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A01 soln

# A01 soln - Math 138 Assignment 1 Solutions 1 If we are...

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Math 138 Assignment 1 Solutions 1. If we are trying to make this true for all x then we are essentially trying to create an identity between the two functions f ( x ) = 8 + Z x a g ( t ) t 2 dt and h ( x ) = 8 3 x that is, we are trying to solve for a and g ( t ) so that f ( x ) h ( x ). If two functions are equal then their derivatives are equal (note the converse is not neces- sarily true) and so we can say f 0 ( x ) h 0 ( x ) g ( x ) x 2 8 3 x - 2 3 by FTC so g ( x ) 8 3 x 4 3 Using this in f ( x ) gives f ( x ) = 8 + 8 3 Z x a t - 2 3 dt f ( x ) = 8 + 8 3 (3 t 1 3 ) x a f ( x ) = 8 + 8 x 1 3 - 8 a 1 3 Since we want f ( x ) h ( x ) then we must have 8 + 8 x 1 3 - 8 a 1 3 8 3 x meaning a 1 3 = 1 so a = 1 2. (a) Z 1 x ln x dx Let u = ln x then du = dx x so the integral becomes Z 1 x ln x dx = Z 1 u du = ln u + C = ln(ln x ) + C therefore Z 1 x ln x dx = ln(ln x ) + C (b) Z 1 - sin x 1 - sin 2 x dx

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Notice that Z 1 - sin x 1 - sin 2 x dx = Z 1 - sin x cos 2 x dx = Z 1 cos 2 x dx - Z sin x cos 2 x dx = Z sec 2 x dx - Z sin x cos 2 x dx let u = cos x , du = - sin x dx = tan x + Z 1 u 2 du = tan x - 1 u + C = tan x - 1 cos x + C =
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A01 soln - Math 138 Assignment 1 Solutions 1 If we are...

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