Problem 30 - Calculus: Early Transcendentals

Question

Express the integral as a limit of Riemann sums. Do not evaluate the limit. \( \displaystyle \int^5_2 \biggl( x^2 + \frac{1}{x} \biggr) \, dx \)

Step-by-Step Solution

Answer

Express the integral as \( \lim_{n \to \infty} \sum_{i=1}^{n} \left( \left( 2 + \frac{3i}{n} \right)^2 + \frac{1}{2 + \frac{3i}{n}} \right) \cdot \frac{3}{n} \).

Step 1: Understanding the Integral

We have a definite integral \( \int^5_2 \left( x^2 + \frac{1}{x} \right) \, dx \). We need to express this integral as a limit of Riemann sums.

Step 2: Setting up the Partition

Consider the interval \([2, 5]\). Divide this interval into \(n\) subintervals of equal width. The width of each subinterval will be \( \Delta x = \frac{5 - 2}{n} = \frac{3}{n} \).

Step 3: Choosing Sample Points

For each subinterval, choose the right endpoint as the sample point. Thus, the \(i\)-th sample point, \(x_i^{*}\), is given by \( x_i^{*} = 2 + i\Delta x = 2 + \frac{3i}{n} \), where \(i = 1, 2, ..., n\).

Step 4: Writing the Function in Terms of Sample Points

The function to integrate is \( f(x) = x^2 + \frac{1}{x} \). Substitute \( x_i^{*} = 2 + \frac{3i}{n} \) into \( f(x) \) to get the function value at the sample points: \( f(x_i^{*}) = \left( 2 + \frac{3i}{n} \right)^2 + \frac{1}{2 + \frac{3i}{n}} \).

Step 5: Formulating the Riemann Sum

The Riemann sum \( S_n \) is expressed as \[ S_n = \sum_{i=1}^{n} f(x_i^{*}) \Delta x = \sum_{i=1}^{n} \left( \left( 2 + \frac{3i}{n} \right)^2 + \frac{1}{2 + \frac{3i}{n}} \right) \cdot \frac{3}{n} \].

Step 6: Expressing the Integral as a Limit

The integral \( \int^5_2 \left( x^2 + \frac{1}{x} \right) \ dx \) is then expressed as \[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{i=1}^{n} \left( \left( 2 + \frac{3i}{n} \right)^2 + \frac{1}{2 + \frac{3i}{n}} \right) \cdot \frac{3}{n} \].

Key Concepts

Definite Integrals

Definite integrals allow us to calculate the accumulated value of a function over an interval. In simpler terms, they help us find the total effect of a function within specific bounds. When you see a definite integral, it is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively. Here, they represent the start and end of the interval on the x-axis.

For example, in our exercise, the integral \( \int_{2}^{5} \big( x^2 + \frac{1}{x} \big) \, dx \) calculates the area under the curve of \( x^2 + \frac{1}{x} \) from \( x = 2 \) to \( x = 5 \). It combines the values of the function within this interval.

This whole process sums up all tiny parts from \( x = 2 \) to \( x = 5 \), providing a single value that captures the function's behavior over the interval.

Limit of Sums

The concept of limit of sums is fundamental to understanding how definite integrals work. By evaluating a function at various points within an interval and summing these values, we can approximate the area under the curve. This approach uses Riemann sums, which are named after the mathematician Bernhard Riemann.

The Riemann sum resembles a series of rectangles beneath the curve. Each rectangle's height is determined by the function's value at a chosen point, and the width by the distance between points. As we increase the number of rectangles (subintervals), our approximation becomes more accurate, closely matching the true area under the curve.

To express a definite integral as a limit of a Riemann sum, we take the limit of these sums as the number of intervals \( n \) approaches infinity. In our specific exercise, this becomes:

• \( \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{i=1}^{n} \left( \left( 2 + \frac{3i}{n} \right)^2 + \frac{1}{2 + \frac{3i}{n}} \right) \cdot \frac{3}{n} \).

This limit of sums yields the exact value of the integral.

Partitioning Intervals

Partitioning intervals involves dividing the integral's range into smaller sections, making it easier to calculate the Riemann sum. In our exercise, the interval from \( x = 2 \) to \( x = 5 \) is divided into \( n \) equal parts.

The width of each subinterval, \( \Delta x \), is calculated by dividing the total range by the number of parts, resulting in \( \Delta x = \frac{5 - 2}{n} = \frac{3}{n} \).

This partitioning forms the backbone of our Riemann sums approach. By making these parts smaller and more numerous, the Riemann sum offers a closer approximation to the actual area under the curve. As \( n \) becomes very large, the total sum approaches the exact value defined by the integral.

Choosing Sample Points

In the partitioning process, we must select a point within each subinterval to evaluate the function. These are called sample points, and they are critical to forming a Riemann sum.

The choice of sample points affects the accuracy of our sum. Common strategies include using the left endpoint, right endpoint, or midpoint of each subinterval. In our exercise, the right endpoint is chosen. This means for the \( i \)-th subinterval, our sample point \( x_i^* \) is calculated as:

• \( x_i^* = 2 + i \Delta x = 2 + \frac{3i}{n} \)

This approach helps determine the height of the corresponding rectangle, contributing to the Riemann sum. As we refine our partition by increasing \( n \), and compute more sample points, the resulting sum converges to the definite integral's exact value.