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Magazine Chapter 6: Problem 49
Express the definite integrals as limits of Riemann sums. $$ \int_{2}^{6}(x+1)^{1 / 3} d x $$
Short Answer
Expert verified
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 by step solution
01
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.
02
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 \( x_0 = 2, x_1, x_2, \ldots, x_n = 6 \).
03
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 \( i = 1, 2, \ldots, n \).
04
Write the Function at Sample Points
Evaluate the function \( (x+1)^{1/3} \) at each sample point: \( f(x_i) = (x_i + 1)^{1/3} = \left(2 + i\frac{4}{n} + 1\right)^{1/3} = \left(3 + i\frac{4}{n}\right)^{1/3} \).
05
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.
06
Limit as a Definite Integral
The definite integral is the limit of the Riemann sum as \( n \to \infty \): \[ \int_{2}^{6}(x+1)^{1 / 3} \ dx = \lim_{n \to \infty} \sum_{i=1}^{n} \left(3 + i\frac{4}{n}\right)^{1/3} \cdot \frac{4}{n} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:\[\lim_{n \to \infty} \sum_{i=1}^{n} \left(3 + i\frac{4}{n}\right)^{1/3} \cdot \frac{4}{n}\]
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.
- Sample points: \(x_i = 2 + i\frac{4}{n}\)
- Right endpoints simplify calculation since they are aligned with each partition's boundary.
