Math Problem Statement

Solve question 22

Solve question 22

Solution

Let’s solve question 22 step by step.

Problem Restatement:

The function f(x)=x2f(x) = x^2 is defined on the interval [0,1][0, 1]. We are tasked with:

  1. Showing that the left sum for the partition of the interval is given by: Ln=1n3(02+12+22++(n1)2).L_n = \frac{1}{n^3} \left( 0^2 + 1^2 + 2^2 + \dots + (n-1)^2 \right).
  2. Using the summation formula: k=0n1k2=(n1)n(2n1)6,\sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}, to simplify LnL_n to the form: Ln=16n2(n1)(2n1).L_n = \frac{1}{6n^2}(n-1)(2n-1).
  3. Computing the limit of LnL_n as nn \to \infty.

Step 1: Left Sum Formula

The interval [0,1][0, 1] is divided into nn subintervals of equal width: Δx=1n.\Delta x = \frac{1}{n}.

The left-hand sum is calculated as: Ln=k=0n1f(xk)Δx,L_n = \sum_{k=0}^{n-1} f(x_k) \Delta x, where xkx_k represents the left endpoint of the kk-th subinterval. For this problem: xk=kn,f(xk)=(kn)2.x_k = \frac{k}{n}, \quad f(x_k) = \left(\frac{k}{n}\right)^2.

Substituting into the formula: Ln=k=0n1(kn)21n.L_n = \sum_{k=0}^{n-1} \left(\frac{k}{n}\right)^2 \cdot \frac{1}{n}.

Simplify the expression: Ln=1n3k=0n1k2.L_n = \frac{1}{n^3} \sum_{k=0}^{n-1} k^2.

Thus, we have shown: Ln=1n3(02+12+22++(n1)2).L_n = \frac{1}{n^3} \left( 0^2 + 1^2 + 2^2 + \dots + (n-1)^2 \right).


Step 2: Simplifying Using the Summation Formula

Using the given summation formula: k=0n1k2=(n1)n(2n1)6,\sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}, we substitute this into LnL_n: Ln=1n3(n1)n(2n1)6.L_n = \frac{1}{n^3} \cdot \frac{(n-1)n(2n-1)}{6}.

Simplify: Ln=(n1)(2n1)6n2.L_n = \frac{(n-1)(2n-1)}{6n^2}.

Thus, the simplified expression for LnL_n is: Ln=16n2(n1)(2n1).L_n = \frac{1}{6n^2}(n-1)(2n-1).


Step 3: Compute the Limit of LnL_n as nn \to \infty

To find: limnLn,\lim_{n \to \infty} L_n, use the expression: Ln=16n2(n1)(2n1).L_n = \frac{1}{6n^2}(n-1)(2n-1).

Expand (n1)(2n1)(n-1)(2n-1): (n1)(2n1)=2n23n+1.(n-1)(2n-1) = 2n^2 - 3n + 1.

Substitute back: Ln=16n2(2n23n+1).L_n = \frac{1}{6n^2} (2n^2 - 3n + 1).

Split into terms: Ln=16(2n2n23nn2+1n2).L_n = \frac{1}{6} \left( \frac{2n^2}{n^2} - \frac{3n}{n^2} + \frac{1}{n^2} \right).

Simplify: Ln=16(23n+1n2).L_n = \frac{1}{6} \left( 2 - \frac{3}{n} + \frac{1}{n^2} \right).

Take the limit as nn \to \infty: limnLn=16(20+0)=26=13.\lim_{n \to \infty} L_n = \frac{1}{6} (2 - 0 + 0) = \frac{2}{6} = \frac{1}{3}.


Final Answer:

The limit of LnL_n as nn \to \infty is: 13\boxed{\frac{1}{3}}


Would you like a detailed explanation of any specific step, or do you have further questions? Here are related questions for practice:

  1. How is the left-hand sum formula derived for general functions?
  2. How can you compute the right-hand sum for the same function f(x)=x2f(x) = x^2?
  3. How does the summation formula for k2k^2 relate to the derivation of LnL_n?
  4. What is the integral of f(x)=x2f(x) = x^2 over [0,1][0, 1], and how does it compare to LnL_n?
  5. How do Riemann sums converge to the definite integral?

Tip: When simplifying expressions with limits, always consider factoring terms to isolate dominant terms for large nn.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Riemann Sums
Summation
Limits
Definite Integrals

Formulas

L_n = (1/n^3) * Σ(k=0 to n-1) k^2
Summation formula: Σ(k=0 to n-1) k^2 = (n-1)n(2n-1)/6
Expansion: (n-1)(2n-1) = 2n^2 - 3n + 1
Limit as n -> ∞: lim(L_n) = 1/3

Theorems

Summation Formula for k^2
Riemann Sum Definition

Suitable Grade Level

Grades 11-12 (Advanced Calculus/Pre-University)