Evaluate Definite Integral Using Right Hand Rule: Show Work

  • Thread starter Thread starter DemiMike
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on evaluating the definite integral of the function \(3x^2 - 5x - 6\) from -4 to 1 using the right-hand Riemann sum method. Participants emphasize the importance of avoiding shortcut methods and instead applying the definition of the definite integral. The correct approach involves calculating the Riemann sums by partitioning the interval and using the right endpoints for evaluation. The user initially attempted a shortcut method but was guided back to the long process required for this evaluation.

PREREQUISITES
  • Understanding of definite integrals and their properties
  • Familiarity with Riemann sums, specifically right-hand Riemann sums
  • Basic algebraic manipulation skills
  • Knowledge of polynomial functions and their integration
NEXT STEPS
  • Study the concept of Riemann sums in detail
  • Learn how to partition intervals for definite integrals
  • Practice evaluating definite integrals using various methods
  • Explore the differences between left-hand, right-hand, and midpoint Riemann sums
USEFUL FOR

Students and educators in calculus, particularly those focusing on integral calculus and Riemann sums. This discussion is beneficial for anyone looking to deepen their understanding of definite integrals without relying on shortcut methods.

DemiMike
Messages
9
Reaction score
0
Use the definition of the definite inegral (with right hand rule) to evaluate the following integral. Show work please

Can NOT use shortcut method.. but be the long process


1
S (3x^2 - 5x - 6) dx
-4

edit:
i got this so far
1
S (3x^2 - 5x - 6) dx = [3x^3 /3 - 5x^2/2 - 6x], (1, -4)
-4
= 3*1/ 3 - 5*1 /2 - 6*1 - (3*4^3 - 5*4^2 - 6*2)

but iam lost in the simplification


edit: NVm i have no idea how to do the long method.. any 1 help <3 lpz
 
Last edited:
Physics news on Phys.org
That looks like the 'shortcut' method, but it's not a bad idea to do that real quick to check yourself.

What you want to do (as per the directions) is evaluate it using Riemann sums. Specifically, right-hand Riemann sums.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
4
Views
3K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K