Problem 31 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{6 d x}{x \sqrt{25 x^{2}-1}} $$

Step-by-Step Solution

Answer

Use substitution \(u = 25x^2 - 1\) to simplify the integral.

Step 1: Identify the substitution

Notice the function inside the square root: \(25x^2 - 1\). A useful substitution can be based on the expression inside the square root. We will use \(u = 25x^2 - 1\). This substitution will simplify the integral considerably.

Step 2: Find the differential

Differentiate \(u\) with respect to \(x\) to find the differential \(du\). Since \(u = 25x^2 - 1\), we have \(\frac{du}{dx} = 50x\). Therefore, \(du = 50x \, dx\).

Step 3: Solve for dx

Rearrange \(du = 50x \, dx\) to express \(dx\) in terms of \(du\) and \(x\). This gives \(dx = \frac{du}{50x}\).

Step 4: Substitute into the integral

Substitute the expressions for \(u\) and \(dx\) into the original integral: \( \int \frac{6}{x \sqrt{u}} \cdot \frac{du}{50x} \). This simplifies to \( \int \frac{6}{50x^2 \sqrt{u}} \cdot du\).

Step 5: Simplify the integrand

Replace \(x^2\) using the substitution \(u = 25x^2 - 1\), which implies \(25x^2 = u + 1\) or \(x^2 = \frac{u + 1}{25}\). Substitute to get \( \int \frac{6}{50 \cdot \frac{u + 1}{25} \sqrt{u}} \, du\).

Step 6: Simplify the fraction

Simplify \( \int \frac{6 \cdot 25}{50(u+1) \sqrt{u}} \, du \) to \( \int \frac{3}{(u+1) \sqrt{u}} \, du \).

Step 7: Re-evaluate the integral with the substitution

The form of the integral is now \( \int \frac{3}{(u+1) \sqrt{u}} \, du \). Continue simplifying and solve using standard integral formulas.

Key Concepts

Definite Integrals

A definite integral is a fundamental concept in calculus which provides the net area under a curve. It is typically expressed with bounds of integration, transforming it from an indefinite integral to a precise numerical value.
In the context of solving the given integral, we aren't directly evaluating a definite integral as there are no explicit bounds mentioned. However, understanding this concept is crucial for when specific limits are provided. The evaluation of such integrals involves determining the exact area under the curve defined by the function within the given interval. This is significant in many real-world applications, such as calculating distances, areas, and total accumulated quantities.
When dealing with definite integrals, the result depends on both the antiderivative of the function and the evaluation at specific bounds. It's essential to understand that definite integrals combine the process of finding an antiderivative with real-world contextual applications.

• Involve limits of integration, providing specific numerical results.
• Expressed as \( \int_{a}^{b} f(x) \, dx \)
• Calculate net areas under curves within given intervals.
• Used to solve practical problems requiring total quantities or change.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are essentially the reverse process of differentiation. Finding an antiderivative means identifying a function whose derivative yields the original function.
In the provided exercise, simplifying the integrand using substitution helps us to find the antiderivative more easily. Once the substitution is made, and the new integral is set up, we aim to reach a form where we can readily determine the antiderivative using standard integral formulas.
Understanding antiderivatives is key to solving both definite and indefinite integrals, as it represents the essence of finding a function's accumulated change or total area under its curve.

• Opposite process of differentiation.
• Key in both definite and indefinite integrals.
• Allows for the calculation of accumulated change or areas.
• Simplified by substituting different variables to match standard forms.

Calculus Exercises

Calculus exercises often involve finding both derivatives and integrals, allowing students to apply theoretical concepts to practical problems. These exercises are designed to enhance understanding and proficiency in calculus by practicing simplification strategies, such as substitution, as in the current problem step-by-step solution.
Substitution is a key technique for solving integrals that simplifies complex integrands to more standard forms. By altering the variable of integration, it reduces challenging expressions into simpler ones that align with known antiderivative forms.
Mastery of calculus exercises through consistent practice allows students to improve problem-solving skills and adapt to various calculus scenarios. It equips them with the competence needed to tackle associated fields, such as physics, engineering, and statistics.

• Involves solving integrals and derivatives.
• Applies theoretical concepts to solve real-world problems.
• Uses techniques like substitution for simplification.
• Enhances problem-solving skills and mathematical adaptability.