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

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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
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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...
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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.

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

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