Problem 12 - Calculus and its applications

Question

Find the area under the given curve over the indicated interval. $$ y=\frac{2}{x} ; \quad[1,4] $$

Step-by-Step Solution

Answer

The area under the curve is approximately \(2.77\).

Step 1: Understand the Problem

We are asked to find the area under the curve of the function \( y = \frac{2}{x} \) over the interval \([1, 4]\). This involves finding the definite integral of the function over the given interval.

Step 2: Set Up the Integral

The area under the curve from \(x = 1\) to \(x = 4\) can be found by setting up the definite integral: \[ \int_{1}^{4} \frac{2}{x} \, dx \].

Step 3: Integrate the Function

We need to find the antiderivative of \( \frac{2}{x} \). The antiderivative of \( \frac{1}{x} \) is \( \ln|x| \). Thus, \( \int \frac{2}{x} \, dx = 2 \ln|x| + C \), where \( C \) is the constant of integration.

Step 4: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the area is given by evaluating the antiderivative at the bounds of the interval: \[ \int_{1}^{4} \frac{2}{x} \, dx = \left[ 2 \ln|x| \right]_1^4 = 2 \ln(4) - 2 \ln(1) \].

Step 5: Simplify the Expression

Since \( \ln(1) = 0 \), the expression simplifies to \( 2 \ln(4) \). Using properties of logarithms, this can be further simplified: \( 2 \ln(4) = 2 \cdot 2 \ln(2) = 4 \ln(2) \).

Step 6: Calculate the Numerical Value

Approximating the value using \( \ln(2) \approx 0.693 \), gives: \( 4 \times 0.693 = 2.772 \). Thus, the area under the curve is approximately \(2.77\).

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus that represents the accumulation of quantities, like areas under curves, over a given interval. For the problem at hand, we need to find the definite integral of the function \[ \int_{1}^{4} \frac{2}{x} \, dx \]This expression calculates the precise area under the curve of the function \( y = \frac{2}{x} \) from \( x = 1 \) to \( x = 4 \). It is important to note that definite integrals provide not just symbolic solutions, but also the actual numerical value representing this area.

• The limits of integration, 1 and 4 in this case, define the interval over which the integration occurs.
• Definite integrals always involve evaluating the antiderivative at these boundaries to obtain the final area.
• The definite integral has no integration constant, unlike the indefinite integral, since it yields a specific number.

Breaking down the problem into understanding and setting up this integral is crucial for computation.

Antiderivative

An antiderivative is the reverse process of differentiation. It is a function whose derivative is the original function given. In computing definite integrals, finding the antiderivative is a critical step.For the problem raised, our function is \[ \frac{2}{x} \]The antiderivative of a basic function \( \frac{1}{x} \) is well-known to be \( \ln|x| \), thus making the antiderivative of \( \frac{2}{x} \)\[ \int \frac{2}{x} \, dx = 2 \ln|x| + C \]where \( C \) is the constant of integration. However, in the context of definite integrals, this constant is not needed, as it cancels out when evaluating the integral over a specific interval.

• The antiderivative helps us move from a rate of change to a total quantity.
• Recognizing standard antiderivatives can expedite the solution to integral problems.
• Understanding properties and rules of logarithms can simplify complex expressions.

Being familiar with common antiderivatives is an essential skill when tackling varying integral challenges.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration, showcasing the deep connection between these two operations. Essentially, it provides a way to evaluate definite integrals using antiderivatives.For the current function, the theorem tells us to find the antiderivative first, which we have as\[ 2 \ln|x| \]Then, evaluate it at the limits of the interval \[ \left[ 2 \ln|x| \right]_1^4 = 2 \ln(4) - 2 \ln(1) \]

• The theorem simplifies calculating areas and values over intervals without needing to approximate with sums.
• The antiderivative evaluated at the boundaries substitutes the area calculation to numeric values.
• This theorem also explains why \( \ln(1) = 0 \), simplifying further computations.

Understanding this duality expressed by the Fundamental Theorem is essential for efficiently solving problems and working through complex integral calculations.