Chapter 7: Problem 50
The left and right Riemann sums of a function \(f\) on the interval [2,6] are denoted by LEFT \((n)\) and \(\operatorname{RIGHT}(n),\) respectively, when the interval is divided into \(n\) equal parts. In Problems \(46-56,\) decide whether the statements are true for all continuous functions, \(f\). Give an explanation for your answer. $$\operatorname{LEFT}(n)-\operatorname{RIGHT}(n)=(f(2)-f(6)) \Delta x$$
Short Answer
Step by step solution
Understanding the Problem
Definition of Riemann Sums
Formulating LEFT(n) and RIGHT(n)
Deriving the Difference
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Left Riemann Sum
- Each subinterval has a width of \( \Delta x \), calculated by dividing the interval length by the number of subintervals, \( n \).
- The formula for Left Riemann Sum is given by: \[ \text{LEFT}(n) = \sum_{i=0}^{n-1} f(a + i\Delta x) \cdot \Delta x \]
Right Riemann Sum
- The calculation for Right Riemann Sum also involves \( \Delta x \), the width of each subinterval.
- The formula is written as: \[ \text{RIGHT}(n) = \sum_{i=1}^{n} f(a + i\Delta x) \cdot \Delta x \]
Continuous Functions
- A continuous function can be graphically represented as an unbroken curve.
- It is crucial for the successful application of Left and Right Riemann Sums to approximate integrals accurately.
Interval Subdivision
- This equal division is crucial as it sets up consistent units to apply either the left or right endpoint Riemann Sum technique.
- Choosing more subintervals (i.e., a larger \( n \)) leads to a finer approximation of the area, as each subinterval becomes narrower, capturing the function's behavior with increased precision.