Problem 8 - Calculus

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 \( \frac{45}{4} \).

Step 1: Identify the integrand

The integrand in the given integral is \( \sqrt[3]{w} \), which can be rewritten as \( w^{1/3} \).

Step 2: Use antiderivative rules

We know from calculus that the antiderivative of \( w^{n} \) is \( \frac{w^{n+1}}{n+1} + C \). Here \( n = \frac{1}{3} \), so the antiderivative of \( w^{1/3} \) is \( \frac{w^{1/3 + 1}}{1/3 + 1} = \frac{w^{4/3}}{4/3} \).

Step 3: Simplify the antiderivative

Simplifying the antiderivative \( \frac{w^{4/3}}{4/3} \) gives \( \frac{3}{4}w^{4/3} \). This is the function we will use for evaluation.

Step 4: Apply the Second Fundamental Theorem of Calculus

According to the Second Fundamental Theorem of Calculus, to evaluate the definite integral \( \int_{a}^{b} f(w) \, dw \), we calculate \( F(b) - F(a) \) where \( F \) is the antiderivative of \( f \).

Step 5: Evaluate the antiderivative at the bounds

Substitute the upper bound and lower bound into the antiderivative: \( F(8) = \frac{3}{4}(8)^{4/3} \) and \( F(1) = \frac{3}{4}(1)^{4/3} \).

Step 6: Compute numerical values

Calculate the values specifically: \( (8)^{4/3} = (2^3)^{4/3} = 2^4 = 16 \) and \( (1)^{4/3} = 1 \). So, \( F(8) = \frac{3}{4} \times 16 = 12 \) and \( F(1) = \frac{3}{4} \times 1 = \frac{3}{4} \).

Step 7: Subtract to find the result

Compute the difference: \( F(8) - F(1) = 12 - \frac{3}{4} = \frac{48}{4} - \frac{3}{4} = \frac{45}{4} \).

Key Concepts

Definite Integral

A definite integral is a way to calculate the accumulation of quantities, such as areas under a curve, over a specified interval. In notation, a definite integral is shown as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration, representing the interval from \( a \) to \( b \). The function \( f(x) \) is known as the integrand.In the example giventhe integrand is \( \sqrt[3]{w} \), or equivalently \( w^{1/3} \). The process involves finding an antiderivative of the function and then evaluating it at the upper and lower bounds,

• Observe the limits of integration: \( 1 \) and \( 8 \).
• The integral \( \int_1^8 w^{1/3} \, dw \) aims to find the net area or accumulated value between these bounds.

This gives us a "definite" result, as the values \( a \) and \( b \) are specifically defined.

Antiderivative

An antiderivative is a function that "undoes" the derivative of a given function. When integrating, finding an antiderivative helps us identify the function whose derivative is the integrand. This is crucial for solving definite integrals as outlined in the Second Fundamental Theorem of Calculus.For a power function like \( w^{n} \), the antiderivative can be calculated using the power rule:

• \( A(w) = \frac{w^{n+1}}{n + 1} + C \) where \( C \) is the constant of integration.

In the exercise, we transformed \( w^{1/3} \) into its antiderivative form:

• \( A(w) = \frac{w^{4/3}}{4/3} = \frac{3}{4}w^{4/3} \).

This expression gives us a continuous function that simplifies the evaluation of the definite integral by applying the bounds.

Evaluation of Integrals

To evaluate a definite integral, the Second Fundamental Theorem of Calculus is applied. This theorem states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then:

• \( \int_a^b f(x) \, dx = F(b) - F(a) \).

This means we compute the difference between the values of the antiderivative at the upper bound (\( b \)) and the lower bound (\( a \)).For our example:

• Calculate \( F(8) = \frac{3}{4} \times 16 = 12 \), since \( (8)^{4/3} = 16 \).
• Calculate \( F(1) = \frac{3}{4} \times 1 = \frac{3}{4} \).

Finally, we subtract these two values:

• \( 12 - \frac{3}{4} = \frac{48}{4} - \frac{3}{4} = \frac{45}{4} \).

This yields the value of the definite integral and represents the net area or accumulation from \( w = 1 \) to \( w = 8 \). The step-by-step breakdown makes it easier to see how we utilize antiderivatives for definite evaluations.