Fall 2006 - Takeda's Class - Exam 1

Fall 2006 - Takeda's Class - Exam 1 - c = 2-1 3 c √ c c...

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Math 20B (Calculus of Science and Engineering), Midterm 1 Oct. 16, 2006 Name: Section: This exam consists of 7 pages including this front page. Ground Rules 1. No calculator is allowed. 2. Show your work for every problem. A correct answer without any justification will receive no credit. 3. You may use one 4-by-6 index card, both sides. Score 1 10 2 10 3 10 4 10 5 10 6 10 Total 60 1
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1. (a) Evaluate Z 1 0 (6 x 3 + e x ) dx Z 1 0 (6 x 3 + e x ) dx = ( 6 4 x 4 + e x ) i 1 0 = 3 2 + e - 1 = 1 2 + e (b) Find the indefinite integral Z 1 x dx Z 1 x dx = Z x - 1 / 2 dx = 1 - 1 / 2 + 1 x - 1 / 2+1 + C = 2 x 1 2 + C 2
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2. Evaluate Z π 0 t cos( t 2 ) dt Let u = t 2 . Then du = 2 tdt , and t 0 π u 0 π . Then Z π 0 t cos( t 2 ) dt = Z π 0 1 2 cos u du = 1 2 sin u i Π 0 = 1 2 (0 - 0) = 0 . 3
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3. Find the number c > 0 so that the area of the region enclosed by y = 2 x 2 and y = x 2 + c becomes 32 / 3. Let 2 x 2 = x 2 + C . Then x = ± c . So the area of the region is Z c - c ( x 2 + c - 2 x 2 ) dx = Z c - c ( - x 2 + c ) dx = 2 Z c 0 ( - x 2 + c ) dx = 2( - 1 3 x 3 + cx ) i
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Unformatted text preview: c = 2(-1 3 c √ c + c √ c ) = 4 3 c √ c. Set 4 3 c √ c = 32 3 . i.e. c √ c = 8 . Thus c = 4 . 4 4. Find the volume of the solid obtained by rotating about the x-axis the region in the first quadrant enclosed by x = 1 , x = 2 , y = 1, and y = 1 /x . Volume = Z 2 1 π (1 2-± 1 x ² 2 dx = π Z 2 1 (1-1 x 2 dx = π ( x + 1 x ) i 2 1 = π (2 + 1 2-(1 + 1)) = π 2 5 5. Find the average value of the function f ( x ) = √ x on the interval [0 , 1]. Average = 1 1-Z 1 √ x dx = 2 3 x 3 2 i 1 = 2 3 6 6. Sketch the curve with the polar equation r = sin( θ/ 2). 7...
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Fall 2006 - Takeda's Class - Exam 1 - c = 2-1 3 c √ c c...

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