A Riemann sum estimates a quantity such as the area under a curve by dividing a region into smaller pieces and adding their areas. It is a foundation for numerical integration and for understanding the definite integral.
Imagine estimating the area beneath a curve. Divide the horizontal interval into subintervals, then use the function's value at a point in each subinterval to set the height of a rectangle. The width times the height gives that rectangle's signed area contribution. Adding the contributions produces a finite-sum estimate.
All flashcards
What is a Riemann sum?
A finite sum that approximates an integral by adding products of sampled function values and subinterval widths.
What is a partition?
An ordered set of points a=x0<x1<⋯<xn=b that splits an interval into subintervals.
What does Δxi mean?
It is the width of the ith subinterval: Δxi=xi−xi−1.
What is a sample point xi∗?
A point chosen inside the ith subinterval; its function value is used as the height contribution for that piece.
Why refine a partition?
As the mesh shrinks, sums for an integrable function approach its integral.
Quiz questions
What is the main purpose of a Riemann sum?Answer: To estimate an accumulated quantity by summing contributions from partitioned pieces.
For an increasing, nonnegative function, how do left- and right-endpoint sums compare with the integral?Answer: The left sum underestimates, and the right sum overestimates.
In a three-dimensional Riemann sum, what does each cell-size factor represent?Answer: The volume of the corresponding cell.
For a continuous function on a closed interval, what happens to Riemann sums as the partition mesh tends to zero?Answer: They converge to the same definite integral for any sample-point choices.
Under the stated derivative bounds, how do midpoint and trapezoidal error bounds compare in order with left and right sums?Answer: They decrease on the order of 1/n2, rather than 1/n.
Based on the Wikipedia article Riemann sum, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.