Spring 2003 Math 152 03/20/03 Common Exam 2 A. Solutions 1. R x 3 x+1 dx = R x2 ¡x+1¡ 1 x+1 dx = x3 3 ¡ x2 2 +x¡lnjx+1j+CAnswer. B 2. Which of the following integrals is an improper integral? Answer. C 3. R12 1x(lnx)2 dx = ¡ 1lnxflfl12 = 1ln2 Answer. C 4. The masses m1 = 3 and m2 = 2 are located at the points P1(¡1;1) and P2(3;1) re- spectively. Since ¯x = 3(¡1)+2(3)2+3 = 35 and ¯y = 3(1)+2(1)2+3 = 1, the center of mass of the system is (35; 1). Answer. E 5. The length of the parametric curve x = cos2t; y = sin2t; …4 • t • …3; is equal to R…=3 …=4 p(2sin2t)2 +(2cos2t)2 dt = R…=3 …=4 2dt = … 6Answer. D 6. The curve y = x2 +1; ¡1 • x • 1 is revolved about the x¡axis. The surface area obtained is given by the integral Z 2…yds = Z 1 ¡1 2…(x2 +1)p1+4x2 dx Answer. E 7. The velocity v(t) (in feet per second) of a moving object is given below at one second increments. Use Simpson’s Rule with n = 4 to approximate the total distance travelled from t = 0 to t = 4 seconds. t 0 1 2 3 4 v(t) 3 2 1 -1 0 Distance travelled is equal to Z 4 0 v(t)dt … ∆x3 (f(x0)+4f(x1)+2f(x2)+4f(x3)+f(x4)) = 13(3+4(2)+2(1)+4(¡1)+0) = 3 Answer. A. 8. Consider the integral R10 ex2 dx. There is no elementary antiderivative, so the integral is approximated. Which of the approximate integration formulas underestimates the integral value? Answer. B. (Left Point Formula) 9. A curve passes through the point (0;1) and has the slope at any point (x;y) equal to y2. Which of the following points is on the curve? A) (2; 12) B) (2; 13) C) (2;¡1) D) (2;1) E) (2;e) Solution. The curve is described by the initial value problem dydx = y2; y(0) = 1. Separate the variables and integrate to get dy y2 = dx; ¡1 y = x+C Use the initial condition y(0) = 1 to obtain C = ¡1. Thus, ¡1y = x¡1y = ¡1x¡1; and hence, y(2) = ¡1. Answer. C. 10. Which of the following is the direction field for the equation dydx = ¡xy? Answer. C. 11. Solve the initial value problem dydx ¡ 1x y = xe¡x; y(1) = ¡1e. a) An integrating factor is „(x) = e R p(x) dx = e¡lnx = 1x. b) Multiplying both sides by „(x) = 1x and integrating („y)0 = e¡x, we get 1 xy = ¡e ¡x +C; y = ¡xe¡x +Cx This is the general solution of the equation. c) The particular solution satisfying the given initial condition, y(1) = ¡1e, is y = ¡xe¡x, since C = 0. d) limx!1¡xe¡x = limx!1¡ xex = 0 (use L’Hospital’s Rule). 12. Find the centroid of the region bounded by y = cos2x; y = 0; x = 0; x = …4. a) The area of the region=A = R…=40 cos2xdx = 12. b) ¯x = 1A R…=40 xcos2xdx = …4 ¡ 12 (use integration by parts with u = x; dv = cos2xdx;du = dx; v = 12 sin2x). c) ¯y = 1A R…=40 12(cos2x)2 dx = …=8 (use the power reduction formula=half angle formula, cos2x = 1+cos4x2 ) 13. Consider the function f(x) = 1x4 +x2 a) The partial fraction decomposition of the function f(x) = 1x4 +x2 = 1x2(x2 +1) has the form Ax + Bx2 + Cx+Dx2 +1 , where A = C = 0; B = 1; D = ¡1: b) R 1x4 +x2 dx = R 1x2 ¡ 1x2 +1 dx = ¡1x ¡arctanx+C c) SinceR11 1x4 +x2 dx = limt!1(¡1t¡arctant)¡(¡1¡…2) = (0¡…2)¡(¡1¡…4) = 1¡…4, the integral converges. You can also use a comparison with R11 1x4 dx or with R11 1x2 dx to make the conclusion about the convergence. But here you were asked also to compute the value of the integral, so just the Comparison Theorem is not enough. 14. A plate of a triangular shape is submerged vertically in water so that its base lies at the surface of the water. The plate’s base is 4 meters, its height is 3 meters. Find the hydrostatic force on one side of the plate. (Use the facts that the density of water is ‰ = 1000 kg/m3 and that the acceleration due to gravity is g = 9:8 m/s2.) Solution. Consider the x¡axis going down, with the origin at the surface of water. Then the depth of the slice is x, the length of the slice is y = 43(3 ¡ x) (use the similarity of the triangles), the area is y = 43(3 ¡ x)dx and the hydrostatic force isR 3 0 ‰gx 4 3(3¡x)dx = 6‰g.Comment: If you use a different coordinate system, your integral may be different, but the answer should be the same. 15. Newton’s law of cooling says that the rate of change of a body’s temperature is pro- portional to the difference between the temperature of the body and the temperature of the surrounding air. A room is kept at 20– C. A boiling pot of soup was brought into the room, and it has cooled from 100– C to 60– C in 10 minutes. Formulate the initial value problem for T(t), the temperature of the pot at time t. Solution. Since the temperature of the pot is proportional to the difference between the temperature of the pot and the temperature of the surrounding air, the differential equation for T(t) is dTdt = k(T ¡ 20) (actually, k < 0, so you may wish to write the equation as dTdt = ¡k(T ¡20); k > 0). The initial condition is T(0) = 100 (the pot is boiling, so the initial temperature is 100–).