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Math 142, Exami, Fall 2009 Solutions Write your answers as legibly as you can. There are 11 problems on 6 pages. Problem 1 is worth 10 points. Each of the other problems is worth 9 points. SHOW your work. Make your work be coherent and clear. Write in complete sentences whenever this is possible. | CIRCLE your answer. CHECK your answer whenever possible. No Calculators or Cell phones. I will post the solutions on my website a few hours after the exam is finished. 1. Define the definite integral. Give a complete definition. Be sure to explain all of your notation. Let f(x) be a function defined on the closed interval a < x < b. For each partition P of the closed interval [a,b] (so, P is a=2p <2, <---
f(x7)A;. i=l 2. State both parts of the Fundamental Theorem of Calculus. Be sure to explain all of your notation. Let f be a continuous function defined on the closed interval [a, 5] . (a) If A(a) is the function A(x) = f” f(¢)dé, for all x € [a,b], then A(x) = f(z) for all a € [a, 6]. (b) If F(z) is any antiderivative of f(a), then ie f(z)dz = F(b) — F(a). 3. Find [ secos(a?)de. Check your answer. Let u=a?. So, du =2zedz. The original integral is equal to 5 5. we | O cceniel 5 | cosudu = 5 sinu+C = 9 sine )+C. Check: The derivative of the proposed answer is 320 cos(x?) Vv. 7 i 2 2 q dy 4. Find / eT . Check your answer. We do a change of variables. Suppose that the expression under the radical was just a variable, then we could do algebraic tricks to finish the problem! We make it so! Let u = 3— 4y. It follows that du = —4dy. The original problem is equal to (34)? du _ 1 ff (B—u)2du / —4Ju “Say vu 2 _ = f (9- ae jdu _ =a fo —6ul/? 4. u9/)du ak (902 - ou? 4. wr?) 40 fl —64 a ~ @ptls ee a, 3/22 4p 8/22 = =a (98 y)'/2 — 6(3 — 4y) 75 + (3 - 4y)°?=) +0 Check: The derivative of the proposed answer is 1 Haq (98 — 49) 27-4) - 6(8 - 4y)*7(~4) + (8 — 4y)/(-4)) pio | es (9 — 6(3 — 4y) + (3 — 4y)”) iE 1 y = == (9 - 18+ 24 — 24y + 16y?) = 16y*) = v. Teja ay (0 ~ 18+ Bay + 9— Day + 16y?) = Tae (16y") = Ge @ 5. Find i —>—— dz. Check your answer. V1—4z4 * We plan to maneuver the given integral into the form =arcsinu+C. Let u = 227. It follows that du = 4adz. The original problem is equal to 1 di iL 1 . Z / eee = qaresin u +0= ri arcsin(2z7) + C. Check: The derivative of the proposed answer is 1 1 oa 4” /1— 222 4 h=«?. The volume of the solid is: 1 1 3 A ar f (4-a)a?dx = an f (42? = 2°) dx = 20 (F = +) 0 0 3 4 1 4 1 = on (5-3): a 3.4 9. Consider the region in the first quadrant bounded by y = e” and x =1n3. Revolve this region about the z-axis. Find the volume of the resulting solid. 4 Spin the rectangle. Get a disk of volume mr?t where r =e? and t = dx. The volume of the solid is . " In3 2x \Ins 21n3 rf dr = ne =a § as =i eee =[47|. 6 2 2 2° 2 2 Io 10. Consider the solid whose base in the first quadrant of the zy plane is bounded by y = x? and y = 1. Each cross section of this solid perpendicular to the y-axis is a square. Find the volume of the resulting solid. 2 % The slice with y-coordinate y has volume side?t where t = dy and the side is the x-coordinate of the point on y = x? where the y-coordinate is y. So the side is \/y. The volume of our slice is Vy dy = ydy. The volume of the solid is 1 2 y dys [vis 2 11. Suppose that a conical tank is filled with oil which has a density of 50 lb/ft? . The radius at the top of the tank is 5 ft and the tank is 15 ft high. How much work is done in pumping the oil over the edge of the tank? 4 =|1/2] 1 0 : distance Be sure to draw and label an axis. We have chosen the y-axis to point upward with y = 0 being the bottom of the tank. The work to lift the layer of oil with y-coordinate y is Force distance = weight - distance = volume: density - distance = rt-50- distance, where t = dy, r is computed from similar triangles z = 3 : vj So, r= ay . The distance to lift the layer of oil whose y-coordinate is y is 15—y. The work to lift one layer of oil is: m(5y)50(15 — y)dy. 
