midterm2_10b_sols

# midterm2_10b_sols - Name Sec No Math 10B Midterm 2 PID Sec...

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Name: PID: Sec. No: Sec. Time: Math 10B Midterm 2 August 26, 2010 Please turn off and but away all electronic devices, except for calculators. You may use 1 page of handwritten notes, but no other resources during the exam. Show all of your work, no credit will be given for unsupported answers. Please write neatly and make sure the answer is clear. 1. (6 points) Consider the function h ( x ) = 25 - 2 x 2 on [0 , 3] (a) Compute TRAP(3). Solution: First note that Δ x = ( b - a ) /n = (3 - 0) / 3 = 1. Therefore TRAP(3) = (Δ x/ 2)( h (0) + h (1)) + (Δ x/ 2)( h (1) + h (2)) + (Δ x/ 2)( h (2) + h (3)) = (1 / 2)(25 + 23) + (1 / 2)(23 + 17) + (1 / 2)(17 + 7) = (1 / 2)(48) + (1 / 2)(40) + (1 / 2)(24) = 24 + 20 + 12 = 56 (b) Compute MID(3) Solution: Of course Δ x is still equal to 1. TRAP(3) = (Δ x )( h (0 . 5)) + (Δ x )( h (1 . 5)) + (Δ x )( h (2 . 5)) = (25 + 23) + (23 + 17) + (17 + 7) = (1 / 2)(48) + (1 / 2)(40) + (1 / 2)(24) = 24 + 20 + 12 = 56 (c) State whether MID(3) or TRAP(3) overestimates 3 0 h ( x ) dx . Explain your answer. Solution: The key observation is that since h ( x ) = - 4, h ( x ) is concave down. There- fore TRAP(3) provides an underestimate of the definite integral and MID(3) provides an overestimate. No. Points Score 1 6 2 5 3 5 4 12 5 6 Total 34

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2. (5 points) Consider the function f ( x ) = tan 2 ( x ) (a) Use the second fundamental theorem of calculus to compute the antiderivative, F ( x ), of f ( x ) with the property that F (1) = 3. Solution: The second fundamental theorem of calculus asserts that G ( x ) = x 1 f ( t ) dt is an antiderivative of f ( x ) with the property that G (1) = 0, at least on [1 , 3 π/ 2). (This was an oversight on my part, f ( x ) = tan 2 ( x ) is not continuous at x = 3 π/ 2.) Since we want F (1) = 3, add 3 to G ( x ), yielding F ( x ) = x 1 f ( t ) dt + 3 (b) Compute F ( π ).
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