Problem 37 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1 / 2}^{1}\left(x^{-3}-8\right) d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{1 / 2}^{1}\left(x^{-3}-8\right) d x$$ using the Fundamental Theorem of Calculus. Answer: The value of the definite integral is \(\frac{3}{2}\).

Step 1: Find the antiderivative

To find the antiderivative, we need to find the function whose derivative is equal to the integrand. For the given integral $$\int_{1 / 2}^{1}\left(x^{-3}-8\right) d x$$, the integrand is $$x^{-3}-8$$. Now, let's find the antiderivative of the integrand: The antiderivative of \(x^{-3}\) is \(-\frac{1}{2}x^{-2}\) (by applying the power rule of integration) The antiderivative of \(8\) is \(8x\) So, combining both results, we find that the antiderivative of our integrand is: $$F(x) = -\frac{1}{2}x^{-2} + 8x$$

Step 2: Evaluate the antiderivative at the upper limit

Now, we need to evaluate the antiderivative \(F(x)\) at the upper limit of integration, which is \(x = 1\). Plugging this value into \(F(x)\), we get: $$F(1)= -\frac{1}{2}(1)^{-2} + 8(1) = -\frac{1}{2} + 8$$

Step 3: Evaluate the antiderivative at the lower limit

Next, we need to evaluate the antiderivative \(F(x)\) at the lower limit of integration. In our case, this is \(x = \frac{1}{2}\). Plugging this value into \(F(x)\), we get: $$F\left(\frac{1}{2}\right)= -\frac{1}{2}\left(\frac{1}{2}\right)^{-2} + 8\left(\frac{1}{2}\right) = -1 + 4$$

Step 4: Subtract the lower limit evaluation from the upper limit evaluation

Now that we have evaluated the antiderivative at both the upper and lower limits of integration, let's subtract the lower limit evaluation from the upper limit evaluation: $$\int_{1 / 2}^{1}\left(x^{-3}-8\right) d x = F(1) - F\left(\frac{1}{2}\right) = \left(-\frac{1}{2} + 8\right) - \left(-1 + 4\right) = \left(\frac{1}{2} + 8\right) - (1 + 4)$$ Calculating the final value: $$\int_{1 / 2}^{1}\left(x^{-3}-8\right) d x= \frac{1}{2} + 8 - 5 = \frac{3}{2}$$ Hence, the value of the definite integral is \(\frac{3}{2}\).

Key Concepts

Antiderivative

The concept of an antiderivative is foundational in calculus. It represents a function whose derivative is the given function, often referred to as the integrand. When we talk about finding the antiderivative, we are essentially looking for a function that "undoes" the differentiation process.
For instance, if you have a function \( f(x) \), its antiderivative is a function \( F(x) \) such that \( F'(x) = f(x) \).

• The antiderivative process involves recognizing patterns, applying integration techniques like substitution or the power rule, and always remembering that there's actually a family of possible antiderivatives. These functions differ by a constant, denoted as \( C \).
• In the problem at hand, the antiderivative of \( x^{-3} \) is found using the power rule, resulting in \(-\frac{1}{2}x^{-2}\). Similarly, the antiderivative of the constant \( -8 \) becomes \( -8x \).

Discovering the antiderivative is crucial when we want to compute definite integrals, which involve evaluating the antiderivative at certain limits.

Definite Integrals

Definite integrals are used to calculate the net area under a curve within a specific interval on the x-axis. The concept heavily relies on the Fundamental Theorem of Calculus. This theorem bridges the world of differentiation and integration.
A definite integral has two bounds, a lower and an upper bound, say \( a \) and \( b \). The process of evaluating a definite integral \( \int_{a}^{b} f(x) \, dx \) involves:

• Calculating the antiderivative \( F(x) \) of the function \( f(x) \).
• Computing \( F(b) - F(a) \), which is the change in the function \( F(x) \) from \( a \) to \( b \).

In the example, the integral \( \int_{1/2}^{1} (x^{-3} - 8) \, dx \) is evaluated by finding the antiderivative \( F(x) \), then substituting the upper limit and subtracting the result of the lower limit.
This yields the total accumulation of change over the specified interval, quantifying the net effect across the delimiters.

Power Rule of Integration

The power rule of integration is a crucial technique that simplifies the process of finding antiderivatives for polynomial functions. This rule states that for any real number \( n eq -1 \), the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \).
It works by "reversing" the power rule of differentiation.
Let's break down how the power rule is applied:

• Increase the exponent of your term by 1. So for \( x^{-3} \), the exponent \( -3 \) becomes \( -2 \).
• Divide by this new exponent. Thus, \( x^{-3} \) becomes \( -\frac{1}{2}x^{-2} \).

This method provides a straightforward way to integrate most polynomial functions you'll encounter.
In the problem given, this rule is instrumental in determining the antiderivative of \( x^{-3} \), streamlining the solution process and making it approachable for different functions you might work with.