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MATH 16200-33: HONORS CALCULUS-2 HOMEWORK 7: SOLUTIONS Problem 19-3. Integrate by parts. Z 2 (ii) x3 ex dx. Solution. Using the substitutions u = x2 , du = 2x dx, 2 dv = xex , v= 1 x2 e , 2 we have Z 3 x2 x e 2 1 dx = x2 ex − 2 Z 2 xex dx = 1 x2 2 e x −1 . 2 Z (viii) cos(log x). Solution. Let I = R cos(log x). Using the substitutions u = cos(log x), du = − dv = dx, sin(log x) dx, x v = x, we have Z I = x cos(log x) + sin(log x) dx. Applying the substitutions u = sin(log x), du = dv = dx, cos(log x) , x v = x, we compute Z Z sin(log x) dx = x sin(log x) − cos(log x) = x sin(log x) − I. Solving for I yields I= x cos(log x) + x sin(log x) . 2 1 Z (x) x(log x)2 dx. Solution. Using the substitutions u = (log x)2 , du = 2 log x dx, x x2 , 2 Z Z 1 2 2 2 x(log x) dx = x (log x) − x log x dx. 2 dv = x dx, we have v= Applying the substitutions u = log x, dv = x dx, du = dx , x v= x2 , 2 we compute Z 1 1 x log x dx = x2 log x − 2 2 Z x dx = 1 2 x2 x log x − . 2 4 Thus Z 1 x(log x) dx = x2 (log x)2 − 2 2 = 1 2 x2 x log x − 2 4 1 1 x2 + x2 (log x)2 − x2 log x . 4 2 2 Problem 19-4. Integrate using trigonometric substitution. Z dx √ (ii) 1 + x2 Solution. Using the substitution x = tan u, dx = sec2 u du: Z Z Z dx sec2 u du sec2 u du √ √ √ = = 1 + x2 1 + tan2 u sec2 u Z = sec u du = log(sec u + tan u) = log(sec(tan−1 x) + tan(tan−1 x)) p = log( 1 + x2 + x) . 2 Z (iv) dx x x2 − 1 √ Solution. Using the substitution x = sec u, dx = sec u tan u du: Z Z Z Z sec u tan u du sec u tan u du dx √ √ √ = = = du = sec−1 x . x x2 − 1 sec u sec2 u − 1 sec u tan2 u (viii) Z p 1 − x2 dx Solution. Using the substitution x = sin u, dx = cos du: Z p Z p Z 2 2 1 − x dx = 1 − sin u cos u du = cos2 u du Z 1 + cos 2u 1 sin 2u = du = u+ 2 2 2 −1 sin(2 sin x) 1 sin−1 x + = 2 2 2 sin(sin−1 x) cos(sin−1 x) 1 −1 sin x + = 2 2 p 1 = sin−1 x + x 1 − x2 . 2 Problem 19-5. Integrate. Z dx (ii) 1 + ex Solution. Using the substitution u = 1 + ex , du = ex dx: Z Z Z dx (1 + ex ) − ex ex = dx = 1 − dx 1 + ex 1 + ex 1 + ex Z Z ex du =x− =x− 1 + ex u = x − ln(u) = x − ln(1 + ex ) . Z (iv) √ dx 1 + ex 3 Solution. Using the substitution u = Z dx √ = 1 + ex Z 2 du = ex Z 2 du = u2 − 1 √ 1 + ex , du = Z x √e 2 1+ex 1 1 − u−1 u+1 dx, √ u − 1 1 + ex − 1 . = ln √ du = ln u + 1 1 + ex + 1 Problem 19-26. Let φ be a nonnegative integrable function such that φ(x) = 0 R1 for |x| > 1 and such that −1 φ = 1. For h > 0, let φh (x) = 1 φ(x/h). h (a) Show that φh (x) = 0 for |x| ≥ h and that Rh φh = 1. −h Proof. If |x| ≥ h then |x/h| ≥ 1, so φ(x/h) = 0, and φh (x) = 1 φ(x/h) = 0. h Using the substitution u = x/h, du = dx/h: Z h φh (x) dx = −h 1 h Z h Z 1 φ(x/h) dx = −h φ(u) du = 1. −1 (b) Let f be an integrable on [−1, 1] and continuous at 0. Show that Z lim+ h→0 1 −1 Z φh f = lim+ h→0 h φh f = f (0). −h Proof. Part (a) shows that the first two integrals are equal. Now let ε > 0 be given, and choose δ > 0 so that |f (x) − f (0)| < ε whenever |x| < δ. Then for 0 < h < δ, Z Z Z h Z h h h φh f = φh f (0) − φh f = φh (f (x) − f (0)) dx f (0) − −h −h −h −h Z h ≤ φh |f (x) − f (0)| dx −h Z h ≤ε φh dx = ε −h 4 Problem 19-42. Find the value of R∞ 2 e−x dx. 0 (a) Show that 1 Z Z 0 2 4 2n · · ··· · , 3 5 2n + 1 (1 − x2 )n dx = 0 ∞ π 1 3 1 2n − 3 dx = · · · · · · · . (1 + x2 )n 2 2 4 2n − 2 Proof. Using the substitution x = cos u, dx = − sin u du: Z 1 Z 2 n 0 2n (1 − x ) dx = 0 (sin Z π/2 = π/2 u)(− sin u) du = sin2n+1 u du 0 2 4 2n · ··· . 3 5 2n + 1 Using the substitution x = cot u, dx = − csc2 u du: Z ∞ Z 0 1 −1 2n dx = (sin u) du (1 + x2 )n sin2 u 0 π/2 Z π/2 = sin2(n−1) u du 0 2n − 3 π 1 3 . = · · ··· 2 2 4 2n − 2 (b) Prove, using the derivative, that 2 1 − x2 ≤ e−x 2 1 e−x ≤ 1 + x2 for 0 ≤ x ≤ 1, for 0 ≤ x. 2 Proof. Let f (x) = e−x + x2 − 1. Then f (0) = 0 and 2 2 f 0 (x) = 2xe−x ex − 1 ≥ 0, 2 so f (x) ≥ 0 for all 0. In particular, 1 − x2 ≤ e−x for 0 ≤ x ≤ 1. For the second inequality, note that 1 + y ≤ ey for all y ≥ 0. Setting y := x2 , 2 2 e−x ≤ 1 + x2 ≤ ex , for x ≥ 0. 5 1 1 + x2 (c) Solution. We have Z 1 Z (1 − x2 )n dx ≤ 0 1 2 e−nx dx ≤ Z 0 ∞ 2 e−nx dx ≤ 0 Z ∞ 0 1 dx, (1 + x2 )n so that Z 1 Z ∞ 2 2 4 2n π 1 3 2n − 3 −nx2 · ··· ≤ e dx ≤ e−nx dx ≤ · · · · · . 3 5 2n + 1 2 2 4 2n − 2 0 0 √ √ Setting y = nx, dy = n dx: √ 2 4 2n 1 · ··· ≤√ 3 5 2n + 1 n Z n 0 2 1 e−y dy ≤ √ n Z ∞ 2 e−y dy ≤ n 0 π 1 3 2n − 3 · · ··· . 2 2 4 2n − 2 Thus √ 2 4 2n n · ··· ≤ 3 5 2n + 1 √ Z n 2 e−y dy ≤ Z 0 ∞ 2 e−y dy ≤ 0 √ π 1 3 2n − 3 n · · ··· . 2 2 4 2n − 2 (d) Now use Problem 41(d) to show that √ Z ∞ π −y 2 . e dy = 2 0 Proof. Problem 15-2(vi). Find lim x→0 1 1 − x sin x using l’Hôpital’s Rule. Solution. Note that 1 1 sin x − x − = . x sin x x sin x Evaluating the expression at 0 gives the indeterminate 0/0, so we may apply l’Hôpital’s Rule. 1 1 sin x − x cos x − 1 lim − = lim = lim . x→0 x x→0 x→0 x cos x + sin x sin x x sin x Evaluation at 0 again gives the indeterminate 0/0, so cos x − 1 − sin x lim = lim = 0. x→0 2 cos x − x sin x x→0 x cos x + sin x 6 Problem 15-33. (a) Show that 1 1 x sin(k + )x − sin(k − )x = 2 sin cos kx. 2 2 2 Proof. Applying the double angle formulas, 1 1 sin(k + )x − sin(k − )x 2 2 x x x x = sin kx cos + cos kx sin − sin kx cos − cos kx sin 2 2 2 2 x = 2 sin cos kx 2 (b) Conclude that 1 sin(n + 1/2)x + cos x + cos 2x + · · · + cos nx = . 2 2 sin x2 Proof. From part (a), n 1 1 x X 1 sin(n + )x − sin = sin(k + )x − sin(k − )x 2 2 2 2 = k=1 n X 2 sin k=1 x cos kx 2 n = 2 sin xX cos kx. 2 k=1 Rearranging gives n X sin(n + 1/2)x − sin x2 = cos kx, x 2 sin 2 k=1 sin(n + 1/2)x 1 = + 2 sin x2 2 as desired. 7 n X k=1 cos kx, Problem 19-43. (b) Use Problem 15-33 to show that Z π sin(n + 12 )t dt = π. sin 2t 0 Proof. We compute, Z 0 π sin(n + 12 )t dt = sin 2t Z π 1+ 0 n X ! 2 cos kt k=1 dt = π+ n X 2(sin(πk) − sin(0k)) k=1 k = π. (c) Prove that Z lim λ→∞ 0 π 1 2 1 sin(λ + )t − dt = 0. 2 t sin 2t Proof. From Problem 15-2, the expression inside the parentheses can be extended to a continuous function on [0, π]. Apply the Riemann-Lebesgue lemma. (d) Use the substitution u = (λ + 12 )t and use part (b) to show that Z ∞ sin x π = . x 2 0 Proof. From part (c), Z π Z π sin(λ + 12 )t 2 sin(λ + 12 )t dt = 0, dt − lim λ→∞ t sin 2t 0 0 and by part (b), the limit of the latter integral as λ → ∞ exists and equals π. Hence Z π Z π 2 sin(λ + 12 )t sin(λ + 21 )t lim dt = lim dt = π. λ→∞ 0 λ→∞ 0 t sin 2t Using the substitution u = (λ + 21 )t, du = λ + 1/2 dt on the former, Z π = lim λ→∞ 0 π Z (λ+1/2)π (λ + 21 ) 2 sin(λ + 12 )t du dt = lim 2 sin u · · λ→∞ 0 t u λ + 12 Z ∞ sin u =2 du. u 0 8