Problem 42 - Calculus Single Variable

Question

Each of the integrands involves an expression of the form \(a^{2}-b^{2} x^{2}, a^{2}+b^{2} x^{2},\) or \(b^{2} x^{2}-a^{2} .\) Use an indirect substitution of the form \(x=(a / b) \sin (\theta), x=(a / b)\) \(\tan (\theta),\) or \(x=(a / b) \sec (\theta)\) to calculate the given integral. $$ \int_{4 / 3}^{\sqrt{2}} \sqrt{9 x^{2}-16} d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately 0.475.

Step 1: Identify the Form

The integrand is in the form \( \sqrt{b^{2} x^{2} - a^{2}} \), where \( b^{2} = 9 \) and \( a^{2} = 16 \). Hence, this matches the form \( b^{2} x^{2} - a^{2} \).

Step 2: Choose the Substitution

For integrals of the form \( b^{2} x^{2} - a^{2} \), we use the substitution \( x = \frac{a}{b} \sec(\theta) \). Thus, let \( x = \frac{4}{3} \sec(\theta) \).

Step 3: Determine \( dx \)

Differentiate \( x = \frac{4}{3} \sec(\theta) \) to find \( dx \). \( dx = \frac{4}{3} \sec(\theta) \tan(\theta) \, d\theta \).

Step 4: Change the Limits of Integration

Substitute the original limits into \( x = \frac{4}{3} \sec(\theta) \) to find the angular limits. For \( x = \frac{4}{3} \), \( \sec(\theta) = 1 \rightarrow \theta = 0 \). For \( x = \sqrt{2} \), \( \sec(\theta) = \frac{3\sqrt{2}}{4} \rightarrow \theta = \sec^{-1}\left(\frac{3\sqrt{2}}{4}\right) \approx 0.6155 \text{ radians}. \)

Step 5: Simplify the Integral

Substitute into the integral:\[\int \sqrt{9\left(\frac{4}{3}\sec(\theta)\right)^2 - 16} \frac{4}{3} \sec(\theta) \tan(\theta) \, d\theta \]Simplify to:\[\int 4 \tan^{2}(\theta) \frac{4}{3} \sec(\theta) \tan(\theta) \, d\theta \] Then:\[\frac{16}{3} \int \tan^{3}(\theta) \sec(\theta) \, d\theta.\]

Step 6: Evaluate the Integral

We perform integration by parts or substitution on \( \int \tan^{3}(\theta) \sec(\theta) \, d\theta \). Let \( u = \tan(\theta), \; du = \sec^{2}(\theta) \, d\theta.\)Then, rewrite:\[\frac{16}{3} \int \tan^{2}(\theta) \sec^{2}(\theta) \, d\theta \]Using \( \tan^{2}(\theta) = \sec^{2}(\theta) - 1 \), we have:\[\frac{16}{3} \int (\sec^{2}(\theta) - 1) \sec^{2}(\theta) \, d\theta \] This simplifies to:\[\frac{16}{3} \int (\sec^{4}(\theta) - \sec^{2}(\theta)) \, d\theta \]Perform the integral:\[\frac{16}{3} \left( \frac{\tan(\theta)\sec(\theta)}{3} - \tan(\theta) \right) + C\] Evaluate the definite integral from \(\theta = 0 \) to \(\theta = \sec^{-1}\left(\frac{3\sqrt{2}}{4}\right)\).

Step 7: Substitute Back and Apply the Limits

Use the substitution, back calculate to convert \( \theta \) back to \( x \) using the identities specific to \( x = \frac{4}{3} \sec(\theta) \). Evaluate expressions at the earlier calculated limits for \( \theta \) and subtract the results to find the definite integral value.

Key Concepts

Definite Integral

In integral calculus, a *definite integral* is used to calculate the area under a curve within a specific interval. It provides a precise value depicting the accumulation of quantities, such as areas, volumes, and other concepts related to the integral. The definite integral \[ \int_{a}^{b} f(x) \, dx \] is calculated over the interval from \( a \) to \( b \). In the given exercise, we are tasked with evaluating the definite integral: \[ \int_{4/3}^{\sqrt{2}} \sqrt{9x^2 - 16} \, dx \] This integral determines the exact area under the curve of the function \( \sqrt{9x^2 - 16} \) between \( x = 4/3 \) and \( x = \sqrt{2} \). By solving this integral, we acquire an understanding of the behavior and properties of the function within the specified limits.

Integration Techniques

To solve definite integrals, particularly ones with complex forms like the one in our exercise, various *integration techniques* come into play. These techniques simplify the integral into a more manageable form. One effective method is *trigonometric substitution*, which we employ for integrating functions that contain expressions such as \( \sqrt{b^2 x^2 - a^2} \).When dealing with an expression of the form \( \sqrt{b^2 x^2 - a^2} \), we use the substitution \( x = (a/b) \sec(\theta) \). Here, we chose \( x = \frac{4}{3} \sec(\theta) \) for our integral.This substitution transforms the integral into a new variable \( \theta \) and makes it easier to handle with standard trigonometric identities. The process involves several key steps:

• Identifying the correct substitution for the form.
• Transforming the limits of integration from x-values to \( \theta \)-values.
• Substituting back the original variable after integration and applying the limits.

Each of these steps requires careful calculations to ensure that the integral correctly represents the area under the curve.

Integral Calculus

*Integral calculus* is a fundamental branch of mathematics focused on the concepts of integrals, which represent the accumulation of quantities. It plays a crucial role in understanding the properties of functions integrated over defined intervals. Within integral calculus, there are two main types of integrals: definite and indefinite. In the context of our exercise, we focus on definite integrals, which provide a specific value signifying the total accumulation between two points. By employing trigonometric substitution in solving our problem, we engaged in a common technique in integral calculus, applying transformations that leverage trigonometric identities to evaluate the integral in a simpler form. Integral calculus not only helps compute areas and volumes, but also solves real-world problems across physics, engineering, and economics. It enables us to:

• Determine the total distance traveled by an object.
• Calculate work done by a force.
• Find the center of mass of an object.

By mastering the tools and methodologies of integral calculus, students can solve complex integrals, interpret physical phenomena, and achieve significant insights into the behavior of dynamic systems. In our solved exercise, the selected techniques and substitutions embody the powerful capabilities of integral calculus in action.