Problem 3 - Active Calculus

Question

On a sketch of \(y=e^{x},\) represent the left Riemann sum with \(n=2\) approximating \(\int_{2}^{3} e^{x} d x\). Write out the terms of the sum, but do not evaluate it: Sum \(=\) ___________________ + ___________________ On another sketch, represent the right Riemann sum with \(n=2\) approximating \(\int_{2}^{3} e^{x} d x\). Write out the terms of the sum, but do not evaluate it: Sum \(=\) ___________________ + ___________________ Which sum is an overestimate? Which sum is an underestimate?

Step-by-Step Solution

Answer

Left Riemann sum: \(e^{2} \times 0.5 + e^{2.5} \times 0.5\). Right Riemann sum: \(e^{2.5} \times 0.5 + e^{3} \times 0.5\). Left sum is an underestimate. Right sum is an overestimate.

Step 1: Understand the interval and number of rectangles

The interval of integration is from 2 to 3 and we are to use 2 rectangles (n=2).

Step 2: Determine the width of each rectangle

Calculate the width of each rectangle \(\text{width} = \frac{b-a}{n}\) where \(a=2\), \(b=3\), and \(n=2\). The width \( \text{width} = \frac{3-2}{2} = 0.5 \)

Step 3: Set up the left Riemann sum

The left Riemann sum uses function values at the left endpoints of the subintervals. The left endpoints are \( x_0 = 2 \) and \( x_1 = 2 + 0.5 = 2.5 \). Therefore, the sum is \(\text{sum} = f(x_0)\text{width} + f(x_1)\text{width} = e^{2} \times 0.5 + e^{2.5} \times 0.5 \)

Step 4: Set up the right Riemann sum

The right Riemann sum uses function values at the right endpoints of the subintervals. The right endpoints are \( x_1 = 2.5 \) and \( x_2 = 2.5 + 0.5 = 3 \). Therefore, the sum is \(\text{sum} = f(x_1)\text{width} + f(x_2)\text{width} = e^{2.5} \times 0.5 + e^{3} \times 0.5 \)

Step 5: Determine which sum is an overestimate and which is an underestimate

The function \(y=e^x\) is increasing. The left Riemann sum underestimates the integral, while the right Riemann sum overestimates the integral.

Key Concepts

Left Riemann Sum

The left Riemann sum is a method to approximate the value of an integral by using the function values at the left endpoints of subintervals. This method is particularly useful when you need a quick estimation of the integral, especially if you don't have access to the tools needed for more precise calculations.

For the given problem, we divide the interval \( [2, 3] \) into \( n = 2 \) subintervals. The width of each subinterval is: \[ \text{width} = \frac{3-2}{2} = 0.5 \]

Since we use the left endpoints for our calculations: \[ x_0 = 2, \ x_1 = 2 + 0.5 = 2.5 \]

The left Riemann sum is then: \[ \text{sum} = f(x_0) \times \text{width} + f(x_1) \times \text{width} = e^2 \times 0.5 + e^{2.5} \times 0.5 \]

This approach will give an underestimate of the actual integral because the function \( y = e^x \) is increasing throughout the interval. The rectangles formed will always lie below the curve.

Right Riemann Sum

The right Riemann sum offers another method for approximating the value of an integral, differing from the left Riemann sum by using the function values at the right endpoints of subintervals. In situations involving increasing functions, this method typically provides an overestimate of the integral.

For the same interval \( [2, 3] \) divided into \( n = 2 \) subintervals, the right endpoints are: \[ x_1 = 2.5, \ x_2 = 2.5 + 0.5 = 3 \]

The right Riemann sum is calculated as: \[ \text{sum} = f(x_1) \times \text{width} + f(x_2) \times \text{width} = e^{2.5} \times 0.5 + e^{3} \times 0.5 \]

This method generally results in an overestimate of the integral because the rectangles will extend above the curve for an increasing function like \( y = e^x \).

Integration Approximation

The methods of left Riemann sum and right Riemann sum fall under the broader category of integration approximation. These techniques provide valuable ways to estimate the value of definite integrals, and are especially helpful when dealing with complicated functions or when exact integration is difficult.

There are several key points to remember about these methods:

• Both left and right Riemann sums rely on dividing the interval of integration into smaller subintervals (rectangles).
• The width of each subinterval is calculated as: \[ \text{width} = \frac{b-a}{n}, \ \text{where } a \text{ and } b \text{ are the limits of integration and } n \text{ is the number of subintervals.} \]
• For the left Riemann sum, the function value at the left endpoint of each subinterval is used.
  For the right Riemann sum, the function value at the right endpoint of each subinterval is used.
• When dealing with an increasing function like \( y = e^x \), the left Riemann sum provides an underestimate, while the right Riemann sum provides an overestimate of the actual integral.

These methods are foundational in calculus and serve as stepping stones to more accurate techniques such as Simpson's Rule and the Trapezoidal Rule. Whether you are estimating an integral by hand or using software, understanding these basic concepts is essential for more advanced studies.