Problem 49 - Calculus for Biology and Medicine

Question

Express the definite integrals as limits of Riemann sums. $$ \int_{2}^{6}(x+1)^{1 / 3} d x $$

Step-by-Step Solution

Answer

The integral is \( \lim_{n \to \infty} \sum_{i=1}^{n} \left(3 + i\frac{4}{n}\right)^{1/3} \cdot \frac{4}{n} \).

Step 1: Understanding the Problem

We need to express the given integral \( \int_{2}^{6}(x+1)^{1 / 3} \, dx \) as a limit of a Riemann sum. A Riemann sum approximates the integral by dividing the area under the curve into rectangles.

Step 2: Partition the Interval

The interval from 2 to 6 is split into \( n \) equal subintervals. The width of each subinterval is \( \Delta x = \frac{6-2}{n} = \frac{4}{n} \). The endpoints of the subintervals are \( x_0 = 2, x_1, x_2, \ldots, x_n = 6 \).

Step 3: Choose Sample Points

For simplicity, choose the right endpoints of each subinterval as the sample points. These will be \( x_i = 2 + i\frac{4}{n} \) for \( i = 1, 2, \ldots, n \).

Step 4: Write the Function at Sample Points

Evaluate the function \( (x+1)^{1/3} \) at each sample point: \( f(x_i) = (x_i + 1)^{1/3} = \left(2 + i\frac{4}{n} + 1\right)^{1/3} = \left(3 + i\frac{4}{n}\right)^{1/3} \).

Step 5: Formulate the Riemann Sum

The Riemann sum for this integral can be expressed as: \[ S_n = \sum_{i=1}^{n} \left(3 + i\frac{4}{n}\right)^{1/3} \cdot \frac{4}{n} \] where \( \frac{4}{n} \) is the width of each subinterval.

Step 6: Limit as a Definite Integral

The definite integral is the limit of the Riemann sum as \( n \to \infty \): \[ \int_{2}^{6}(x+1)^{1 / 3} \ dx = \lim_{n \to \infty} \sum_{i=1}^{n} \left(3 + i\frac{4}{n}\right)^{1/3} \cdot \frac{4}{n} \].

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus, representing the signed area under a curve of a function between two points, known as the limits of integration. In our exercise, the definite integral is represented by \[\int_{2}^{6}(x+1)^{1 / 3} \, dx\]which is used to calculate the area under the curve of the function \(f(x) = (x+1)^{1/3}\) from \(x = 2\) to \(x = 6\).

• The lower limit of the integral is 2, and the upper limit is 6.
• The integrand is the function \((x+1)^{1/3}\), which is continuous on the interval \([2, 6]\).
• This integral helps find the accumulated value or "total change" over the interval.

The process of integrating involves finding an antiderivative, followed by evaluating it at the upper and lower limits and taking the difference.

Limit of a Riemann Sum

A Riemann sum provides an approximation for the integral of a function over an interval by summing up the areas of several rectangles. The concept of a limit of a Riemann sum is pivotal in understanding how these approximations converge to the exact value of a definite integral.In our problem, we approximate the integral by using:\[\sum_{i=1}^{n} \left(3 + i\frac{4}{n}\right)^{1/3} \cdot \frac{4}{n}\]where each rectangle has width \(\Delta x = \frac{4}{n}\) and height determined by the function value at the selected sample point.

• The width \(\Delta x\) is smaller for larger \(n\), increasing the number of rectangles used.
• The height is based on \(\left(3 + i\frac{4}{n}\right)^{1/3}\) for right endpoint selection.
• As \(n\) approaches infinity, the approximation of the Riemann sum converges to the actual value of the definite integral:\[\lim_{n \to \infty} \sum_{i=1}^{n} \left(3 + i\frac{4}{n}\right)^{1/3} \cdot \frac{4}{n}\]

This concept helps illustrate the relationship between discrete approximations and continuous accumulation.

Partitioning Intervals

Partitioning an interval is a key step in the process of evaluating integrals using Riemann sums. This involves dividing the interval of integration into smaller segments, called subintervals. In our case, the interval [2, 6] is divided into \(n\) equal parts.

• The length of each subinterval, denoted by \(\Delta x\), is given by:\[\frac{6 - 2}{n} = \frac{4}{n}\]
• The endpoints of these subintervals are \(x_0 = 2\), \(x_n = 6\), with \(x_i = 2 + i\frac{4}{n}\) for the intermediate points.

Choosing sample points within these intervals is crucial for computing the Riemann sum. Often, these are taken as the left endpoints, right endpoints, or midpoints of each subinterval. Here, we chose the right endpoints:

• Sample points: \(x_i = 2 + i\frac{4}{n}\)
• Right endpoints simplify calculation since they are aligned with each partition's boundary.

Partitioning allows the function's behavior over the whole interval to be explored in smaller, more manageable parts, facilitating the approximation of the integral.