Question

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

Short answer

Answer: The area of the region is \(\ln 2\).

Step-by-step solution

Set up the integral

The area bounded by the graph of the function and the x-axis is represented by the definite integral: $$A = \int_{-2}^{-1} f(x) dx = \int_{-2}^{-1} \frac{1}{x} dx $$

Evaluate the integral

The function \(f(x)=\frac{1}{x}\) can be rewritten as \(x^{-1}\) and we can then use the power rule for integration to evaluate the integral: The power rule states that $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ Here, our function is \(x^{-1}\). So, the integral can be evaluated as $$\int x^{-1} dx = \int \frac{1}{x} dx = \frac{x^0}{0} + C $$ However, we cannot divide by zero. Therefore, instead of the power rule, we have to find the antiderivative of \(\frac{1}{x}\) and compute the definite integral: $$\int \frac{1}{x} dx = \ln|x| + C$$ Now, we need to evaluate the definite integral for the bounds \(-2\) and \(-1\):

Compute the area

To find the area represented by the definite integral, we plug in the bounds and compute the difference: $$A = \int_{-2}^{-1} \frac{1}{x} dx = \left[ \ln|-1| - \ln|-2|\right] = \left[ \ln 1 - \ln 2\right] = -\ln 2$$ Since the area is a positive value, we take the absolute value of the result: $$A = |\,-\ln 2\,| = \ln 2$$ So, 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 \(\ln 2\).

Key concepts

Integration by Power Rule

When integrating functions in calculus, the power rule is a fundamental technique that simplifies the process. It specifically applies to polynomials, where each term is a power of the variable x. The rule is expressed mathematically as
\[\int x^n dx = \frac{x^{n+1}}{n+1} + C,\]
where n is any real number except for -1, because, as in our exercise, division by zero would occur—a clearly undefined operation.

• If the exponent n is a positive integer, you simply add one to the exponent and divide by this new number.
• If n is negative, the same process is followed, ensuring the exponent doesn’t equal -1.
• If n is a fraction, it indicates the presence of a root, but you can still apply the power rule similarly.

It is crucial to remember that obtaining an antiderivative includes the constant of integration C, which is required for indefinite integrals. In the context of the given exercise, this rule cannot be used directly because we encounter the exceptional case (n being -1). Therefore, we must use a special property for integrating functions that resembles x^{-1}.

Antiderivatives

In the realm of calculus, antiderivatives are the reverse process of derivatives. While differentiation determines the rate at which a function changes, antiderivatives provide the original function before it was differentiated, up to a constant. Antiderivatives are crucial for finding the area under a curve, as they represent the accumulation of values.

• The antiderivative of a function is not unique because it includes an arbitrary constant C, known as the constant of integration.
• When dealing with definite integrals, which calculate the area under a function within specified bounds, that constant cancels out because you are taking the difference between two antiderivative values.
• In our exercise, finding an antiderivative of \(\frac{1}{x}\) does not follow the standard power rule and is represented by the natural logarithm function.

The process of finding antiderivatives is integral to many applications in physics, engineering, and other fields where the accumulation of quantities is essential.

Natural Logarithm Properties

The natural logarithm, denoted as ln, is a mathematical function that is the inverse of the exponential function with base e. Here are some fundamental properties that proved helpful in solving the given exercise:

• The natural logarithm of 1 is always 0, that is, \(\ln(1) = 0\).
• For any positive number a, the logarithm of an inverse, \(\frac{1}{a}\), is the negative logarithm of a: \(\ln\left(\frac{1}{a}\right) = -\ln(a)\).
• The difference between two logarithms is the logarithm of their quotient: \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\).

These properties allow us to transform and simplify the expressions involving logarithms. In the solution to our exercise, we used these attributes to find the area bounded by \(\frac{1}{x}\) and the x-axis for the interval [-2,-1]. Ultimately, we utilized the fact that \(\ln(1)=0\) to simplify the answer to simply \(\ln 2\).

Calculus Applications

Calculus is a branch of mathematics that allows us to study change and motion, and it has countless applications across various disciplines. One of the most significant applications is in computing areas and volumes.

• In physics, calculus is used to model systems and predict outcomes, from gravitational fields to electrical currents.
• In biology, it is applied to model population growths and understand changes within ecosystems.
• In economics, calculus assists in optimizing functions like cost, revenue, and profit to make informed financial decisions.

Our problem involved calculating the area under a curve, which is a classical application of definite integrals in calculus. By finding the antiderivative of the given function and evaluating it at the bounds of the interval, we determined the precise area of the region bounded by the function's graph and the x-axis.