Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

8. Definite Integrals

Riemann Sums

Calculus

8. Definite Integrals

Riemann Sums

Previous problemNext problem

2:11 minutes

Problem 5.2.33d

Textbook Question

{Use of Tech} Approximating definite integrals Complete the following steps for the given integral and the given value of n.
(d) Determine which Riemann sum (left or right) underestimates the value of the definite integral and which overestimates the value of the definite integral.

∫₁⁷ 1/𝓍 d𝓍 ; n = 6

Verified step by step guidance

Step 1: Understand the problem. The integral ∫₁⁷ (1/𝓍) d𝓍 represents the area under the curve of the function f(𝓍) = 1/𝓍 from x = 1 to x = 7. We are tasked with determining which Riemann sum (left or right) underestimates or overestimates the value of this definite integral, given n = 6 subintervals.

Step 2: Divide the interval [1, 7] into n = 6 subintervals. The width of each subinterval, Δ𝓍, is calculated as Δ𝓍 = (7 - 1)/6 = 6/6 = 1. This means the subintervals are [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], and [6, 7].

Step 3: Recall the definitions of the left and right Riemann sums. The left Riemann sum uses the function values at the left endpoints of each subinterval, while the right Riemann sum uses the function values at the right endpoints of each subinterval. For f(𝓍) = 1/𝓍, the left endpoints are {1, 2, 3, 4, 5, 6}, and the right endpoints are {2, 3, 4, 5, 6, 7}.

Step 4: Analyze the behavior of the function f(𝓍) = 1/𝓍. Since f(𝓍) is a decreasing function on the interval [1, 7], the left Riemann sum will underestimate the area (because it uses smaller function values at the left endpoints), and the right Riemann sum will overestimate the area (because it uses larger function values at the right endpoints).

Step 5: Conclude that for the given integral ∫₁⁷ (1/𝓍) d𝓍 and n = 6, the left Riemann sum underestimates the value of the definite integral, while the right Riemann sum overestimates the value of the definite integral.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Riemann Sums

Riemann sums are a method for approximating the value of a definite integral by dividing the area under a curve into rectangles. The sum of the areas of these rectangles provides an estimate of the integral's value. Depending on whether the left or right endpoints of the subintervals are used, the Riemann sum can either overestimate or underestimate the actual area, which is crucial for understanding the behavior of the integral.

Definite Integrals

A definite integral represents the net area under a curve between two specified limits, providing a precise value for the accumulation of quantities. It is denoted as ∫ₐᵇ f(x) dx, where 'a' and 'b' are the limits of integration. Understanding how to evaluate definite integrals is essential for applying Riemann sums effectively, as it allows for comparison between the approximated and actual values.

Underestimation and Overestimation

In the context of Riemann sums, underestimation occurs when the sum of the areas of rectangles is less than the actual area under the curve, while overestimation occurs when it is greater. For a decreasing function, the left Riemann sum underestimates and the right Riemann sum overestimates the integral. Conversely, for an increasing function, the left sum overestimates and the right sum underestimates, making it important to analyze the function's behavior to determine which sum to use.

Watch next

Master Sigma Notation with a bite sized video explanation from Patrick

Start learning

Related Videos

Related Practice

Textbook Question