Spring 2006 - Nagy's Class - Exam 1

Spring 2006 - Nagy's Class - Exam 1 - Name Section Number...

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Unformatted text preview: Name: Section Number: TA Name: Section Time: Math 20B. Midterm Exam 1 April 28, 2006 No calculators or any other devices are allowed on this exam. Write your solutions clearly and legibly; no credit will be given for illegible solutions. Read each question carefully. If any question is not clear, ask for clarification. Answer each question completely, and show all of your work. 1. (10 points) Evaluate the following integrals. π/3 (a) 0 (b) sin(θ) dθ cos2 (θ) 1+x dx 1 + x2 (a) Substitute u = cos(θ), so du = − sin(θ) dθ, and u(0) = 1, u(π/3) = 1/2. Then π/3 0 sin(θ) dθ = cos2 (θ) 1/2 1 − du = u2 1 u−2 du = 1/2 1 (−1) u−1 1 1/2 = −(1 − 2) = 1. (b) 1+x dx = 1 + x2 1 dx + 1 + x2 x dx 1 + x2 Integrate the first term directly, and in the second term do the substitution u = 1 + x 2 , with du = 2x dx, then 1+x dx = arctan(x) + 1 + x2 1 du , 2 u 1 ln(u) + c, 2 = arctan(x) + ln( 1 + x2 ) + c. = arctan(x) + # 1 2 3 4 Σ Score 2. (10 points) Find the area of the shaded region y sin(x) 1 cos(x) pi/2 x Split the integration region in two intervals, [0, π/4] and [π/4, π/2]. Then, the area A of the shaded region is π/4 A = π/2 sin(x) dx + 0 cos(x) dx, π/4 π/4 π/2 − cos(x)|0 + sin(x)|π/4 , √ √ 2 2 −1 + 1− , = − 2 2 √ = 2 − 2. = 0 V = h Then, ab 2 ab 1 y dy = 2 h3 2 h h 3 A(y) = Then x(y) = ⇒ V = 1 abh. 3 ab 2 y . h2 (a/2) y. h Analogously, the x(y) function is a line given by z(y) = (b/2) y. h Concentrate in the x = 0 plane, then the top of the pyramid is a line in the zy-plane, passing through the origin, so the function z(y) is given by This area is given by A(y) = 2z(y) 2x(y). ¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦§ § § § § § § § § § § § § § § § § § § § § § § § § § § § § § § ¦§¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤ ¦§¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¦¦ § § § § § § § § § § § § § § § § § § § § § § § § § § § § § § § ¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦¤¦§ ¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤¦§¤ 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(10 points) Find the volume of a pyramid with rectangular base of sides a and b, and height h. 4. (8 points) Compute both (1 + i)8 and (1 + i)10 . z = 1+i ⇒ √ √ r = 1 + 1 = 2, tan(θ) = 1 ⇒ θ = π/4. z = 21/2 [cos(π/4) + i sin(π/4)] ⇒ z 8 = 28/2 [cos(8π/4) + i sin(8π/4)] = 24 [cos(2π) + i sin(2π)] = 16. (1 + i)8 = 16. z = 21/2 [cos(π/4) + i sin(π/4)] ⇒ ⇒ ⇒ z 10 = 210/2 [cos(10π/4) + i sin(10π/4)] = 25 [cos(5π/2) + i sin(5π/2)] ⇒ z 10 = 32[cos(π/2) + i sin(π/2)] = 32 i. (1 + i)10 = 32 i. ⇒ ...
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This note was uploaded on 04/29/2008 for the course MATH 20B taught by Professor Justin during the Spring '08 term at UCSD.

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