This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: f ( 3 i n ) 3 n = n X i =1 ± i 2 n 2 + 1 ² 3 n = 3 + 3 n 3 n X i =1 i 2 Remark: We may further simply them using the formula n X k =1 k 2 = n ( n + 1)(2 n + 1) 6 . Lower sum = 3 + 3 n 3 n X i =1 ( i1) 2 = 3 + 3 n 3 n1 X i =1 i 2 = 3 + 3 n 3 · ( n1) n (2 n1) 6 = 3 + 2 n 23 n + 1 2 n 2 = 4 + 13 n 2 n 2 Upper sum = 3 + 3 n 3 n X i =1 i 2 = 3 + 3 n 3 · n ( n + 1)(2 n + 1) 6 = 3 + 2 n 2 + 3 n + 1 2 n 2 = 4 + 1 + 3 n 2 n 2 1 Problem 3. Solve the initial value problem dy dx = 3 x with the initial value y(0) = 1. Solution 3. One can check that 3 x = e x ln 3 so that antiderivatives have the form 1 ln 3 3 x + C where C ∈ R . The initial value condition gives y (0) = 1 ln 3 + C = 1 so that C = 11 ln 3 and y = 1 ln 3 3 x + (11 ln 3 ) . 2...
View
Full
Document
This note was uploaded on 10/08/2010 for the course MATH 21B taught by Professor Vershynin during the Spring '08 term at UC Davis.
 Spring '08
 Vershynin
 Calculus, Antiderivatives, Derivative, Slope

Click to edit the document details