Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\int_{0}^{3} \frac{x^{3} d x}{\sqrt{9+x^{2}}}\) by making the substitutions \(u=\sqrt{9+x^{2}}, \quad u^{2}=9+x^{2}, \quad 2 u d u=2 x d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( 18(1 - \sqrt{2}) \).

Step 1: Identify the integral

We need to evaluate the integral \( \int_{0}^{3} \frac{x^{3} dx}{\sqrt{9+x^{2}}} \).

Step 2: Make the substitution

Given that \( u = \sqrt{9 + x^2} \), then \( u^2 = 9 + x^2 \). Thus, \( x^2 = u^2 - 9 \). The derivative is \( 2x dx = 2u du \), hence \( x dx = u du \).

Step 3: Substitute and simplify the integral

Substituting \( x dx = u du \) and \( x^2 = u^2 - 9 \) into the integral gives:\[ \int \frac{x^3}{\sqrt{9 + x^2}} dx = \int \frac{x \cdot x^2}{u} dx = \int \frac{x (u^2 - 9)}{u} dx. \]Now substitute \(x dx = u du \):\[ \int \frac{(u^2 - 9) x dx}{u} = \int \frac{(u^2 - 9)}{u} u du = \int (u^2 - 9) du. \]

Step 4: Evaluate the integral

Integrate the expression: \( u^2 - 9 \):\[ \int u^2 du - \int 9 du = \frac{u^3}{3} - 9u + C. \]

Step 5: Transform back to original variables and evaluate the definite integral

Transform back using \( u = \sqrt{9 + x^2} \) and evaluate from 0 to 3:* When \( x = 0 \), \( u = \sqrt{9 + 0^2} = 3 \).* When \( x = 3 \), \( u = \sqrt{9 + 3^2} = \sqrt{18} = 3\sqrt{2} \).Thus, the definite integral from 3 to \(3\sqrt{2} \):\[ \left[ \frac{u^3}{3} - 9u \right]_3^{3\sqrt{2}} \]\[ = \left( \frac{(3\sqrt{2})^3}{3} - 9(3\sqrt{2}) \right) - \left( \frac{3^3}{3} - 9 \times 3 \right) \]\[ = \left( \frac{27\sqrt{2}}{3} - 27\sqrt{2} \right) - (9 - 27) \].

Step 6: Compute the values

Continue with the arithmetic:\[ = \left( 9\sqrt{2} - 27\sqrt{2} \right) - (-18) \]\[ = -18\sqrt{2} + 18 \]\[ = 18(1 - \sqrt{2}) \].

Key Concepts

Substitution Method

The substitution method in calculus integration is a crucial technique to simplify integrals by transforming them into a different, often easier, form to integrate. We often substitute a part of the integrand with a new variable. This can simplify complex algebraic expressions into standard forms that are easier to work with.

Here's how it works:

• Identify a substitution that will make the integral easier. This is often a part of the integrand that matches a derivative form.
• Express the original integral in terms of the new variable.
• Don't forget to adjust both the function being integrated and the limits of integration if it's a definite integral.
• After integrating, change back to the original variable to find the solution.

This method is particularly useful in solving integrals where a direct integration approach seems complicated or impossible.

Definite Integral Evaluation

In this exercise, we perform definite integral evaluation, which involves finding the exact value of an integral over a specific interval. Unlike indefinite integrals, which have a constant of integration, definite integrals calculate the net area under the curve from one point to another.

Steps involved in evaluating a definite integral:

• Transform the original function into a simpler form using techniques like integration by substitution.
• Integrate the function with respect to the new variable.
• Evaluate the integrated result at the upper and lower limits of the original integral.
• Subtract the value at the lower limit from the value at the upper limit.

This process gives the net area between the curve and the x-axis over the specified range, considering any parts of the curve below the x-axis as negative area.

Square Root Substitution

Square root substitution is a powerful technique when dealing with integrals containing terms inside square roots. By carefully selecting a substitution that involves the square root, the integral can be transformed into a simpler form.

In this exercise, we used the substitution:

• Let \( u = \sqrt{9 + x^2} \).
• This leads to \( u^2 = 9 + x^2 \), which simplifies our problem.

By substituting and rearranging, you can often eliminate the square root and create a more manageable integral,
such as polynomials, which are far easier to integrate. It requires adjusting both the integrand and the differential variable accordingly, as seen in the original solution.

Arithmetic Computation in Integrals

After substituting and integrating the function, arithmetic computation helps derive the final value. This includes evaluating at the limits and finding the difference as per definite integral rules.

The main focus here is on performing careful arithmetic operations:

• Ensure that you correctly substitute back into the original variable before calculation.
• Accurately perform arithmetic operations when evaluating the integrated function at the limits.
• Subtract the lower limit value from the upper limit value to find the net area or result.

This step is crucial because a small error in arithmetic can lead to incorrect solutions. Careful checking of calculations ensures accuracy and correctness in the final result.