Integration Techniques

Transcription

1 Review for the Final Exam - Part - Solution Math Name Quiz Section The following problems should help you review for the final exam. Don t hesitate to ask for hints if you get stuck. Integration Techniques. Evaluate the following integrals. a b π/ cos xdx Let u = sinx and du = cosxdx. π/ cos x cosxdx = 9x dx x π/ sin x cosxdx = u du = 8 Let x = cosθ. Then dx = sinθdθ and 9x = sinθ. sinθ sin cosθ θ cos sinθdθ = cosθ dθ = θ dθ = secθ cosθdθ cosθ = sinθ ln secθ + tanθ +C = 9x 9x ln x + x +C c e x lnxdx Use Integration by Parts: u = lnx dv = x dx du = x dx v = x e x lnxdx = e x lnx e x dx = e + 6 d x + x + dx Use the Rationalizing Substitution u = x +, udu = dx u u udu = u u du = + u du =* Use Partial Fractions: u = u + u + *= + u du = u + ln u u + u + +C = x + + ln x + +C x + +

2 Improper Integrals. Compute the integral sum of two integrals. First compute xx + dx: x + x dx. Hint: write it as the.. Now, Let u = x and udu = dx Then dx = xx + dx = lim x + x a + and dx = lim x + x b so dx = x + x b u + du = tan u +C = tan x +C. a π dx = lim x + x a + tan a = π dx = lim tan b π = π x + x b dx + x + x. x + x dx = π + π = π. 6 Volumes. The region to the right of the y-axis, above the x-axis, and below the curve y = x is revolved about the y-axis. Set up an integral that represents the volume of the resulting solid a by the method of slices discs b by the method of cylindrical shells. Then find the volume, by either method you like a x = y V = π y y dy = π ydy = π = 8π. b V = πx x dx = π x x = π8 = 8π.

3 Work. A cable hanging from the top of a building is m long and has a mass of kg. A kg weight is attached to theend of the rope. How much work is required to pull m of the cable up to the top? Let x be the distance from the top in meters. To lift the top meters of rope: g 8 x i Δx here, ρ = kg m. So Wtop = g 8 xdx = g x = g. W i where W i = F i d i = m i gd i = ρ Δxgx i = x i To lift the bottom meters of rope and the weight: W bot = Fd = mg = 8 + }{{ } cable g. Total work: + g = = Joules. }{{} weight g = Area Between Curves. Find the area of the region bounded by y = x + x and y = x +.7 between x = and x =. x + x = x +.7 gives x =.,. so they cross at x = x +.7 x + xdx =.7... x + x x +.7dx =.67 The area is.7 + =

4 Review for the Final Exam - Part - Solution Math The following problems should help you review for the final exam. Don t hesitate to ask for hints if you get stuck. Arclength and Approximation. Write an integral that computes the arclength of the curve y = e x/ between x = and x =. Use Simpson s Rule with n = subintervals to estimate the value of the integral. y = ex/ and y = ex so the integral is Do Simpson s Rule with x = /, fx = + ex dx / The actual answer is about ex dx. + ex and x =, x =, x =, x =, x = : [ ] f + f + f + f + f.668 Center of Mass. Find the center of mass of a plate with constant density that occupies the region π x, y sec x.. The area is sec x dx = tanx =. So x = M y = x sec x dx ȳ = Use Integration by Parts: u = x dv = sec x dx du = dx v = tanx x = x tanx tanx dx = π ln secx sec x dx = + tan x sec x dx. Let u = tanx so du = sec x dx. Then ȳ = The center of mass is at about.88, Net and Total Distance = π + ln.88 + u du = u + 6 u. You throw a ball straight up into the air with velocity ft/sec and catch it at the same height when it comes back down. What is the total distance traveled by the ball? =. Choose coordinates so that the ball travels along the y-axis and so that its initial height y =.

5 The acceleration of the ball, due to gravity, is at = ft/sec. The velocity is vt = at = = t + C where = v = C. The ball reaches the top of its flight when = vt = t +, or t = seconds. The distance the ball travels going up is The total distance is = feet. / t + =. Differential Equations t. Let ft be a continuous function and let a be a constant. Show that y = e at e as fs ds satisfies the differential equation dy + ay = ft. Use the Product Rule and the Fundamental Theorem of Calculus to compute Then dy t = a e at dy + ay = t e as fs ds + e at e at ft = a e at e as fs ds + ft. a e at t e as fs ds + ft + a e at t e as fs ds = ft.. An electric circuit with resistance ohms and inductance henrys is powered by a -volt battery. The current I in amperes at time t in seconds in such a circuit satisfies the differential equation + I =. Suppose that I = when the circuit is activated at time t =. a Find the current I at all times t >. = 6 I 6 I = 6 I = ln 6 I = t + C So 6 I = A e t where A = ±e C. Plug in t = I = to get 6 = A and I = 6 b Find the limiting value of I as t. 6 lim e t = 6 =. amps. t c After what time is the current within. ampere of its limiting value? e t. Solve. = It = 6 e t to get t = ln.9698 seconds.