this is the case, then \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\). An odd function is one that satisfies \(f(-x) = -f(x)\). If a function is odd, then \(\int_{-a}^{a} f(x) dx = 0\).

Step 2: Determining the nature of the first function

The function in part (a) is \(f(x) = 1/(1+x^{2})\). Substituting \(-x\) for \(x\) gives \(f(-x) = 1/(1+(-x)^{2}) = 1/(1+x^{2}) = f(x)\). Therefore, the function is even.

Step 3: Calculating the integral of the first function

Since the function in part (a) is even, use the property of even functions to calculate the integral: \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\). Therefore, the answer to part (a) is \(\int_{-a}^{a} \frac{1}{1+x^{2}} dx = 2 \int_{0}^{a} \frac{1}{1+x^{2}} dx\), which matches with option (iii) in the exercise.

Step 4: Determining the nature of the second function

The function in part (b) is \(f(x) = x/(1+x^{2})\). Substituting \(-x\) for \(x\) gives \(f(-x) = -x/(1+(-x)^{2}) = -x/(1+x^{2}) = -f(x)\). Therefore, the function is odd.

Step 5: Calculating the integral of the second function

Since the function in part (b) is odd, use the property of odd functions to calculate the integral: \(\int_{-a}^{a} f(x) dx = 0\). Therefore, the answer to part (b) is \(\int_{-a}^{a} \frac{x}{1+x^{2}} dx = 0\), which matches with option (i) in the exercise.

Key Concepts

Definite Integrals

When solving mathematical problems, definite integrals are a common encounter. They represent the net area under a curve between two specified points on the x-axis. To find this integral, you look at the function provided and evaluate it from the lower to the upper limit of integration. Consider the expression \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are your limits, and \(f(x)\) is the function being integrated.

In practice, the Fundamental Theorem of Calculus links the concept of differentiation with integration, offering a way to evaluate definite integrals by using antiderivatives. For an antiderivative \(F\) of \(f\), the definite integral can be calculated as \(F(b) - F(a)\).

Definite integrals are especially interesting when the function involved has certain symmetries, which simplifies computation. This aspect is crucial when we discuss even and odd functions and their integrals over symmetric intervals.

Even and Odd Functions

Understanding the symmetry of functions is essential when working with integrals. Even functions are symmetric about the y-axis. Mathematically, a function \(f\) is even if \(f(-x) = f(x)\) for all \(x\) in the function's domain. The graphs of even functions often resemble mirrored images on either side of the y-axis.

Odd functions, on the other hand, have rotational symmetry around the origin. Their defining characteristic is that \(f(-x) = -f(x)\) for all \(x\). If you rotate the graph of an odd function 180 degrees around the origin, it will look the same.

These properties not only help in graphing but also play a pivotal role in simplifying integration processes, particularly when the interval of integration is symmetric about the origin, which leads us to exploring symmetry in integration.

Parity of Functions

Parity of functions refers to the attribute of being even or odd as mentioned above. It's the 'behavior' of functions when their inputs are replaced by their negations. The concept of parity becomes instrumental when calculating integrals over symmetric intervals, as it can greatly reduce the complexity of the calculation.

Application in Integrals

For an even function \(f\), integrating over a symmetric interval \(\[-a, a\]\) results in double the integral from \(0\) to \(a\) because the area under the curve from \(0\) to \(a\) is exactly the same as the area from \(\[-a, 0\]\). Conversely, the integral of an odd function over a symmetric interval is always zero because the areas above and below the x-axis cancel each other out.

This unique property provides a shortcut to evaluating definite integrals and is a key reason why recognizing the parity of a function can make a significant difference when tackling integration problems.

Symmetric Intervals Integration

Symmetric intervals play a special role when integrating even and odd functions. When you encounter an integral with limits that are equal and opposite in sign—like \(\int_{-a}^{a} f(x) \, dx\)—you can leverage the function's parity to evaluate the integral more easily. This is because the symmetry in the intervals complements the symmetry in the functions.

Even Function Over Symmetric Interval

For even functions, since the function values are mirrored across the y-axis, each slice of area between \(0\) and \(a\) has a matching slice between \(0\) and \(\[-a\]\). This means you can calculate the integral from \(0\) to \(a\) and simply double the result.

Odd Function Over Symmetric Interval

Odd functions, by contrast, have slices of area that are equal but opposite in sign across the y-axis. Therefore, the positive area is exactly negated by the negative area, leading to an integral value of zero over symmetric intervals.

Recognizing these patterns is more than a mathematical curiosity—it's a powerful tool to streamline computations and deepen the understanding of how functions behave under integration.