Evaluating Riemann Integrals of f(x)=x^k where k>1 is an Integer

Click For Summary
SUMMARY

The discussion focuses on evaluating the Riemann integral of the function f(x) = x^k over the interval [0,1], where k is an integer greater than 1. Participants outline the steps to compute the lower and upper Riemann sums, L(f,P_m) and U(f,P_m), using a specific partition P_m defined by q_m = m^(-1/m). The limits of these sums converge to 1/(k+1), confirming the integrability of f on [0,1] and establishing that the integral evaluates to 1/(k+1).

PREREQUISITES
  • Understanding of Riemann integrals
  • Familiarity with limits and convergence
  • Knowledge of partitions in calculus
  • Basic algebraic manipulation of functions
NEXT STEPS
  • Study the properties of Riemann integrals in detail
  • Learn about different types of partitions and their impact on integrals
  • Explore the concept of integrability criteria for functions
  • Investigate the application of the Fundamental Theorem of Calculus
USEFUL FOR

Students and educators in calculus, mathematicians focusing on real analysis, and anyone interested in understanding the evaluation of Riemann integrals for polynomial functions.

abm
Messages
3
Reaction score
0
Please Help... Riemann

Please Help!
To compute the Riemann integral of f:[0,1]->R given f(x)=x^k where k>1 is an integer
1. Let m>2 and define q_m= m^(-1/m) Let P_m be the partition of [0,1] given by P_m=(0< q_m^m < q_m^(m-1)< ...< q_m <1)
Explicitly evalute L(f,P_m) and U(f,P_m)
2. Show that lim n->inf. L(f,P_m)= 1/(k+1),and lim n->inf. U(f,P_m)= 1/(k+1)
3. Show that f is integrable on [0,1]
4. Show that integra[0,1] f(x) dx= 1/(k+1)
 
Physics news on Phys.org
We'll help, but you have to work with us. Show how you started and where you got stuck.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K