Problem 24 - Calculus of a Single Variable

Question

Evaluating a Definite Integral In Exercises \(21-32\) evaluate the definite integral. $$ \int_{\sqrt{3}}^{3} \frac{1}{x \sqrt{4 x^{2}-9}} d x $$

Step-by-Step Solution

Answer

The evaluated definite integral is \(3\sqrt{3}\)

Step 1: Identify Substitution Variable

Analyze the integral and notice that \(u = 4x^{2}-9\) is a good choice for substitution since its derivative is in the integral, and it is within a square root function. The derivative \(du = 8x dx\), which aligns with the presence of \(x dx\), except the coefficient of \(x\) in the integral is \(1\), not \(8\). That discrepancy will be fixed by dividing \(du\) by \(8\).

Step 2: Perform the Substitution

Rewrite \(du\) in terms of \(dx\), and substitute \(u\) and \(du\) into the integral: \[dx = du/8x\]Substituting \(du\) and \(u\) into given equation results in\[\int_{\sqrt{3}}^{3} \frac{1}{2 \sqrt{u}} du\]

Step 3: Update Limits of Integration

Now, substitute also the original boundaries from x-values to u-values: when \(x=\sqrt{3}\), \(u = 4(\sqrt{3})^{2}-9 = 3\), and when \(x=3\), \(u = 4(3)^{2}-9 = 27\). The updated integral is\[\int_{3}^{27} \frac{1}{2 \sqrt{u}} du\]

Step 4: Evaluate the Integral

The antiderivative of \(1/2\sqrt{u}\) is \(\sqrt{u}\). So, applying the Fundamental Theorem of Calculus:\[\left[\sqrt{u} \right]_3^{27} = \sqrt{27} - \sqrt{3} = 4\sqrt{3} - \sqrt{3}\]

Step 5: Final Simplification

Simplify the answer: \(4\sqrt{3} - \sqrt{3} = 3\sqrt{3}\)

Key Concepts

Substitution Method

The Substitution Method is a key technique used to evaluate integrals, especially when faced with complex expressions. This method simplifies the integral by changing variables. Think of it as replacing a complicated part with a simpler placeholder.
In the given exercise, we cleverly selected the substitution variable as \( u = 4x^2 - 9 \). The reason for this choice is that its derivative \( du = 8x \, dx \) reflects the structure within the square root and the integral. To align with \( du \), we adjusted by introducing a factor to match the terms perfectly. This involves dividing by 8 because the coefficient in \( du \) is 8, while in the integral it was 1.

• This method helps in cases where the integral seems daunting or not easily solvable by basic integration rules.
• Choosing the right substitution is critical, often involving expressions whose derivatives are present in the integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between derivatives and integrals. It makes evaluating definite integrals straightforward once we have an antiderivative.
In this exercise, after performing substitution, we obtained an integral in terms of \( u \). The theorem tells us that if we find an antiderivative of our integrand, we can easily compute the definite integral by substituting in our limits.

• The theorem suggests that integration reverses differentiation.
• It allows the evaluation of definite integrals using antiderivatives.

By applying this powerful theorem, we were able to evaluate the expression \([\sqrt{u}]_3^{27}\), which translates to substituting 27 and 3 into our antiderivative \( \sqrt{u} \) and finding the difference.

Antiderivative

An Antiderivative is a function whose derivative is the original function you started with. Finding an antiderivative is crucial for solving integrals.
In the problem, once the substitution had simplified the integral to \( \int \frac{1}{2 \sqrt{u}} \, du \), the task became finding the antiderivative. This was \( \sqrt{u} \), a function whose derivative matches the integrand. Understanding this allows us to compute the definite integral through evaluation.

• Knowing common antiderivatives is key to quickly solving integrals.
• In this problem, recognizing that \( \frac{1}{2 \sqrt{u}} \) relates directly to the derivative of \( \sqrt{u} \) made the calculation simpler.

Integration Limits

Integration Limits define the scope over which you are integrating. They crucially determine the values at which you evaluate any antiderivative during the computation of a definite integral.
Upon performing a substitution, the limits of integration transform as well. In our exercise, x-limits of \( \sqrt{3} \) and 3 were transformed into u-limits of 3 and 27. This step is vital, as it ensures you are integrating over the correct interval.

• Always update limits when making a variable substitution, reflecting how the original variables transform.
• Correct limits lead to accurate evaluation of the integral.

By ensuring our limits matched the new variable \( u \), we directly applied the Fundamental Theorem of Calculus to find the difference \( \sqrt{27} - \sqrt{3} \), simplifying to the solution \( 3\sqrt{3} \).