Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{1}^{4} \frac{1}{w^{2}} d w $$

Step-by-Step Solution

Answer

The value of the definite integral is 0.75.

Step 1: Understand the Problem

The problem requires evaluating the definite integral \( \int_{1}^{4} \frac{1}{w^{2}} \, dw \) using the Second Fundamental Theorem of Calculus.

Step 2: Rewrite the Integrand

Rewrite the integrand. The function \( \frac{1}{w^2} \) can be written as \( w^{-2} \).

Step 3: Find the Antiderivative

The antiderivative of \( w^{-2} \) is found using the power rule for integration: \( \int w^{n} \, dw = \frac{w^{n+1}}{n+1} + C \). Therefore, the antiderivative of \( w^{-2} \) is \( \frac{w^{-1}}{-1} = -w^{-1} = -\frac{1}{w} \).

Step 4: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( F(w) = -\frac{1}{w} \).

Step 5: Evaluate the Antiderivative at the Upper Limit

Evaluate \( F(4) \): \(-\frac{1}{4} = -0.25 \).

Step 6: Evaluate the Antiderivative at the Lower Limit

Evaluate \( F(1) \): \(-\frac{1}{1} = -1 \).

Step 7: Calculate the Definite Integral

Using the result from the Fundamental Theorem of Calculus, compute \( F(4) - F(1) = -0.25 - (-1) \). Simplifying gives \( -0.25 + 1 = 0.75 \).

Key Concepts

Definite Integral

A definite integral essentially measures the total accumulation of a quantity over an interval, such as area under a curve. In this context, it is symbolized as \( \int_{a}^{b} f(x) \, dx \), representing the integration of a function \( f(x) \) from \( a \) to \( b \).
These endpoints \( a \) and \( b \) define the interval over which you integrate. The result is a specific number, which represents the net area between the function's curve and the x-axis, taken account of if the curve is above or below the x-axis.

To calculate definite integrals, the Second Fundamental Theorem of Calculus proves invaluable as it connects differentiation with integration, making the process simpler. In practice, once you find an antiderivative of the function, you simply evaluate it at the endpoints of the interval and subtract the values to find the total integral.

Antiderivative

An antiderivative is a function that reverses differentiation, also known as the indefinite integral. If you have a function \( f(x) \), an antiderivative \( F(x) \) is a function whose derivative gives you back the original \( f(x) \) (i.e., \( F'(x) = f(x) \)).
For example, if \( f(w) = \frac{1}{w^2} \), then \( F(w) = -\frac{1}{w} \) is its antiderivative. This is derived from applying the power rule for integration.

Having an antiderivative is crucial for solving definite integrals. Once you find it, this function will be evaluated over specified limits to determine the integral's value. This connection between antiderivatives and definite integrals is the core idea behind the Second Fundamental Theorem of Calculus.

Power Rule Integration

The power rule for integration is a vital tool when dealing with polynomial functions or functions that can be expressed as a power of a variable. The rule states that to integrate \( w^n \), we use the formula \( \int w^n \, dw = \frac{w^{n+1}}{n+1} + C \), where \( n eq -1 \).
By using this rule, finding antiderivatives becomes straightforward. For our example \( \int \frac{1}{w^2} \, dw \), rewriting the integrand as \( w^{-2} \) makes this rule applicable.

Applying the power rule, the antiderivative of \( w^{-2} \) is \(-\frac{1}{w}\). Note that the \( + C \) term often seen in indefinite integrals is omitted in definite integration because the constant cancels out in the subtraction of \( F(b) - F(a) \).

Integral Evaluation

Evaluating an integral means calculating its numeric value, particularly a definite integral, by applying the Second Fundamental Theorem of Calculus. Once the antiderivative \( F(x) \) is determined, the next step is plugging in the upper and lower limits of the integral into this antiderivative.
Specifically, you compute \( F(b) - F(a) \), where \( a \) and \( b \) are the bounds of your integral. This simple subtraction provides the net area described by the definite integral.

In our example, the antiderivative \( F(w) \) is \(-\frac{1}{w}\). By calculating \( F(4) = -\frac{1}{4} \) and \( F(1) = -1\), the integral evaluates to \( -0.25 + 1 = 0.75 \). This value represents the total accumulated area from the function \( \frac{1}{w^2} \) between 1 and 4 on the x-axis.