Question

Find \(c\) so that \(\int_{-c}^{c} \frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x=0.95 .\) Hint: Use symmetry.

Short answer

Use symmetry and tables to find c, where \(2F(c) = 0.95\).

Step-by-step solution

Understanding Symmetry

The function \( f(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} \) is an even function. This means that \( f(x) = f(-x) \). As a result, the integral from \(-c\) to \(c\) is twice the integral from 0 to \(c\). Mathematically, this can be represented as: \( \int_{-c}^{c} f(x) \, dx = 2 \int_{0}^{c} f(x) \, dx \).

Key concepts

Symmetry in Functions

Symmetry in functions is a fascinating topic that helps make complex integrals more manageable. A function can be symmetric in various ways: it can be even, odd, or neither. Symmetric functions offer computational ease because they reduce the range over which you need to integrate. Let's break this down further.

• Even functions: These are functions where the value at any point \(x\) is the same as at \(-x\). Mathematically, this means \(f(x) = f(-x)\). Even functions have symmetry about the y-axis.


• Odd functions: These are functions where the value at \(x\) is the negative of its value at \(-x\). Mathematically, \(f(x) = -f(-x)\), and they exhibit symmetry about the origin.

If you're dealing with a symmetric function, you can often simplify the integral, particularly when solving definite integrals over symmetric intervals (like \([-c, c]\)). This approach significantly reduces the complexity of the calculation.

Even Functions

Even functions are extremely useful for solving definite integrals over symmetric intervals. The defining property \(f(x) = f(-x)\) implies that the function mirrors itself on the y-axis.

A classic example of an even function is the Gaussian function, \[\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2},\] which you often encounter in probability and statistics. This function's symmetry allows you to leverage the property that \(\int_{-c}^{c} f(x)\) is simply \(2 \int_{0}^{c} f(x) \).

This characteristic reduces computational effort as you only need to compute the integral from \(0\) to \(c\), then double it. This is powerful when you are looking for specific values of \(c\), like in the original exercise where you need to find a \(c\) that makes the integral span 0.95 of the distribution.

Definite Integrals

Definite integrals are a method for calculating the area under a curve within a given interval. These integrals are expressed in the form \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration. Definite integrals provide valuable information about the accumulation of quantities, such as probabilities.

The beauty of using definite integrals with symmetric functions like even functions lies in their ability to simplify computation. If a function is even, and you're integrating over a symmetric interval such as \([-c, c]\), you can utilize the property that:

• \(\int_{-c}^{c} f(x) \, dx = 2 \int_{0}^{c} f(x) \, dx\)

This is incredibly helpful when you need to find a specific integral value, like 0.95 in the original exercise. The symmetry simplifies the process to just focusing on half of the interval and doubling the result, saving both time and effort in calculation.