Problem 24 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^8_1 x^{-2/3} \,dx \)

Step-by-Step Solution

Answer

The integral evaluates to 3.

Step 1: Identify the Integral Type

The given integral \( \int_{1}^{8} x^{-2/3} \, dx \) is a definite integral. We need to find an antiderivative of the function \( x^{-2/3} \).

Step 2: Apply the Power Rule for Integrals

To integrate \( x^{-a} \), we use the formula for power integrals: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), assuming \( n eq -1 \). Here, \( n = -2/3 \), so \( n + 1 = 1/3 \).

Step 3: Integrate the Function

Apply the power rule: \( \int x^{-2/3} \, dx = \frac{x^{1/3}}{1/3} + C = 3x^{1/3} + C \).

Step 4: Evaluate the Definite Integral

Now, we evaluate this antiderivative from 1 to 8: \[ \int_{1}^{8} x^{-2/3} \, dx = \left[ 3x^{1/3} \right]_{1}^{8} = 3(8)^{1/3} - 3(1)^{1/3}. \]

Step 5: Calculate the Numerical Values

Calculate: * \( 8^{1/3} = 2 \) because \( 2^3 = 8 \).Substitute:\[ 3 \times 2 - 3 \times 1 = 6 - 3 = 3. \]

Key Concepts

Power Rule for Integrals

The power rule for integrals is a fundamental concept in calculus that helps us to find the area under a curve. When dealing with a function of the form \( x^n \), the power rule provides us with a simple formula to integrate these types of functions. This rule states that to find the integral of \( x^n \), you can use the formula:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

This formula is applicable as long as \( n eq -1 \). The reason behind this exception is tied to the behavior of the function \( x^{-1} \), whose integral is the natural logarithm function \( \ln|x| + C \). To use the power rule effectively:

• Identify the exponent \( n \).
• Add 1 to \( n \) to adjust the exponent of \( x \).
• Divide \( x^{n+1} \) by \( n+1 \).
• Add the constant of integration \( C \), though this is not needed for definite integrals.

In the original problem, \( n = -2/3 \), leading to an antiderivative \( 3x^{1/3} + C \). This is a practical application of the power rule.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function given for integration. It essentially reverses the process of differentiation. If you have a function \( f(x) \), an antiderivative \( F(x) \) is such that:

• \( F'(x) = f(x) \)

This concept is vital because it connects differentiation and integration:

• Differentiation gives you the slope or rate of change.
• Integration, or finding the antiderivative, provides the accumulated area under the curve (or the total effect of the rates of change).

Finding the antiderivative involves applying various rules, such as the power rule, to reverse the derivative operations. In the original exercise, the antiderivative we found using the power rule was \( 3x^{1/3} + C \). For definite integrals, while \( C \) usually appears in the indefinite integral, it cancels out when computing the definite integral from one point to another.

Calculus

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. Its two main parts are:

• Differential calculus, which studies how things change (derivatives).
• Integral calculus, which focuses on accumulation, or finding the total effect over an interval (integrals).

Integral calculus, which we use to solve definite and indefinite integrals, is particularly concerned with:

• Finding areas under curves, lengths, and other related measurements.
• Solving problems involving the total accumulation of quantities.

The definite integral, like the one in the exercise \( \int_{1}^{8} x^{-2/3} \, dx \), provides a numerical value representing the total area under the curve from one limit to another. It is achieved by evaluating the antiderivative at the upper and lower limits of integration and finding their difference.In essence, calculus equips us with tools to handle continuous change and to derive meaningful values from complex, variable equations.