hw1sol

# hw1sol - f ( 3 i n ) 3 n = n X i =1 ± i 2 n 2 + 1 ² 3 n =...

This preview shows pages 1–2. Sign up to view the full content.

Fall 2010, MAT 21B, Paper Homework # 1 Problem 1. Find the curve y = f ( x ) in the xy -plane that passes through the point (9 , 4) and whose slope at each point is 3 x . Solution 1. Observe that all antiderivatives of the function 3 x have slope 3 x at each point x . These antiderivatives have the form 2 x 3 2 + C where C is any constant in R . The particular function we want passes through (9 , 4) , which means that f (9) = 4 . Thus f (9) = 2 · 9 3 2 + C = 4 so that C = - 50 and f ( x ) = 2 x 3 2 - 50 . Problem 2. Using sigma notation, express the upper sum an the lower sum of f ( x ) = x 2 9 +1 obtained by dividing the interval [0 , 3] into n equal subintervals. Solution 2. Here Δ x = 3 n and x i = i Δ x = 3 i n for each i = 0 , 1 , ··· ,n . Note that f ( x ) = x 2 9 + 1 increases monotonically in [0 , 3] , so in each subinterval, the left endpoint ( 3( i - 1) n for i ∈ { 1 , 2 ,...,n } ) attains the minimum, and the right endpoint ( 3 i n for i ∈ { 1 , 2 ,...,n } ) attains the maximum. Thus Lower sum = n X i =1 f ( 3( i - 1) n ) 3 n = n X i =1 ± ( i - 1) 2 n 2 + 1 ² 3 n = 3 + 3 n 3 n X i =1 ( i - 1) 2 Upper sum = n X i =1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the 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 ( i-1) 2 = 3 + 3 n 3 n-1 X i =1 i 2 = 3 + 3 n 3 · ( n-1) n (2 n-1) 6 = 3 + 2 n 2-3 n + 1 2 n 2 = 4 + 1-3 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 = 1-1 ln 3 and y = 1 ln 3 3 x + (1-1 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.

### Page1 / 2

hw1sol - f ( 3 i n ) 3 n = n X i =1 ± i 2 n 2 + 1 ² 3 n =...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online