Problem 45 - Calculus Single Variable

Question

Calculate \(\mathcal{R}\left(f, \mathcal{L}_{N}\right),\) the Riemann sum \(\mathcal{R}(f, \mathcal{S})\) using the midpoint of each subinterval for the choice of points, and \(\mathcal{R}\left(f, \mathcal{U}_{N}\right)\). (You will notice that the inequalities of line \((5.2 .2)\) hold. \()\) $$ f(x)=e^{x} \quad I=[0,1], N=1 $$

Step-by-Step Solution

Answer

The midpoint Riemann sum is \(e^{0.5}\), the lower sum is 1, and the upper sum is \(e\).

Step 1: Define the interval and partition

The interval given is \([0, 1]\). We need to partition this interval into smaller subintervals. Given \(N=1\), this means there will be only one subinterval: \([0, 1]\).

Step 2: Calculate the width of each subinterval

The width of an interval \( \Delta x \) is calculated as \( \frac{b-a}{N} \), where \(a\) and \(b\) are the endpoints of the interval and \(N\) is the number of subintervals. Here it will be \( \Delta x = \frac{1-0}{1} = 1 \).

Step 3: Determine the midpoint

For the midpoint Riemann sum, the point to evaluate the function is the midpoint of each subinterval. Since there is only one interval \([0, 1]\), the midpoint is \( m = \frac{0+1}{2} = 0.5 \).

Step 4: Evaluate the function at the midpoint

Calculate \(f(m)\) at the midpoint: \(f(0.5) = e^{0.5}\).

Step 5: Calculate the Riemann sum using midpoint

The Riemann sum \( \mathcal{R}(f, \mathcal{L}_N) \) using the midpoint will be \( \Delta x \cdot f(0.5) = 1 \cdot e^{0.5} = e^{0.5}\).

Step 6: Upper and Lower Riemann Sums

For \(N=1\), the function is increasing on \([0,1]\), so the lower sum \(\mathcal{R}(f, \mathcal{L}_N) = f(0) = e^0 = 1\), and the upper sum \(\mathcal{R}(f, \mathcal{U}_N) = f(1) = e^1 = e\).

Key Concepts

Midpoint Rule

The midpoint rule is a technique for approximating the integral of a function over a given interval. It falls under the category of Riemann sums, which are used to approximate the area under a curve. In this method, instead of evaluating the function at the endpoints of the subintervals or using the left or right endpoints, we use the midpoint of each subinterval. This method can lead to more accurate results because the midpoint often provides a better average value of the function over the interval. For example, if we need to approximate the integral of an increasing function, the midpoint tends to balance out the underestimation that might occur if only the left endpoints were used, and the overestimation if only the right endpoints were used. To apply the midpoint rule:

• Divide the interval of interest into equal subintervals.
• Calculate the midpoint of each subinterval.
• Evaluate the function at each of these midpoints.
• Multiply each function value by the width of the subinterval, and sum them up.

Partition of Interval

In the context of Riemann sums, partitioning an interval involves breaking it down into smaller, equally spaced subintervals. This process is crucial as it determines where on the function curve we will take our sample points for approximation. For instance, when the problem states a partition of an interval \(N=1\), it means the interval isn't further subdivided, and you are looking at the entire interval as a single piece. More generally:

• For an interval [a, b], you'd calculate the width of each subinterval as \(\Delta x = \frac{b-a}{N}\), where \(N\) is the number of subintervals.
• The choice of \(N\) affects accuracy: higher \(N\) gives better approximation and a finer partition.

By choosing different values of \(N\), you can check how closely your Riemann sum approaches the true value of the integral, with smaller intervals generally giving a more precise approximation.

Upper and Lower Sums

Upper and lower Riemann sums provide bounds on the exact area under a curve. These sums offer a way to understand how close our Riemann approximation is to the actual integral. Specifically:- **Lower Sum:** This is calculated by taking the minimum function value on each subinterval, essentially assuming a rectangle with height equal to the lowest point of the function within that subinterval. For increasing functions, it often refers to the height at the left endpoint.- **Upper Sum:** Conversely, the upper Riemann sum uses the maximum function value on each subinterval. For increasing functions, this often refers to the height at the right endpoint.In practice:

• For the interval [0, 1] with exponential function \(f(x)=e^x\), the lower sum for \(N=1\) is \(f(0)=1\).
• The upper sum is \(f(1)=e\).

These bounds provide a helpful way to check that your midpoint or other Riemann sums are reasonable. They should always fall between these two values.

Exponential Function

An exponential function is a function in which an independent variable appears in the exponent. The general form of an exponential function is \(f(x) = a\cdot e^{bx}\), where \(e\) approximately equals 2.71828, a constant known as Euler's number. In simpler terms, these functions grow rapidly as the variable increases.In this context, the function \(f(x)=e^x\) is a classic exponential function:

• It's continuously increasing as \(x\) increases.
• Evaluated over [0, 1] for Riemann sums, it demonstrates a key property of exponential growth.
• This function is smooth and continuous, making it a perfect candidate for approximation techniques like Riemann sums.

Understanding exponential functions helps provide clarity when calculating integrals, particularly for problems involving growth modeling, compound interest, and natural logarithms.