Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, use the Second Fundamental Theorem of Calculus to evaluate each definite integral. \(\int_{1}^{4} \frac{s^{4}-8}{s^{2}} d s\)

Step-by-Step Solution

Answer

The value of the definite integral is 15.

Step 1: Simplify the Integrand

First, simplify the expression inside the integral. The integrand \( \frac{s^4 - 8}{s^2} \) can be separated into two separate terms by dividing each term by \( s^2 \). This gives: \[ \frac{s^4}{s^2} - \frac{8}{s^2} = s^2 - 8s^{-2}. \]

Step 2: Integrate the Simplified Function

Now integrate the simplified expression, \( s^2 - 8s^{-2} \). The antiderivative of \( s^2 \) is \( \frac{s^3}{3} \), and the antiderivative of \( -8s^{-2} \) is \( 8s^{-1} \). So the antiderivative is: \[ \int (s^2 - 8s^{-2}) \ ds = \frac{s^3}{3} + 8s^{-1}. \]

Step 3: Evaluate the Definite Integral Using Limits

According to the Second Fundamental Theorem of Calculus, the definite integral from \( a \) to \( b \) of a function \( f(s) \) is given by \( F(b) - F(a) \), where \( F \) is the antiderivative of \( f \). For the integral \( \int_{1}^{4} (s^2 - 8s^{-2}) \ ds \), substitute \( s = 4 \) and \( s = 1 \) into the antiderivative: \[ \left[ \frac{s^3}{3} + \frac{8}{s} \right]_{1}^{4} = \left( \frac{4^3}{3} + \frac{8}{4} \right) - \left( \frac{1^3}{3} + \frac{8}{1} \right). \]

Step 4: Calculate the Values

Carry out the calculations for each function value at the specified limits: \( \frac{4^3}{3} + \frac{8}{4} \) simplifies to \( \frac{64}{3} + 2 = \frac{70}{3} \) and \( \frac{1^3}{3} + \frac{8}{1} \) simplifies to \( \frac{1}{3} + 8 = \frac{25}{3} \).

Step 5: Compute the Final Result

Subtract the lower limit result from the upper limit result to find the value of the definite integral: \[ \frac{70}{3} - \frac{25}{3} = \frac{45}{3} = 15. \]

Key Concepts

Definite Integral

A definite integral gives the accumulated area under a curve between two specific points, often denoted as \( \int_{a}^{b} f(x) \, dx \). In simple terms, it measures the total quantity where the function magnitude is significant—like area, distance, or even total change, from point \(a\) to point \(b\).

• The limits at the bottom \(a\) and top \(b\) define the range along the x-axis where the area is calculated.
• It provides a numerical value that represents the sum of all tiny areas, called infinitesimal areas, under the curve from \(a\) to \(b\).

The use of a definite integral in calculus serves as a key way to quantify the total impact of a function between two given limits, making it a fundamental tool in mathematical analysis.

Antiderivative

The antiderivative is the reverse operation of differentiation. It finds a new function whose derivative is the original function, termed the integrand. In our solution, we seek the antiderivative that leads us back to the associated expression.

• For a function \( f(s) \), if \( F(s) \) is an antiderivative of \( f(s) \), then \( F'(s) = f(s) \).
• Finding the antiderivative allows us to evaluate definite integrals easily using the Second Fundamental Theorem of Calculus.

Integration is thus a crucial process in calculus, where finding the antiderivative forms the core step to resolving the overall definite integral across specific limits. Understanding the structure of derivative functions plays a pivotal role.

Integrand Simplification

In solving integrals, simplifying the integrand can make the integration process easier. For instance, the expression \( \frac{s^4 - 8}{s^2} \) can initially look complex, but simplification can make it more tractable.

• Breaking down a complex expression into simpler terms usually makes integration simpler and more straightforward.
• In our specific case, dividing each term by \( s^2 \) results in the simpler form \( s^2 - 8s^{-2} \).

Simplification lays the groundwork for easier integration by effectively reducing the complexity of terms within the integrand.It can also reveal standard types of functions and facilitate recognition of basic antiderivative forms.

Limit Evaluation

The final step in calculating a definite integral involves the evaluation of limits, applying the Second Fundamental Theorem of Calculus. Here's how it works:

• This theorem suggests calculating the difference \( F(b) - F(a) \), where \( F \) is the antiderivative of the integrand.
• After computing \( F \) at the upper limit \( b \) and the lower limit \( a \), subtract the results to get the definite integral's value.

For our problem, substituting \( s = 4 \) and \( s = 1 \) into the antiderivative and taking the difference gave us a final result of 15. By understanding the importance of the Second Fundamental Theorem, we recognize how calculus can solve real problems, measuring total changes over intervals.