By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
A Riemann sum is a mathematical concept used to approximate the area under a curve by dividing it into small rectangles or trapezoids. This technique is essential in calculus, particularly in integration, and is used to estimate the area under curves, volumes of solids, and other quantities.
Riemann sums have numerous real-world applications in various fields, including physics, engineering, economics, and data analysis. For instance, in physics, Riemann sums can be used to estimate the work done by a variable force over a given distance. In economics, Riemann sums can be used to estimate the total revenue generated by a company over a specific time period.
A Riemann sum is defined as the sum of the areas of small rectangles or trapezoids that approximate the area under a curve.
There are four types of Riemann sums: * Left Riemann sum: divides the area into rectangles that touch the left side of the curve. * Right Riemann sum: divides the area into rectangles that touch the right side of the curve. * Midpoint Riemann sum: divides the area into rectangles that touch the midpoint of the curve. * Trapezoidal Riemann sum: divides the area into trapezoids instead of rectangles.
The formula for a Riemann sum is given by:
$$\sum_{i=1}^{n} f(x_i) \Delta x$$
where $f(x_i)$ is the function value at the $i$th point, $\Delta x$ is the width of each rectangle or trapezoid, and $n$ is the number of subintervals.
As the number of subintervals ($n$) approaches infinity, the Riemann sum approaches the definite integral:
$$\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \int_{a}^{b} f(x) dx$$
To solve a problem involving Riemann sums, follow these steps:
Given the function $f(x) = 2x + 1$ and the interval $[0, 3]$, approximate the area under the curve using a left Riemann sum with $n = 4$ subintervals.
$$\Delta x = \frac{3 - 0}{4} = 0.75$$ $$f(x_1) = f(0) = 1$$ $$f(x_2) = f(0.75) = 2(0.75) + 1 = 2.5$$ $$f(x_3) = f(1.5) = 2(1.5) + 1 = 4$$ $$f(x_4) = f(2.25) = 2(2.25) + 1 = 5.5$$ $$\text{Left Riemann sum} = (1)(0.75) + (2.5)(0.75) + (4)(0.75) + (5.5)(0.75) = 7.5$$
The area under the curve is approximately 7.5 square units.
Given the function $f(x) = x^2$ and the interval $[0, 2]$, approximate the area under the curve using a right Riemann sum with $n = 3$ subintervals.
$$\Delta x = \frac{2 - 0}{3} = 0.67$$ $$f(x_1) = f(0.67) = (0.67)^2 = 0.45$$ $$f(x_2) = f(1.33) = (1.33)^2 = 1.77$$ $$f(x_3) = f(2) = 2^2 = 4$$ $$\text{Right Riemann sum} = (0.45)(0.67) + (1.77)(0.67) + (4)(0.67) = 4.51$$
The area under the curve is approximately 4.51 square units.
Given the function $f(x) = 3x - 2$ and the interval $[1, 4]$, approximate the area under the curve using a midpoint Riemann sum with $n = 5$ subintervals.
$$\Delta x = \frac{4 - 1}{5} = 0.6$$ $$f(x_1) = f(1.3) = 3(1.3) - 2 = 2.9$$ $$f(x_2) = f(1.9) = 3(1.9) - 2 = 4.7$$ $$f(x_3) = f(2.5) = 3(2.5) - 2 = 6.5$$ $$f(x_4) = f(3.1) = 3(3.1) - 2 = 8.3$$ $$f(x_5) = f(3.7) = 3(3.7) - 2 = 10.1$$ $$\text{Midpoint Riemann sum} = (2.9)(0.6) + (4.7)(0.6) + (6.5)(0.6) + (8.3)(0.6) + (10.1)(0.6) = 24.3$$
The area under the curve is approximately 24.3 square units.
Question: What is the formula for a Riemann sum? A) $\sum_{i=1}^{n} f(x_i) \Delta x$ B) $\sum_{i=1}^{n} f(x_i) \Delta x^2$ C) $\sum_{i=1}^{n} f(x_i) \Delta x^3$ D) $\sum_{i=1}^{n} f(x_i) \Delta x^4$ Correct Answer: A Explanation: The formula for a Riemann sum is given by $\sum_{i=1}^{n} f(x_i) \Delta x$. Why the Distractors Are Tempting: The distractors are tempting because they are close to the correct answer, but they are not the correct formula for a Riemann sum.
Question: What is the type of Riemann sum that divides the area into rectangles that touch the left side of the curve? A) Left Riemann sum B) Right Riemann sum C) Midpoint Riemann sum D) Trapezoidal Riemann sum Correct Answer: A Explanation: The left Riemann sum divides the area into rectangles that touch the left side of the curve. Why the Distractors Are Tempting: The distractors are tempting because they are close to the correct answer, but they are not the correct type of Riemann sum.
Question: What is the limit of a Riemann sum as $n$ approaches infinity? A) $\sum_{i=1}^{n} f(x_i) \Delta x$ B) $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \int_{a}^{b} f(x) dx$ C) $\sum_{i=1}^{n} f(x_i) \Delta x^2$ D) $\sum_{i=1}^{n} f(x_i) \Delta x^3$ Correct Answer: B Explanation: The limit of a Riemann sum as $n$ approaches infinity is the definite integral $\int_{a}^{b} f(x) dx$. Why the Distractors Are Tempting: The distractors are tempting because they are close to the correct answer, but they are not the correct limit of a Riemann sum.
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