midterm-exam1-sol

midterm-exam1-sol - Math 21B UC Davis, Winter 2010 Dan...

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Math 21B UC Davis, Winter 2010 Dan Romik Solutions to Midterm Exam 1 1 Write the sum 1 + 2 + 3 + 4 + ... + 29 + 30 in notation, and evaluate it. Solution : 1 + 2 + ... + 30 = 30 k =1 k = 30 × 31 2 = 465 . 2 Find the total area of the regions bounded between the graph of y = sin( x ) and the x -axis (both below and above the axis) in the interval [0 , 2 π ]. Solution : Recall that sin x 0 if x is in [0 ], and sin x 0 if x is in [ π, 2 π ]. So the total area is Z 2 π 0 | sin x | dx = Z π 0 sin xdx + Z 2 π π ( - sin x ) dx = - cos x ± ± ± π 0 + cos x ± ± ± 2 π π = (1 - ( - 1)) + (1 - ( - 1)) = 4 . A common mistake was to integrate sin x instead of | sin x | , which gives 0. This clearly does not make sense, since the area has to be positive. The moral is that it is always good to do a “sanity check” on your answers. 3 If f ( x ) is integrable and 4 R 0 f ( x ) dx = 10, 4 R 2 f ( x ) dx = 6, what is 1 R 0 f (2 x ) dx ? Solution
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midterm-exam1-sol - Math 21B UC Davis, Winter 2010 Dan...

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