Problem 18 - 5 Steps to A 5: AP Calculus AB 2021

Question

Evaluate the following definite integrals. Evaluate \(\int_{0}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}} .\)

Step-by-Step Solution

Answer

Question: Evaluate the definite integral \(\int_{0}^{1/2} \frac{1}{\sqrt{1-x^2}} dx\). Answer: \(\frac{\pi}{6}\)

Step 1: Identify the antiderivative of the given integrand function

The given function is \(\frac{1}{\sqrt{1-x^2}}\). By recognizing that the function is the derivative of the arcsin function, we have: \(\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin(x) + C\) Now, we have to find the definite integral by applying the limits of integration: \(\int_{0}^{1/2} \frac{1}{\sqrt{1-x^2}} dx\)

Step 2: Substitute the limits of integration and calculate the difference

We will substitute the limits of integration into the antiderivative function and find the difference: \(\arcsin(\frac{1}{2}) - \arcsin(0) = \arcsin(\frac{1}{2})\)

Step 3: Simplify the result

Now, we simplify the result using the property of the arcsin function: \(\arcsin(\frac{1}{2}) = \frac{\pi}{6}\) Therefore, the value of the definite integral is: \(\int_{0}^{1/2} \frac{1}{\sqrt{1-x^2}} dx = \frac{\pi}{6}\).

Key Concepts

Antiderivative

Understanding the concept of the antiderivative is crucial when dealing with definite integrals. An antiderivative is essentially the reverse of a derivative. If a function is a derivative, then finding its antiderivative means obtaining the original function, up to a constant.In the original problem, the function given is \( \frac{1}{\sqrt{1-x^2}} \), which is recognized as the derivative of the arcsin function. Therefore, the antiderivative of \( \frac{1}{\sqrt{1-x^2}} \) is \( \arcsin(x) + C \), where \( C \) is the constant of integration. When solving definite integrals, the constant \( C \) becomes unnecessary because you evaluate bounded intervals, leaving us with just \( \arcsin(x) \) as the practical antiderivative for this purpose. This step sets the stage for applying the limits of integration and obtaining a precise numerical result.

Arcsin Function

The arcsin function, written as \( \arcsin(x) \), is the inverse of the sine function over its principal range.

• The function \( \arcsin(x) \) is defined as the angle whose sine is \( x \).
• Its range is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), making it a unique function for values between -1 and 1.

In this exercise, the integrand \( \frac{1}{\sqrt{1-x^2}} \) directly relates to the derivative of the \( \arcsin(x) \) function. This step occurs by identifying that the provided function fits the format of the derivative of arcsin due to its components and behavior.The task involves evaluating \( \arcsin \left( \frac{1}{2} \right) \), which simplifies to \( \frac{\pi}{6} \). Understanding this simplification comes from knowing that sine of \( \frac{\pi}{6} \) equals \( \frac{1}{2} \), hence reinforcing how the arcsin and sine functions relate.

Limits of Integration

Definite integrals have set boundaries, known as limits of integration, guiding us on where to evaluate the antiderivative.

• The lower and upper limits of integration, denoted as \( 0 \) and \( \frac{1}{2} \) respectively in this example, determine the segment of the function to be assessed.
• These limits essentially "cap" the function, transforming an indefinite problem into a definite quantity.

In this solution process, substituting these values into the antiderivative \( \arcsin(x) \) gives us \( \arcsin(\frac{1}{2}) - \arcsin(0) \). Then:\[ \arcsin(\frac{1}{2}) = \frac{\pi}{6} \text{ and } \arcsin(0) = 0.\]Calculating their difference delivers the answer for the definite integral: \( \frac{\pi}{6} \). Thus, the integration process is completed by understanding how limits of integration frame and determine the scope of the result.