HW9sol - = 1 √ 2 t 4 and by FTC1 g x = F x 2-F(tan x...

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M408C, Homework 9 — Solution Instructor: Guoliang Wu 1. If f ( x ) = x 2 - 2 x, 0 x 3, evaluate the Riemann sum with n = 6, taking the sample points to be the middle endpoints of sub-intervals. Illustrate with a diagram. Solution: The graph f ( x ) = ( x - 1) 2 - 1 is as follows. The Riemann sum, with n = 6 and the sample points 1 / 4 , 3 / 4 , 5 / 4 , 7 / 4 , 9 / 4 , 11 / 4, is given by the net sum of the areas of the rectangles: M 6 = 1 2 f (1 / 4) + 1 2 f (3 / 4) + 1 2 f (5 / 4) + 1 2 f (7 / 4) + 1 2 f (9 / 4) + 1 2 f (11 / 4) = 71 16 . 2. Find the derivative g ( x ) of the function g ( x ) = x 2 tan x 1 2 + t 4 dt. 1
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Solution: Let F ( t ) be an antiderivative of 1 2+ t 4 , then F ( t
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Unformatted text preview: ) = 1 √ 2+ t 4 , and by FTC1, g ( x ) = F ( x 2 )-F (tan x ). Therefore, by chain rule, g ′ ( x ) = F ′ ( x 2 )2 x-F ′ (tan x ) sec 2 x = 2 x √ 2 + x 8-sec 2 x √ 2 + tan 4 x 3. Evaluate the integral ∫ π/ 4 1 1 + cos 2 θ cos 2 θ dθ. Solution: ∫ π/ 4 1 1 + cos 2 θ cos 2 θ dθ = ∫ π/ 4 1 ( 1 cos 2 θ + 1 ) dθ = ∫ π/ 4 1 ± sec 2 θ + 1 ² dθ = tan θ + θ ] π/ 4 1 = (1 + π/ 4)-(tan 1 + 1) = π/ 4-tan 1 . 2...
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HW9sol - = 1 √ 2 t 4 and by FTC1 g x = F x 2-F(tan x...

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