Problem 61 - Calculus Single Variable

Question

Let \(\Delta x=(b-a) / 2 .\) Use the Mean Value Theorem to find points \(s_{1}\) in \([a,(a+b) / 2]\) and \(s_{2}\) in \([(a+b) / 2, b]\) such that the Riemann sum $$ \mathcal{R}\left(f,\left\\{s_{1}, s_{2}\right\\}\right)=f\left(s_{1}\right) \cdot \Delta x+f\left(s_{2}\right) \cdot \Delta x $$ is equal to the Riemann integral of \(f\) over \([a, b]\). $$ f(x)=1 / x \quad [1,2] $$

Step-by-Step Solution

Answer

The points are \(s_1 = \frac{1}{2\ln(1.5)}\) in \([1, 1.5]\) and \(s_2 = \frac{1}{2(\ln(2) - \ln(1.5))}\) in \([1.5, 2]\).

Step 1: Understanding the Problem

We need to find points using the Mean Value Theorem for integration, such that the Riemann sum is equal to the definite integral of a function over an interval. Here, the function is given as \(f(x) = \frac{1}{x}\) over the interval \([1, 2]\). The interval is split into two parts, \([1, 1.5]\) and \([1.5, 2]\), and we need to use the Mean Value Theorem to find the points \(s_1\) and \(s_2\) where the function equals the average value over these subintervals.

Step 2: Calculating \(\Delta x\)

The term \(\Delta x\) is defined as \((b-a)/2\). Here, \(a = 1\) and \(b = 2\), so we have \(\Delta x = \frac{(2 - 1)}{2} = 0.5\).

Step 3: Setting Up the Mean Value Theorem for Integration

The Mean Value Theorem for integration states that for a continuous function \(f(x)\), there exists a point \(c\) in \([a, b]\) where \(f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx\). We apply this to the subintervals \([1, 1.5]\) and \([1.5, 2]\) to find \(s_1\) and \(s_2\).

Step 4: Calculating the Integral over Subinterval \([1, 1.5]\)

Calculate \(\int_1^{1.5} \frac{1}{x} \, dx\). The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\). So, we have \(\int_1^{1.5} \frac{1}{x} \, dx = [\ln x]_1^{1.5} = \ln(1.5) - \ln(1) = \ln(1.5)\).

Step 5: Calculating \(s_1\) Using the Mean Value Theorem

For the subinterval \([1, 1.5]\), set \(f(s_1) \cdot 0.5 = \ln(1.5)\). Therefore, \(\frac{1}{s_1} \cdot 0.5 = \ln(1.5)\), which implies \(s_1 = \frac{1}{2\ln(1.5)}\).

Step 6: Calculating the Integral over Subinterval \([1.5, 2]\)

Calculate \(\int_{1.5}^2 \frac{1}{x} \, dx\). Using the antiderivative \(\ln|x|\), we find \(\int_{1.5}^2 \frac{1}{x} \, dx = [\ln x]_{1.5}^2 = \ln(2) - \ln(1.5)\).

Step 7: Calculating \(s_2\) Using the Mean Value Theorem

For the subinterval \([1.5, 2]\), set \(f(s_2) \cdot 0.5 = \ln(2) - \ln(1.5)\). Therefore, \(\frac{1}{s_2} \cdot 0.5 = \ln(2) - \ln(1.5)\), which implies \(s_2 = \frac{1}{2(\ln(2) - \ln(1.5))}\).

Step 8: Checking the Riemann Sum Equals the Integral

Compute \(f(s_1)\cdot \Delta x + f(s_2)\cdot \Delta x\) using the values of \(s_1\) and \(s_2\) found. Verify if this equals \(\int_1^2 \frac{1}{x} \, dx = \ln(2)\). Since both parts have been calculated such that their evaluated products and sums match the integral over the full interval, the solution is verified.

Key Concepts

Riemann Sum

The Riemann Sum is a fundamental concept in calculus used to approximate the area under a curve or, in other words, the integral of a function. It works by dividing the region under the curve into smaller sections called subintervals.
By using these subintervals, we can create rectangles that approximate the area.
Each rectangle’s height is determined by the function value at a specific point within the interval.

• The Riemann Sum is calculated as the sum of the area of these rectangles: \[ \mathcal{R}(f, \{s_1, s_2, ..., s_n\}) = f(s_1) \cdot \Delta x + f(s_2) \cdot \Delta x + ... + f(s_n) \cdot \Delta x \]
• In our specific example, we were tasked with calculating the sum for two rectangles, which means we needed two function values, \(f(s_1)\) and \(f(s_2)\).

This sum provides us an approximation, but when the number of rectangles becomes infinitely large, the Riemann Sum approaches the exact integral value of the function over the interval.

Definite Integral

The Definite Integral is a crucial idea bridging the Riemann Sum and the exact area under a curve. It gives us the precise accumulation of quantities, such as area, when taking infinitely many infinitely thin slices.
For a continuous function \(f(x)\) over an interval \([a, b]\), the definite integral is represented as:

• \( \int_a^b f(x) \, dx \)
• The integral computes the exact area below the curve and above the x-axis over \([a, b]\).
• In the problem context, we are finding the definite integral of the function \(f(x) = \frac{1}{x}\) over \([1, 2]\).

By using the Mean Value Theorem for Integration, we ensured that the Riemann Sum equals this definite integral, allowing us to find precise function values \(s_1\) and \(s_2\) where this match occurs.

Antiderivative

The Antiderivative, also known as an indefinite integral, is a function \(F(x)\) whose derivative is the given function \(f(x)\). This concept is integral to solving definite integrals as it provides a way to express cumulative values in calculus.
For example, for \(f(x) = \frac{1}{x}\), the antiderivative is:

• \( F(x) = \ln|x| + C \)
• \(C\) is any constant, which typically cancels out in definite integral calculations.

In the exercise, knowing that \( \int \frac{1}{x} \, dx = \ln|x| \) allowed us to calculate the specific definite integral values required over the given subintervals \([1, 1.5]\) and \([1.5, 2]\).
These integral results were then employed to locate \(s_1\) and \(s_2\).

Subintervals

Subintervals are smaller divisions of a given interval, used in the approximation of integrals through Riemann Sums and definite integrals. In problems involving integration, the main interval is divided into several parts, each called a subinterval, which simplifies the calculation method by focusing on smaller, manageable sections.
In the context of our exercise:

• The interval \([1, 2]\) was divided into two subintervals: \([1, 1.5]\) and \([1.5, 2]\).
• Subdividing aids in utilizing the Mean Value Theorem for Integration more effectively, allowing the calculation of specific points and cumulative areas.

By computing integrals and Riemann sums over these subintervals individually, we can get closer to matching the exact integral value of the whole interval. This approach is instrumental in achieving precise solutions in problems like the one provided.