Problem 94 - Calculus Volume 1

Question

In the following exercises, use the identity \(\int_{-A}^{A} f(x) d x=\int_{-A}^{0} f(x) d x+\int_{0}^{A} f(x) d x\) to compute the integrals. \(\int_{-\pi}^{\pi} \frac{\sin t}{1+t^{2}} d t \quad(\text {Hint} : \sin (-t)=-\sin (t))\)

Step-by-Step Solution

Answer

The integral equals 0 because the function is odd over symmetric limits.

Step 1: Apply the Identity

The given integral is \(\int_{-\pi}^{\pi} \frac{\sin t}{1+t^{2}} d t\). We will use the identity \(\int_{-A}^{A} f(x) d x=\int_{-A}^{0} f(x) d x+\int_{0}^{A} f(x) d x\) to split the integral into two parts:\[\int_{-\pi}^{0} \frac{\sin t}{1+t^{2}} d t + \int_{0}^{\pi} \frac{\sin t}{1+t^{2}} d t\].

Step 2: Use the Property of Odd Functions

We know from the hint that \(\sin(-t) = -\sin(t)\), which means \(\sin t\) is an odd function. For odd functions, the integral over symmetric limits around zero equals zero. This implies: \[\int_{-\pi}^{\pi} \frac{\sin t}{1+t^{2}} dt = 0\].

Key Concepts

Odd Functions

Understanding odd functions is crucial in integral calculus. An odd function is a function that satisfies the property:

• \( f(-x) = -f(x) \).

This means that when you plug in the negative of any input into the function, the output is the negative of what you would get from the original input. This behavior reflects symmetry about the origin. For example, the sine function, \( \sin(x) \), is an odd function because \( \sin(-x) = -\sin(x) \).

Odd functions are significant when calculating definite integrals over symmetric intervals. If a function is odd and the limits of integration are symmetric about zero, such as from \(-A\) to \(A\), the total integral of the function over this range equals zero. Reason being, the area under the curve from \(-A\) to \(0\) cancels out the area from \(0\) to \(A\).

Definite Integrals

Definite integrals are a fundamental concept in calculus. They provide the total change, area, or accumulated value represented by a function over an interval. When you calculate a definite integral, you evaluate the function's values between two points on the x-axis, from \(a\) to \(b\). This results in a number, not a function.

Mathematically, a definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), and it represents the "net area" between the function and the x-axis. This means it accounts for areas above the x-axis as positive and areas below the x-axis as negative. Understanding this concept is key when evaluating functions like \( \frac{\sin(t)}{1+t^2} \) over intervals that stretch symmetrically across the origin.

Properties of Functions

Properties of functions are rules that describe how functions behave. In calculus, these properties can significantly simplify solving problems. Familiar properties include:

• Symmetry: Functions can be even, odd, or neither. Odd functions, as mentioned before, satisfy \( f(-x) = -f(x) \).
• Periodicity: Some functions repeat at regular intervals. For example, trigonometric functions like sine and cosine are periodic.
• Continuity: A function is continuous if it can be drawn without lifting the pen off the paper. This property ensures that integrals can be evaluated over an interval.

Being familiar with such properties helps in deciding appropriate techniques for evaluating integrals. In particular, recognizing an odd function allowed us to conclude the definite integral over \(-\pi\) to \(\pi\) is zero.

Symmetric Integration

Symmetric integration comes into play when you are evaluating integrals over intervals that are symmetric around zero, such as from \(-A\) to \(A\). This technique uses the symmetric properties of functions to simplify calculus problems. For instance:

• For an odd function, the integral over \(-A\) to \(A\) is always zero. This occurs because the negative areas cancel out the positive ones.
• For an even function, the integral can be simplified by doubling the integral from 0 to \(A\).

The identity \( \int_{-A}^{A} f(x) \, dx = \int_{-A}^{0} f(x) \, dx + \int_{0}^{A} f(x) \, dx \) leverages these properties efficiently. By knowing these principles, you can more easily tackle complicated integrals without explicit calculations, saving time and effort.