Problem 42 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \sqrt{\frac{x^{4}}{x^{3}-1}} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{2}{3} \sqrt{x^3 - 1} + C \).

Step 1: Simplify the Integrand

Start by simplifying the integrand \( \sqrt{\frac{x^{4}}{x^{3}-1}} \). We can rewrite it as \( \frac{x^{2}}{\sqrt{x^{3}-1}} \). This is because \( \sqrt{\frac{x^{4}}{x^{3}-1}} = \frac{\sqrt{x^4}}{\sqrt{x^{3}-1}} = \frac{x^{2}}{\sqrt{x^{3}-1}} \).

Step 2: Apply Substitution Method

Let's use substitution to simplify further. Set \( u = x^3 - 1 \), so \( du = 3x^2 dx \). Then, the integral becomes \( \int \frac{x^{2}}{\sqrt{u}} \cdot \frac{1}{3x^2} du = \frac{1}{3} \int u^{-1/2} du \).

Step 3: Integrate Using Power Rule

Now, integrate \( \frac{1}{3} \int u^{-1/2} du \) using the power rule. The power rule for integration states that \( \int u^{n} du = \frac{u^{n+1}}{n+1} + C \) for any real number \( n eq -1 \). Applying this, we get \( \frac{1}{3} \cdot \frac{u^{1/2}}{1/2} + C = \frac{2}{3} u^{1/2} + C \).

Step 4: Back-Substitute the Original Variable

Substitute back \( u = x^3 - 1 \) into the integral result. This gives us \( \frac{2}{3} (x^3 - 1)^{1/2} + C \). Hence, the integral evaluates to \( \frac{2}{3} \sqrt{x^3 - 1} + C \).

Key Concepts

Substitution Method

The substitution method is a handy technique in calculus integration. It simplifies complex integrals, making them easier to solve.
This method involves replacing a part of the integrand with a single variable, often denoted by 'u'.
### Performing a Substitution - Identify a part of the integrand that can simplify the integral when replaced by 'u'. In the given problem, the expression inside the square root, \( x^3 - 1 \), is designated as 'u'.- Calculate the differential, \( du \), which is the derivative of 'u', with respect to 'x'. Here, \( du = 3x^2dx \).- Substitute \( u \) and \( du \) back into the original integral to transform it into an easier form. In this case, simplify it to \( \frac{1}{3}\int u^{-1/2} du \).- This transformation also often requires rearranging terms to match \( du \). This simplifies calculations significantly and leads to a straightforward integration later.

Power Rule for Integration

The power rule is one of the simplest yet powerful tools for integration. It provides a formula to integrate functions of the form \( u^n \).
This rule enables us to find the antiderivative, which is crucial for solving integrals obtained after substitution.
**Understanding the Power Rule**- The formula for the power rule is \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n eq -1 \). Here, 'C' stands for the constant of integration.- In this exercise, after substituting \( u \), we applied the power rule with \( n = -1/2 \).- Applying the rule: \( \int u^{-1/2} \, du \), gives \( \frac{u^{1/2}}{1/2} + C \), which simplifies to \( 2u^{1/2} + C \).- It is a straightforward calculation, assuming you know how to handle exponents and fractions. This rule is vital in transforming an indefinite form into a defined function.

Back-Substitution

Back-substitution is the final step in problem-solving when using the substitution method.
Once the integral is solved in terms of 'u', convert it back to the original variable.
### How to Back-Substitute- Recall the original substitution made, which in our example was \( u = x^3 - 1 \).- Replace 'u' with the original expression to transform the solution back into the context of the original problem.- For our integral, back-substituting \( u \) gives \( \frac{2}{3}(x^3 - 1)^{1/2} + C \).- This step ensures that the solution relates to the variable in the original problem, making it meaningful and applicable.Back-substitution may seem minor, but it is crucial for maintaining consistency and correctness in the solution process.