Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

evaluate the following. a) \(\int_{0}^{1} \sqrt{1-x^{2}} d x\) (b) \(\int_{1}^{-1} \sqrt{1-x^{2}} d x\)

Step-by-Step Solution

Answer

The answer for (a) is \( \frac{\pi}{4} \) and for (b), it is \( \frac{\pi}{2} \).

Step 1: Identify the Function and Boundaries

Here, the function to be integrated is \( \sqrt{1-x^{2}} \). For the two integrals, the boundaries to integrate over are [0, 1] and [-1, 1], respectively.

Step 2: Apply Appropriate Integral Property

In case (b), as the integral boundaries are symmetric around x=0, and the function to be integrated is also symmetric around x=0. Thus, the integral can be simplified as \[ \int_{1}^{-1} \sqrt{1-x^{2}} dx = 2 \int_{0}^{1} \sqrt{1-x^{2}} dx \]

Step 3: Recognize and Substitute Integral Form

We recognize that the integral is in the form of \(\int \sqrt{r^{2}-x^{2}}d x\), which corresponds to \( \frac{1}{2} \left[x \sqrt{r^{2}-x^{2}}+r^{2} \arcsin \left(\frac{x}{r}\right)\right]\). By substituting \( r = 1 \), simplifying, and applying the limits of integration, we get \( \frac{1}{2} * (\arcsin{1} - 0) \) for (a) and \( \frac{1}{2} * 2 * (\arcsin{1} - 0) \) for (b).

Step 4: Evaluate the Arcsin Function

We know arcsin(1) = \( \frac{\pi}{2} \). Substituting this value in the above expressions ,

Step 5: Multiply and Simplify

We simplify to get the final answer by multiplying and reducing the fraction.

Key Concepts

Definite Integrals

A definite integral is a fundamental concept in calculus that represents the accumulation of quantities and can be thought of as the area under a curve between two points on the x-axis. The integral sign \( \int \) with the upper and lower limits of integration signifies a definite integral. Consider the integral \( \int_{a}^{b} f(x) dx \) where \( a \) and \( b \) are the lower and upper limits, respectively, and \( f(x) \) is the function to be integrated.

The result of this process is a number representing the net area, taking into account that areas below the x-axis are considered negative. Getting better at evaluating definite integrals involves recognizing patterns and applying the rules of integration, including various properties that can simplify the process, as seen in the textbook solution where the integral of \( \sqrt{1-x^2} \) from 0 to 1 is required.

Integration Properties

Integration has several properties that help us solve integrals more efficiently. These properties reflect the behavior of integrals with respect to different mathematical operations. A crucial property used in the exercise to simplify the work is the symmetry property. If a function \( f(x) \) is even, that is, \( f(x) = f(-x) \), then \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).

Another property is the constant multiple rule, which allows us to take constants outside the integral, simplifying the expression before proceeding with integration. Utilizing such properties not only saves time but also reduces the potential for errors in complicated integrations.

Arcsin Function

The arcsin function, also known as the inverse sine function, is denoted as \( \arcsin(x) \) or \( \sin^{-1}(x) \) and it returns the angle whose sine is \( x \). The domain for \( \arcsin(x) \) is \( [-1, 1] \) and its range is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) in radians. When integrating the square root of a circular function, like \( \sqrt{1-x^2} \) in the exercise, recognizing that it relates to the arcsin function is a valuable insight.

It is common to encounter \( \frac{1}{2} \left[x \sqrt{r^{2}-x^{2}} + r^{2} \arcsin \left(\frac{x}{r}\right)\right] \) when integrating such expressions. For the given function, \( r \) is recognized to be 1, leading to the involvement of the \( \arcsin \) function when applying the antiderivative. Understanding how the arcsin function behaves and how to integrate expressions involving it is crucial for solving trigonometric integrals effectively.