Problem 13 - Calculus: Early Transcendental Functions

Question

Use Part I of the Fundamental Theorem to compute each integral exactly. $$\int_{0}^{1 / 2} \frac{3}{\sqrt{1-x^{2}}} d x$$

Step-by-Step Solution

Answer

The exact value of the integral is \( \frac{\pi}{2} \).

Step 1: Identify the integrand

The function to be integrated is \( \frac{3}{\sqrt{1-x^{2}}} \). The antiderivative of this function is recognized as 3 times the antiderivative of \( \frac{1}{\sqrt{1-x^{2}}} \), that is \( 3 \text{ arcsin}(x) \).

Step 2: Apply limits of integral

Plugging the upper limit (1/2) and then subtracting the result of the lower limit (0) in the antiderivative we get \( \left[3 \text{ arcsin}\left(\frac{1}{2}\right) - 3 \text{ arcsin}(0)\right] \)

Step 3: Simplify the results

Evaluate arcsin(1/2) which equals \( \frac{\pi}{6} \) and arcsin(0) which equals 0, multiply by 3 to reach the conclusion that \( 3\left[\frac{\pi}{6} - 0\right] = \frac{\pi}{2} \).

Key Concepts

Definite Integral

A definite integral is a way to calculate the area under a curve within certain bounds. It's a fundamental concept in calculus and very important for many types of real-world computations. When you see an integral with limits, like \( \int_{0}^{1/2} \frac{3}{\sqrt{1-x^{2}}} \, dx \), it's a definite integral.

Here's how it works in simple steps:

• You find the antiderivative (also called the indefinite integral) of the function inside the integral sign, which is often referred to as the "integrand." In our example, the integrand is \( \frac{3}{\sqrt{1-x^{2}}} \).
• After finding the antiderivative, you evaluate it at the upper and lower limits of the integral. This gives you a number that represents the exact area under the curve between those two points.
• Finally, subtract the result after plugging in the lower limit from the result after plugging in the upper limit to get the value of the definite integral.

This method is part of why the Fundamental Theorem of Calculus is so crucial— it turns the difficult task of finding areas into simple arithmetic.

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially the reverse of differentiation. If you have a function \( f(x) \) and you find another function \( F(x) \) such that \( F'(x) = f(x) \), then \( F(x) \) is the antiderivative of \( f(x) \).

For the integrand \( \frac{3}{\sqrt{1-x^{2}}} \), identifying its antiderivative involves recognizing a common form. The function \( \frac{1}{\sqrt{1-x^{2}}} \) is a standard form whose antiderivative is \( \text{arcsin}(x) \). Therefore, the antiderivative of our integrand is \( 3 \times \text{arcsin}(x) \).

• This recognition of standard forms is a vital skill in calculus, as it allows you to tackle integrals effectively.
• Once the antiderivative is determined, you can use it in the process of evaluating definite integrals.

Having the antiderivative makes it straightforward to apply the limits of integration and find exact areas, as shown in the example from the step-by-step solution.

Arcsine Function

The arcsine function, denoted as \( \text{arcsin}(x) \), is the inverse of the sine function on a restricted domain. It's used to determine the angle whose sine is \( x \). The output of \( \text{arcsin}(x) \) lies within \([-\pi/2, \pi/2]\).

In the context of our exercise, recognizing \( \frac{1}{\sqrt{1-x^{2}}} \) as the derivative of \( \text{arcsin}(x) \) is crucial. This connection allows us to determine that \( \text{arcsin}(x) \) is the antiderivative of \( \frac{1}{\sqrt{1-x^{2}}} \).

• Using the arcsine function involves understanding its properties and constraints, such as its limited range and domain.
• When evaluating, like finding \( \text{arcsin}(\frac{1}{2}) \), we determine which angle yields that particular sine value. Here, it's \( \frac{\pi}{6} \) because \( \sin(\frac{\pi}{6}) = \frac{1}{2} \).

Mastering functions like arcsine enriches your ability to solve integrals involving trigonometric functions and their inverses.