Problem 36 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{18}_1 \sqrt{\frac{3}{z}} \,dz \)

Step-by-Step Solution

Answer

\(6 \sqrt{6} - 2 \sqrt{3}\)

Step 1: Simplify the Integrand

First, let's rewrite the integrand, \( \sqrt{\frac{3}{z}} \), using the property of exponents. The square root can be expressed as a power: \( \sqrt{\frac{3}{z}} = \left(\frac{3}{z}\right)^{1/2} \). This simplifies to \( 3^{1/2} \cdot z^{-1/2} \), or equivalently, \( \sqrt{3} \cdot z^{-1/2} \).

Step 2: Set Up the Integral

The integral now becomes \( \int_{1}^{18} \sqrt{3} \cdot z^{-1/2} \, dz \). We can factor out the constant \( \sqrt{3} \) to simplify the integration: \( \sqrt{3} \int_{1}^{18} z^{-1/2} \, dz \).

Step 3: Find the Antiderivative

Integrate \( z^{-1/2} \). The antiderivative of \( z^{n} \) is \( \frac{z^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Hence, the antiderivative of \( z^{-1/2} \) is \( \frac{z^{1/2}}{1/2} = 2z^{1/2} \).

Step 4: Evaluate the Definite Integral

Replace the antiderivative into the integral to evaluate: \( \sqrt{3} \times \left[ 2z^{1/2} \right]_{1}^{18} \). Calculate this by substituting the limits of integration: \[ \sqrt{3} \times \left( 2 \times 18^{1/2} - 2 \times 1^{1/2} \right) \].

Step 5: Calculate the Final Expression

Evaluate \( 18^{1/2} \) which is the square root of 18. Simplify: \( 18^{1/2} = \sqrt{18} = 3\sqrt{2} \). Now substituting back we have: \[ \sqrt{3} \times (2 \times 3\sqrt{2} - 2 \times 1) \]\[ = \sqrt{3} \times (6\sqrt{2} - 2) \]Simplify the expression: \[ = \sqrt{3} \times 6\sqrt{2} - \sqrt{3} \times 2 \].

Step 6: Simplify the Result

Distribute \( \sqrt{3} \) in the expression:\[ 6\sqrt{2} \times \sqrt{3} = 6 \sqrt{6} \] and \( \sqrt{3} \times 2 = 2 \sqrt{3} \).Thus, the final expression is: \[ 6 \sqrt{6} - 2 \sqrt{3} \].

Key Concepts

Antiderivatives

An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. It can be considered the reverse process of differentiation. In the given problem, we are tasked with finding the antiderivative of a specific function within the integral.

For functions of the form \( z^n \), the antiderivative can be obtained using the formula:

• \( \frac{z^{n+1}}{n+1} + C \)
• Here, \( C \) represents the constant of integration.

In our example, the function to integrate is \( z^{-1/2} \). Applying the formula, we get the antiderivative:

• \( \frac{z^{1/2}}{1/2} = 2z^{1/2} \)

This process allows us to find the function that describes the rate of change that our original function depicted. Understanding how to find antiderivatives is crucial in solving integrals like this one.

Exponents

Exponents often appear in mathematical expressions and play a crucial role in integration. They indicate how many times a number, known as the base, is multiplied by itself. For example, the expression \( z^{-1/2} \) involves a negative exponent and a fractional form.

When dealing with fractional exponents:

• The numerator signifies the power, i.e., how many times the number is used in a multiplication.
• The denominator signifies the root, i.e., the type of root being taken.

In the problem:

• Initially, we have \( \sqrt{\frac{3}{z}} \), which can be rewritten as \((\frac{3}{z})^{1/2}\).
• This simplifies to \( \sqrt{3} \cdot z^{-1/2} \), making it easier to integrate.

Knowing how to manipulate and interpret exponents is vital for simplifying expressions before integrating, as demonstrated here.

Integration by Parts

Integration by Parts is a technique used for integrating products of functions. It's based on the product rule for differentiation and can simplify complex integrals. The general formula is:
\[ \int u \, dv = uv - \int v \, du \]
Here,

• \( u \) is a function of \( z \).
• \( dv \) is the differential of another function of \( z \).

While integration by parts is an essential tool, it wasn't directly required for this particular problem. The integrand was in a straightforward form, \( z^{-1/2} \), that could be integrated using basic antiderivative rules.

However, understanding integration by parts is invaluable. It applies to more complicated products of functions where one can simplify an integral into more manageable terms.