Problem 50 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{0}^{3 \sqrt{2} / 4} \frac{d s}{\sqrt{9-4 s^{2}}}$$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{\pi}{8} \).

Step 1: Identify the Form of the Integral

The integral given is \( \int \frac{ds}{\sqrt{9-4s^2}} \). This matches the form of \( \int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left( \frac{u}{a} \right) + C \), where \( a = 3 \). Here we need to adjust for any constants by substitution.

Step 2: Substitute to Simplify the Integral

To simplify, notice the inner component \( 4s^2 \). Set \( u = 2s \), then \( du = 2 \, ds \) or \( ds = \frac{1}{2} \ du \). The integral becomes \( \int \frac{1}{2} \cdot \frac{du}{\sqrt{9 - u^2}} \).

Step 3: Change the Limits of Integration

For the substitution \( u = 2s \):- When \( s = 0 \), \( u = 0 \).- When \( s = \frac{3\sqrt{2}}{4} \), \( u = \frac{3\sqrt{2}}{2} \).Thus, the new limits are from 0 to \( \frac{3\sqrt{2}}{2} \).

Step 4: Evaluate the Integral

The integral now is \( \frac{1}{2} \int_{0}^{\frac{3\sqrt{2}}{2}} \frac{du}{\sqrt{9-u^2}} \). This evaluates to \( \frac{1}{2} \left[ \sin^{-1} \left( \frac{u}{3} \right) \right]_{0}^{\frac{3\sqrt{2}}{2}} \).

Step 5: Compute the Final Result

Calculate the antiderivative at the bounds:1. For the upper bound, \( u = \frac{3\sqrt{2}}{2} \): \( \sin^{-1}\left( \frac{\frac{3\sqrt{2}}{2}}{3} \right) = \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) = \frac{\pi}{4} \).2. For the lower bound, \( u = 0 \): \( \sin^{-1}\left( \frac{0}{3} \right) = 0 \).3. The result of the integral is \( \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{8} \).

Key Concepts

Definite Integrals

A definite integral is a concept from integral calculus that computes the accumulation of quantities, such as area under a curve, between two specified limits. It provides a way to calculate the exact sum of infinitesimally small data points over an interval.

• The notation used for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. These limits determine the starting and ending points of the region over which the function \( f(x) \) is integrated.
• The result of a definite integral is a pure number, representing a real-world quantity, such as area, volume, or even a probability.
• Definite integrals have a property known as the "Fundamental Theorem of Calculus." This theorem connects the process of differentiation with integration, allowing us to evaluate a definite integral using an antiderivative or an indefinite integral.

The exercise asks us to evaluate a definite integral with specific boundaries. These are often used to find areas under curves represented by a mathematical function.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving certain radical expressions. It changes the variable of integration using trigonometric identities to make the integral more straightforward to solve.

• This technique is handy when dealing with integrals containing expressions like \( \sqrt{a^2 - x^2} \), which don't simplify easily with direct integration. In the problem provided, this structure is present as \( \sqrt{9 - 4s^2} \).
• Trigonometric substitution uses relationships such as \( x = a \sin(\theta) \) or \( x = a \cos(\theta) \), aligning the complex radical expression to a simpler trigonometric model like \( 1 - \sin^2(\theta) = \cos^2(\theta) \).
• In our example, substituting \( u = 2s \) reduces the complexity by transforming the integral boundaries and the expression for the integrand into a simpler form.

Using trigonometric substitution requires careful consideration of the limits of integration as the initial variable changes to match the trigonometric function.

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial in evaluating certain integrals, particularly those arising from trigonometric and hyperbolic substitutions. These functions help in returning to the original variable after simplification.

• Commonly encountered inverse trigonometric functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \). They provide the angle whose trigonometric value is a given number.
• These functions are often used in integration methods, like transforming expressions back to the original variable after employing trigonometric substitutions.
• In the exercise, after manipulating the integral into a form we could integrate directly, we use the inverse sine function, \( \sin^{-1}\), to express the antiderivative.

Inverse trigonometric functions reverse earlier substitutions, yielding exact solutions for the range and nature of the original problem settings.