Problem 57 - Calculus Early Transcendentals

Question

Find the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the given interval. $$f(x)=\frac{1}{x} \text { on }[-2,-1]$$

Step-by-Step Solution

Answer

Answer: The area of the region bounded by the graph of the function \(f(x) = \frac{1}{x}\) and the \(x\)-axis on the interval \([-2, -1]\) is \(\ln 2\) square units.

Step 1: Identify the function and interval

The function to integrate is \(f(x) = \frac{1}{x}\) over the interval \([-2, -1]\). Since the function is negative over this interval, the area will be the absolute value of the integral.

Step 2: Integrate the function

We need to find the integral of \(f(x) = \frac{1}{x}\) over the interval \([-2, -1]\). The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\), so: $$\int_{-2}^{-1} f(x)dx = \int_{-2}^{-1} \frac{1}{x}dx = [\ln|x|]_{-2}^{-1}.$$

Step 3: Evaluate the antiderivative

Evaluate the antiderivative at the bounds of the integral: $$[\ln|x|]_{-2}^{-1} = \ln|-1| - \ln|-2| = \ln 1 - \ln 2 = -\ln 2.$$

Step 4: Take the absolute value

Since the function is negative over the interval \([-2, -1]\), we need to take the absolute value of the resulting integral to find the area: $$|\int_{-2}^{-1} \frac{1}{x}dx| = |- \ln 2|= \ln 2.$$

Step 5: Write the final answer

The area of the region bounded by the graph of \(f(x) = \frac{1}{x}\) and the \(x\)-axis on the interval \([-2, -1]\) is \(\ln 2\) square units.

Key Concepts

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. This concept plays a crucial role in integral calculus, as integrating a function involves finding its antiderivative. To put it simply, if you have a function \(f(x)\), an antiderivative is \(F(x)\) such that \(F'(x) = f(x)\).
In our exercise, the function \(f(x) = \frac{1}{x}\) has the antiderivative \(F(x) = \ln|x|\). This means that the derivative of \(\ln|x|\) brings you back to \(\frac{1}{x}\). This correspondence is used to solve definite integrals.

• The antiderivative helps in computing the definite integral by evaluating the function at the boundaries of the interval.
• For the given problem, the evaluation was done at \([-2, -1]\).

Understanding antiderivatives is essential for integrating functions and finding areas under curves.

Logarithmic Integration

Logarithmic integration is a type of integral involving logarithmic functions. These integrals often show up when the integrand is a function like \(\frac{1}{x}\), which leads to a logarithm in its antiderivative.
When integrating \(\frac{1}{x}\), we find its antiderivative is \(\ln|x|\). This integration step is applied to find the area under curves that have a relation similar to the form \(\frac{1}{x}\).

• Logarithmic integration is used when dealing with functions that simplify to a logarithm in the result.
• The absolute value in \(\ln|x|\) is significant. It ensures the logarithm is defined over both positive and negative values of \(x\).

Logarithmic integration is a fundamental technique necessary for solving a variety of calculus problems that involve these natural logarithm characteristics.

Negative Function Integration

Negative function integration deals with integrating functions that are negative over the interval of interest. When integrating such functions to find areas, the result typically is negative. However, since area cannot be negative, we take the absolute value of the integral result.
In our example, the function \(f(x) = \frac{1}{x}\) is negative over the interval \([-2, -1]\). Hence, the computed integral yields a negative value, which is then converted to its absolute value to determine the actual area.

• Ensure that negative results from integration are adjusted to reflect real, tangible quantities like area.
• This adjustment is a crucial step to accurately represent the area between a curve and the x-axis.

Understanding how to manage negative integrals is vital for applications in calculus involving areas and volumes.