Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{1}^{8} \sqrt[3]{w} d w $$

Step-by-Step Solution

Answer

The definite integral evaluates to 11.25.

Step 1: Identify the function to integrate

The given integral is \( \int_{1}^{8} \sqrt[3]{w} \, dw \). The function we need to integrate is \( f(w) = \sqrt[3]{w} \), which can also be written as \( f(w) = w^{\frac{1}{3}} \).

Step 2: Find the antiderivative

To apply the Second Fundamental Theorem of Calculus, find the antiderivative of \( f(w) = w^{\frac{1}{3}} \). Use the power rule for integration, which states that \( \int w^n \, dw = \frac{w^{n+1}}{n+1} + C \) for any real number \( n eq -1 \). Thus, the antiderivative of \( w^{\frac{1}{3}} \) is \( \frac{w^{\frac{1}{3}+1}}{\frac{1}{3}+1} = \frac{w^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4}w^{\frac{4}{3}} \).

Step 3: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if \( F(w) \) is an antiderivative of \( f(w) \), then \( \int_{a}^{b} f(w) \, dw = F(b) - F(a) \). Here, \( F(w) = \frac{3}{4}w^{\frac{4}{3}} \). Substitute \( a = 1 \) and \( b = 8 \) into the theorem: \( \int_{1}^{8} \sqrt[3]{w} \, dw = \frac{3}{4}(8^{\frac{4}{3}}) - \frac{3}{4}(1^{\frac{4}{3}}) \).

Step 4: Calculate the values at the limits

Calculate \( 8^{\frac{4}{3}} \) and \( 1^{\frac{4}{3}} \). For \( 8^{\frac{4}{3}} \), consider that \( 8 = 2^3 \), so \( 8^{\frac{1}{3}} = 2 \). Therefore, \( 8^{\frac{4}{3}} = (8^{\frac{1}{3}})^4 = 2^4 = 16 \). \( 1^{\frac{4}{3}} = 1 \) because any power of 1 is 1. Substitute these values back into the expression: \( \frac{3}{4}(16) - \frac{3}{4}(1) \).

Step 5: Simplify and find the result

Simplify the expression \( \frac{3}{4}(16) - \frac{3}{4}(1) \). First evaluate \( \frac{3}{4}(16) = 12 \) and then \( \frac{3}{4}(1) = \frac{3}{4} \). Thus, the integral evaluates to \( 12 - \frac{3}{4} = 12 - 0.75 = 11.25 \).

Key Concepts

definite integral evaluation

Evaluating a definite integral might seem daunting at first, but it becomes straightforward once you understand the concepts and steps involved. In straightforward terms, a definite integral represents the net area under a curve from one point (the lower limit) to another (the upper limit) on a graph. In mathematical notation, a definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the function you are integrating from \( x = a \) to \( x = b \).

This process involves:

• Finding the antiderivative of the function: This gives you a new function whose derivative is the original function.
• Applying the limits: You evaluate this antiderivative function at the upper limit and then at the lower limit.
• Subtracting the lower limit result from the upper limit result: This gives the value of the definite integral.

Doing this allows us to accurately calculate the total area, accounting for any parts of the graph that dip below the x-axis, which count as negative areas.

integration by power rule

One of the simplest rules of integration is the power rule. This rule is especially handy when you're dealing with functions of the form \( x^n \), where \( n \) is a real number. The power rule states: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]where \( C \) is the constant of integration.

Let's break it down:

• Add one to the exponent \( n \).
• Divide the term by the new exponent.
• Add the constant of integration \( C \) when dealing with indefinite integrals.

This rule cannot be used when \( n = -1 \), since this would result in division by zero. In our example, the function \( \sqrt[3]{w} = w^{\frac{1}{3}} \) can be integrated by the power rule. By applying the power rule, we find the antiderivative before applying the definite integral limits.

antiderivative calculation

Finding an antiderivative is like reversing the process of differentiation. An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \). Once you find an antiderivative, you can calculate definite integrals by utilizing the Fundamental Theorem of Calculus.

In the given exercise, we aimed to find the antiderivative of \( f(w) = w^{\frac{1}{3}} \). By using the integration by power rule here, you obtain the antiderivative:

• \( \frac{w^{\frac{1}{3}+1}}{\frac{1}{3}+1} = \frac{w^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4}w^{\frac{4}{3}} \)

This function, \( \frac{3}{4}w^{\frac{4}{3}} \), is the antiderivative we needed, enabling us to evaluate the definite integral by substituting the values of the limits into this expression.

solving cubic root integrals

Solving integrals involving cubic roots involves a few extra steps compared to standard polynomial integrals. A cubic root function like \( \sqrt[3]{w} \) translates into an exponent when expressed in terms of power functions. This transformation is key to applying the power rule.

Given that \( \sqrt[3]{w} \) can be written as \( w^{\frac{1}{3}} \), you can utilize the power rule to find an antiderivative by following these steps:

• Identify the expression as a power: \( w^{\frac{1}{3}} \).
• Apply the power rule to find the antiderivative: \( \frac{w^{\frac{4}{3}}}{\frac{4}{3}} \).
• Simplify to achieve the function: \( \frac{3}{4}w^{\frac{4}{3}} \).

After finding the antiderivative, it becomes straightforward to evaluate the definite integral by substituting the limits using the Second Fundamental Theorem of Calculus. This approach makes solving cubic root integrals much more systematic and manageable.