Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, 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 integral evaluates to \(\frac{45}{4}\).

Step 1: Identify the Antiderivative

We need to find the antiderivative of the function under the integral, which is \(\sqrt[3]{w}\) or equivalently \(w^{1/3}\). The antiderivative of \(w^n\) is \(\frac{w^{n+1}}{n+1}\) when \(n eq -1\). Here, \(n = \frac{1}{3}\), so the antiderivative is \(\frac{w^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} = \frac{w^{4/3}}{4/3}\).

Step 2: Simplify the Antiderivative Expression

The simplified form of \(\frac{w^{4/3}}{4/3}\) is \(\frac{3}{4}w^{4/3}\). This is the function we will evaluate at the bounds of the integral, according to the Second Fundamental Theorem of Calculus.

Step 3: Apply the Limits of Integration

Using the Second Fundamental Theorem of Calculus, we evaluate \(F(b) - F(a)\), where \(F(w)\) is the antiderivative \(\frac{3}{4}w^{4/3}\). Here, \(a = 1\) and \(b = 8\). So we compute \(\frac{3}{4}(8)^{4/3} - \frac{3}{4}(1)^{4/3}\).

Step 4: Calculate \(F(8)\) and \(F(1)\)

First, calculate \((8)^{4/3}\). The cube root of 8 is 2, and \(2^4 = 16\), so \((8)^{4/3} = 16\). Hence, \(F(8) = \frac{3}{4} \times 16 = 12\). Next, \((1)^{4/3} = 1\), so \(F(1) = \frac{3}{4} \times 1 = \frac{3}{4}\).

Step 5: Compute the Integral Value

Subtract \(F(1)\) from \(F(8)\): \(12 - \frac{3}{4} = \frac{48}{4} - \frac{3}{4} = \frac{45}{4}\). Thus, the value of the definite integral \(\int_{1}^{8} \sqrt[3]{w} \, dw\) is \(\frac{45}{4}\).

Key Concepts

Definite Integrals

Definite integrals represent a core concept in calculus that deals with the accumulation of quantities, such as areas under curves. When working with definite integrals, you're dealing with an integral that has lower and upper bounds, indicated here by the numbers 1 and 8. These bounds essentially tell you the interval over which you are accumulating the quantity. The outcome is a numerical value, unlike indefinite integrals which yield functions.When solving a definite integral like \( \int_{1}^{8} \sqrt[3]{w} \, dw \), you're calculating the area under the curve of the function \( \sqrt[3]{w} \) from \( w = 1 \) to \( w = 8 \). This value represents the total accumulation between these two points.Definite integrals require the calculation of an antiderivative for the function within the integral, which we'll elaborate on in the next section.

Antiderivatives

Antiderivatives play a crucial role in solving definite integrals. The antiderivative of a function is essentially the "reverse" operation of taking a derivative. It answers the question: "What function could this derivative have originated from?" For our function \( \sqrt[3]{w} \), or equivalently \( w^{1/3} \), the task was to determine the antiderivative.Using the rule which states that the antiderivative of \( w^n \) is \( \frac{w^{n+1}}{n+1} \) (for \( n eq -1 \)), we find:

• With \( n = \frac{1}{3} \), the antiderivative is \( \frac{w^{4/3}}{4/3} \).
• Upon simplifying, this becomes \( \frac{3}{4}w^{4/3} \).

This function, \( \frac{3}{4}w^{4/3} \), will be evaluated at the bounds of the integral, helping us determine the total accumulated value over the specified interval.

Calculus Problem-Solving

Problem-solving in calculus often involves a systematic approach to applying theorems and rules. In this case, the Second Fundamental Theorem of Calculus provides a straightforward method to solving definite integrals. It states that if you have a continuous function \( f \) on an interval \([a, b]\) and \( F \) is an antiderivative of \( f \), then:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]For our definite integral problem, this means we can use the antiderivative we found, \( \frac{3}{4}w^{4/3} \), and calculate \( F(8) - F(1) \). Thus:

• First, evaluate \( F(8) = \frac{3}{4}(8)^{4/3} = 12 \).
• Next, evaluate \( F(1) = \frac{3}{4}(1)^{4/3} = \frac{3}{4} \).
• Subtract to find the final answer: \( 12 - \frac{3}{4} = \frac{45}{4} \).

This demonstrates the power of the theorem in converting complex accumulation problems into a simple subtraction operation.

Integration Techniques

To tackle integration problems effectively, especially those involving definite integrals, it is important to be familiar with various integration techniques. In our exercise, we utilized a basic power rule, a foundational technique crucial for finding antiderivatives:

• The rule for the antiderivative of \( w^n \) is applied when solving integrals of polynomial and power functions.
• Simplification of expressions, such as converting \( \frac{w^{4/3}}{4/3} \) to \( \frac{3}{4}w^{4/3} \) helps achieve a final, usable antiderivative.

By mastering these techniques, students can handle a wide range of integration challenges, transforming complex calculus problems into more approachable tasks.