Problem 41 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \sqrt{\frac{x^{3}-3}{x^{11}}} d x$$

Step-by-Step Solution

Answer

\( \frac{2}{27} (1-3x^{-3})^{3/2} + C \).

Step 1: Simplify the Integrand

The given integral is \( \int \sqrt{\frac{x^3-3}{x^{11}}} \, dx \). Begin by separating the terms in the square root. We can rewrite this as \( \sqrt{x^3 \cdot \frac{1}{x^{11}} - \frac{3}{x^{11}}} \), which is \( \sqrt{x^{-8} - 3x^{-11}} \).

Step 2: Simplify the Expression Further

Inside the square root, use the property \( x^a \cdot x^b = x^{a+b} \). The expression \( \sqrt{x^{-8} - 3x^{-11}} \) can be simplified to \( \sqrt{x^{-8}} \cdot \sqrt{1 - 3x^{-3}} \) because the square root of a product is the product of the square roots.

Step 3: Further Simplify and Set Up for Integration

By simplifying \( \sqrt{x^{-8}} \), we get \( x^{-4} \). So now, the integral becomes \( \int x^{-4} \sqrt{1 - 3x^{-3}} \, dx \).

Step 4: Use Substitution Method

Let \( u = 1 - 3x^{-3} \), which implies \( du = 9x^{-4} \, dx \). Thus, \( x^{-4} \, dx \) can be replaced by \( \frac{1}{9} du \). Substitute back, and the integral becomes \( \frac{1}{9} \int \sqrt{u} \, du \).

Step 5: Integrate

The integral \( \int \sqrt{u} \, du \) is \( \int u^{1/2} \, du \). This evaluates to \( \frac{2}{3} u^{3/2} + C \). Substitute back \( u = 1 - 3x^{-3} \) to get \( \frac{2}{27} (1 - 3x^{-3})^{3/2} + C \) as the integral's value.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify the process of integration. Let's walk through how substitution makes things easier:

If you have an integral that looks complicated, like \[ \int x^{-4} \sqrt{1 - 3x^{-3}} \, dx,\] you can simplify it by introducing a new variable. In this example, we let \( u = 1 - 3x^{-3} \).

Why do we do this?

• It helps transform the integral into a simpler form.
• It can turn a complex expression into a basic polynomial or another simpler function.

With \( u = 1 - 3x^{-3} \), we find the derivative \( du = 9x^{-4} \, dx \), so \( x^{-4} \, dx \) is simply \( \frac{1}{9} \, du \). Now we can rewrite the integral:\[ \int x^{-4}\sqrt{1-3x^{-3}}\, dx = \frac{1}{9} \int \sqrt{u} \, du.\]
By changing variables, the integration becomes a more straightforward process. This is why substitution is a favored technique among calculus students.

Rational Exponents

Understanding rational exponents is key in calculus and particularly in integration, as many problems involve fractional powers.

In our given problem, there is an expression: \[ \sqrt{x^{-8}-3x^{-11}},\] which involves rational exponents.

Note that a square root can be written as an exponent:

• The square root \( \sqrt{a} \) is the same as \( a^{1/2} \).
• For our function, \( \sqrt{u} \) is \( u^{1/2} \).

These notations are interchangeable and can make integration more manageable, as powers are often easier to work with than roots when simplifying or performing calculus operations.

This approach reflects the use of properties leverage in this problem, where simplification of the expression with rational exponents allows for easier integration. Once you get familiar with changing roots to exponents, these problems become less intimidating.

Integral Simplification

Integral simplification is an essential skill that makes complex integrals approachable. Let's explore how you might simplify an integral to make it more workable:

In the initial expression \[ \int \sqrt{\frac{x^3-3}{x^{11}}} \, dx,\] rearranging and simplifying the terms is crucial.

Here’s how you can think about simplification:

• Break down the expression inside the integral into factors or smaller components.
• Simplify complex fractions and terms like \( \frac{x^3-3}{x^{11}} \) into single exponents (\( x^{-8} - 3x^{-11} \)) by splitting and reorganizing terms.

The square root of the product turned into a multiplication of square roots, making it easier to isolate integral parts you can work with separately, leading to \[ \int x^{-4} \sqrt{1 - 3x^{-3}} \, dx.\]

Breaking down integrals and seeing where you can simplify step-by-step is key, allowing tricky integrals to transform into simpler forms that are straightforward to evaluate.