Problem 43 - Applied Calculus

Question

For each function: a. Integrate ("by hand") to find the area under the curve between the given \(x\) -values. b. Verify your answer to part (a) by having your calculator graph the function and find the area (using a command like FnInt or \(\int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\) ). $$ f(x)=\frac{1}{x^{3}} \text { from } x=1 \text { to } x=4 $$

Step-by-Step Solution

Answer

The area under the curve from \( x = 1 \) to \( x = 4 \) is \( \frac{15}{32} \).

Step 1: Set up the integral

We need to find the area under the curve for the function \( f(x) = \frac{1}{x^3} \) from \( x = 1 \) to \( x = 4 \). To do this, we set up the definite integral as follows: \[ \int_{1}^{4} \frac{1}{x^3} \, dx \].

Step 2: Find the antiderivative

The antiderivative of \( \frac{1}{x^3} \) is \( -\frac{1}{2x^2} \). This is because taking the derivative of \( -\frac{1}{2x^2} \) will give you \( \frac{1}{x^3} \).

Step 3: Evaluate the definite integral

Evaluate the antiderivative from \( x = 1 \) to \( x = 4 \): \[ \left[-\frac{1}{2x^2}\right]_1^4 = \left(-\frac{1}{2\times16}\right) - \left(-\frac{1}{2\times1}\right). \] Simplifying gives \( -\frac{1}{32} + \frac{1}{2} = \frac{1}{2} - \frac{1}{32} = \frac{16}{32} - \frac{1}{32} = \frac{15}{32} \).

Step 4: Verify with calculator

Use a graphing calculator with a feature like FnInt or the integral function. Enter the function \( f(x) = \frac{1}{x^3} \) and calculate the area from \( x = 1 \) to \( x = 4 \). The calculator should display a result close to \( \frac{15}{32} \) or approximately \( 0.46875 \). This verifies our manual calculation.

Key Concepts

Integral Calculus

Integral calculus focuses on understanding the concept of integration, which is essentially the reverse process of differentiation. The integral represents the accumulation of quantities like area. In the context of a curve on a graph, we're often interested in the area between the curve and the x-axis. This area gives us valuable information about the total accumulated value over an interval.
To find this area, you use the concept of integrals. There are two primary types of integrals:

• Indefinite Integrals: These do not have specified limits and result in a family of functions. They are essentially the antiderivatives of a function.
• Definite Integrals: These include limits of integration, providing a specific numerical value representing the area under a curve within those limits.

Understanding how to set up these integrals and interpret their results is a crucial part of mastering integral calculus.

Antiderivative

An antiderivative of a function is a function whose derivative is the original function. It is essentially the inverse operation to differentiation. When integrating a function, you need to find its antiderivative to solve the integral.
For example, in the exercise, we are given the function \(f(x) = \frac{1}{x^3}\). To integrate this manually, the antiderivative would be \(-\frac{1}{2x^2}\). This is obtained because the derivative of \(-\frac{1}{2x^2}\) results in \(\frac{1}{x^3}\).
Finding the antiderivative is a vital step in calculating integrals as it helps to determine the area under the curve between specified limits.

Definite Integral

A definite integral calculates the exact area under a curve over a specific interval. It is represented as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the lower and upper limits of the interval. The result is a numerical value, unlike an indefinite integral, which results in a function plus a constant of integration.
In the exercise, the definite integral \(\int_{1}^{4} \frac{1}{x^3} \, dx\) is evaluated to find the area from \(x = 1\) to \(x = 4\). The antiderivative is applied at these bounds, providing us with the total area under the curve. This process involves taking the antiderivative calculated previously, \([-\frac{1}{2x^2}]_1^4\), and calculating the difference between the values plugged into \(x\).
This method provides an exact area measure, ensuring precise calculations in practical applications.

Graphing Calculator

A graphing calculator is a powerful tool that allows you to visualize functions and calculate integrals quickly. These calculators come with features like FnInt or similar integral functions that can compute the area under curves without manual computation. This is particularly useful when verifying hand-calculated solutions.
For the exercise, using a graphing calculator can help double-check the calculated area under the curve of \(f(x) = \frac{1}{x^3}\) from \(x = 1\) to \(x = 4\). By inputting the function and specifying the limits, the calculator will output a numerical result, confirming the accuracy of manual calculations.
Graphing calculators not only confirm solutions but also provide a visual representation of the function, aiding in a deeper understanding of the behavior and properties of the function over the selected interval.