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hw2sol - -1 ln π OR Let u = x 1 then du = dx When x =-1,u...

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Fall 2010, MAT 21B, Written Homework # 2 Problem 1. Draw the graph of the function y = 1 - | x | over [ - 1 , 1], and then use area to evaluate the integral Z 1 - 1 1 - | x | dx Solution 1. The graph of y = 1 - | x | is seen below. x f ( x ) - 1 0 1 1 We observe that the area of the triangle of base 2 and height 1 is 1. Interpreting the integral of a nonnegative function as the area under the graph, we get Z 1 - 1 1 - | x | dx = 1 Problem 2. Evaluate the integral Z 0 - 1 π x +1 dx Solution 2. Note that π x +1 = π x · π , so we have Z 0 - 1 π x +1 dx = π Z 0 - 1 π x dx = π · π x ln π | x =0 x = - 1 = π ln π ± 1 - 1 π ² = π
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Unformatted text preview: -1 ln π . OR: Let u = x + 1 , then du = dx . When x =-1 ,u = 0 , and when x = 0 , u = 1 . Thus, Z-1 π x +1 dx = Z 1 π u du = π u ln π | u =1 u =0 = π-1 ln π . Problem 3. Calculate d dx Z 2 x tan x dt 1 + t 2 1 Solution 3. d dx Z 2 x tan x dt 1 + t 2 = (2 x ) 1 + (2 x ) 2-(tan x ) 1 + (tan x ) 2 = 2 1 + 4 x 2-sec 2 x 1 + tan 2 x = 2 1 + 4 x 2-1 , because of the identity 1 + tan 2 x = sec 2 x . 2...
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hw2sol - -1 ln π OR Let u = x 1 then du = dx When x =-1,u...

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