Week #13 - Integration by Parts & Numerical Integration Section 7.2

Transcription

1 Week #3 - Inegraion by Pars & Numerical Inegraion Secion 7. From Calculus, Single Variable by Hughes-Halle, Gleason, McCallum e. al. Copyrigh 5 by John Wiley & Sons, Inc. This maerial is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS. Inegrae by pars, selecing: u = dv = sin d so du = d v = cos udv = uv vdu, sin d = cos ( cos ) d = cos + sin + C 3. Inegrae by pars, selecing: u = dv = sin d so du = d v = cos sin d = cos ( cos ) d = cos + cos d This new inegral is simpler (he power of wen from squared o linear), and if we inegrae by pars again, we should ge a solvable inegral: selecing: u = dv = cos d so du = d v = sin [ cos d = sin ] sin d = [ sin + cos ] + C Combining his wih he previous resul, he final inegral evaluaes o sin d = cos + sin + cos + C 9. To inegrae by pars, we mus selec u = ln x, because we don know how o inegrae i: selecing: u = ln x dv = x 3 dx so du = x dx x v = x 3 ln x dx = x x ln x x x dx = ln x x 3 dx = x ln x 6 x + C

2 7. Since everyhing seems o be in erms of (θ + ), i migh help o do a subsiuion firs: Le w = (θ +, so dw = dθ, and (θ + )sin(θ + ) dθ = w sin(w)dw This we can evaluae by inegraion by pars, selecing: u = w dv = sin w dw so du = dw v = cos w w sin(w)dw = w cos w ( cos w)dw = w cos w + sinw = (θ + )cos(θ + ) + sin(θ + ) + C. Neiher subsiuion nor inegraion by pars looks o helpful a firs glance. When possible, i can help o separae sums in he numeraor o obain wo simpler inegrals: + 7 d = 5 d 5 }{{} I d } {{ } I The second inegral, I, can be done by subsiuion, or by guess and check: 7 d = 7 5 (5 ) / = (5 ) / + C The firs inegral, I, is more complicaed, bu wih wha we jus found, we should be able o use inegraion by pars: I = d = (5 ) / d 5 selecing: u = dv = (5 ) / d so du = d v = (5 ) / I = (5 ) / d = (5 ) / ( (5 ) / ) d = (5 ) / 3 (5 )3/ + C Combining I and I, and combining he consans ino C, we obain + 7 d = I + I = (5 ) / (5 )3/ (5 ) / +C 5 }{{ 3 }}{{} I I

3 5. This looks like a classical subsiuion problem, given he x inside he arcan, and is derivaive, x ou fron. Le s sar wih ha subsiuion: Le w = x, so dw = x dx, so xarcan x dx = arcan w dw However, we only know he derivaive of arcan, no is inegral, so we need o use inegraion by pars, selecing: u = arcan w dv = dw so du = +w v = w arcan w dw = (w arcan w w + w dw) This new inegral can also be solved by subsiuion, since he derivaive of he denominaor will involve a w erm. Le z = + w, so dz = wdw: w + w dw = z dz = ln z = ln( + w ) Combining all he pars, and hen subsiuing back ino he original x variable, xarcan x dx = (w arcan w w + w dw) = w arcan w ln( + w ) + C = x arcan x ln( + x ) + C 35. We only know how o differeniae ln, so rying inegraion by pars makes sense, selecing: u = ln( + ) dv = d so du = + d v = 5 ln( + ) d = ln( + ) + d 3

4 The new inegral can be easily solved wih he subsiuion w = +, so = w and dw = d: w w + d = w dw = w dw w dw = w ln w + C = + ln( + ) + C so ln( + ) d = ln( + ) + d = ln( + ) ( + ln( + ) + C) Wih he limis, 5 = ( + )ln( + ) + C if we le C = C ln( + ) d = ( + )ln( + ) = (6ln 6 6) (ln() ) = 6ln QUIZ PREPARATION PROBLEMS 38. (a) This inegral can be evaluaed using inegraion by pars wih u = x,dv = sin x dx. (b) We evaluae his inegral using he subsiuion w = + x 3. (c) We evaluae his inegral using he subsiuion w = x. (d) We evaluae his inegral using he subsiuion w = x 3. (e) We evaluae his inegral using he subsiuion w = 3x +. (f) This inegral can be evaluaed using inegraion by pars wih u = x,dv = sin x dx. (g) This inegral can be evaluaed using inegraion by pars wih u = ln x, dv = dx. 39. You should have some sense of a skech of his graph: i will look like he sine graph, bu he ampliude will be growing wih linearly x because he sin(x) is muliplied by x. Because i is a produc of x and sin(x), he funcion will sill have y = values, crossing he x axis, a he same poins when sin(x) =. The firs arch of he curve is herefore beween x = and x = π, as hese are he firs wo poins when f(x) =. This means he area we re looking for will be represened by he inegral To evaluae his, we ll use inegraion by pars: selecing: u = x dv = sin x dx so du = dx v = cos x π π x sin(x) dx = x( cos x) π π cos(x) dx = x cos(x) + sin(x) π = [( π cos(π) + sin(π)] [( cos() + sin()] = π( ) = π xsin(x) dx.

5 7. We inegrae by pars. Le u = x n and dv = cos ax dx, so du = nx n dx and v = a sin ax. Then x n cos axdx = a xn sin ax (nx n )( sin ax) dx a = a xn sin ax n x n sinax dx. a 5. Since f (x) = x, inegraion by pars ells us ha f(x)g (x)dx = f(x)g(x) f (x)g(x)dx = f()g() f()g() xg(x)dx. We can use le and righ Riemann Sums wih x = o approximae xg(x)dx: Lef sum g() x + g() x + g() x + 6 g(6) x + 8 g(8) x = ((.3) + (3.) + (.) + 6(5.5) + 8(5.9)) = 5.6 Righ sum g() x + g() x + 6 g(6) x + 8 g(8) x + g() x = (3.) + (.) + 6(5.5) + 8(5.9) + (6.)) = A good esimae for he inegral is he average of he lef and righ sums, so xg(x)dx Subsiuing values for f and g, we have = f(x)g (x)dx = f()g() f()g() xg(x)dx (6.) (.3) (66.6) =