Quiz_1_Solution

Quiz_1_Solution - Math 20B (Shenk). Quiz 1 Solution. July...

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Unformatted text preview: Math 20B (Shenk). Quiz 1 Solution. July 7, 2008. 4 1. 2. 3. T (4) = T (1) + 1 T (t) dt = 60 − 10.9 + 15.4 = 64.5◦ F 5 sin x − 3 sec2 x dx = −5 cos x − 3 tan x + C −5 −10 1 dx = ln |x| x 5 = ln | − 5| − ln | − 10| = ln(5) − ln(10) (which can be written − ln(2)) −10 4. u = sin x, du = cos x dx • = 1 2 sin x cos x dx = u du = 12 2u +C sin2 x + C 1 2 5. 6. 4 x 2 x +4 dx = 1 2 tan–1 1 2x +C = 1 ⇐⇒ x2 = 4 ⇐⇒ x = ±2 • Figure A6 • 2 [Area] = 1 4 x2 −1 dx = − 4 −x x 2 1 = [− 4 − 2] − [−4 − 1] = 1 2 y y = 4x−2 3 y=1 Figure A6 1 2 x 1 Math 20B. Summer, 2008. Quiz 1 Solution. (7/7/08) p. 2 7. Base: Figure A7a • The cross section at x is a square of width w = ex . • Figure A7b • 1 1 A(x) = w2 = (ex )2 = e2x for 0 ≤ x ≤ 1 • [Volume]= 0 A(x) dx = 0 e2x dx • 1 2x 2e u = 2x, du = 2 dx • [Volume] = 1 2x 2e 1 0 e2x dx = 1 2 1 2 e2x (2x dx) = e2 − 1 1 2 eu du = 1 eu + C = 2 +C • 1 = 2 e2 − 1 e0 = 2 y 3 2 1 x Figure A7a 1 y = ex x=1 w = ex x w = ex Figure A7b p, 3 Math 20B. Summer, 2008. Quiz 1 Solution. (7/7/08) 8. 1 1 2π 2 2 2 = (0) − (−1) = π π π [Average value] = π /2 sin x dx = 0 2 − cos x π π /2 = 0 2 [− cos π 1 2π − 2 [− cos (0)] π ...
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Quiz_1_Solution - Math 20B (Shenk). Quiz 1 Solution. July...

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