Analisi Matematica 3 Esercizi

Serie di Fourier

f(x) = x³
z₀ = 1/π ^π∫ x³ dx - 1/π ∫_π/-π [ x⁺⁴ / π ]^π_-π = 0

c_n = ∫_-π^π x³ cos (nx) dx = ∫ x³ sin (nx) / n - ∫ 3x x cos (nx) / n dx + 1/π
= -1/π [ x³ sin (nx) / n - 3x² sin (nx) / n² - ∫ -π_π 6x sin (nx) / n² dx ]
== -1/π [ x³ sin (nx) / n - 3x² sin (nx) / n² + ∫_-π^π 6x sin (nx) / n² dx ]
== -1/π [ x³ sin (nx) / n - 3x² sin (nx) / n² + (6x sin (nx) / n³) - ∫_-π^π 6 cos (nx) / n³ dx ]
== -1/π [ x³ sin (nx) / n - 3x² sin (nx) / n² - 6x cos (nx) / n³ + ∫_-π^π 6 cos (nx) / n³ dx ]

Serie di Fourier
f(x) = x³
z₀ = ¹/_π ∫_-π^π x³ dx = [^π/_π(^x4/₄)]_-π^π = [^π4/₄ - ^π4/₄] = 0

c_n = ¹/_{π ∫-π^π x³ cos(nx) dx = [x³ sin(nx)/n]-π^π - 3x² ∫-π^π cos(nx)/n dx
= [^-1/π { [x³ sin(nx)/n - 3x² sin(nx)/n²}- ∫-π^π 6x sin(nx)/n² dx]]
= [^-1/π { [x³ sin(nx)/n - 3x² sin(nx)/n²} - ∫-π^π 6x sin(nx)/n² dx]
= [^-1/π [x³ sin(nx)/n - 3x² sin(nx)/n² + (6x cos(nx)/n³]-π^π + ∫-π^π 6 cos(nx)/n³ dx]
= [^-1/π [x³ sin(nx)/n - 3x² sin(nx)/n² - 6x cos(nx)/n³ + ∫-π^{π 6 cos(nx)/n3 dx]}= 0}

Curva e integrali

ϕ(t) = (t, cos(t), cos₂(t))
Dϕ(t) = (1, -sin(t), -2 sin(t) )
∫₀^2π ||Dϕ(t)|| dϕ = ∫₀^2π √1+(-sin(t))² + (-2 sin(t) )² == ∫₀^2π √1+ sin²(t) + 4 sin²(t) = ∫₀^2π√1+1-cos²(t)+4-4cos²(t)

Curva γ(t) = (5cos(t), 5sin(t), -sin(5t)) con 0 < t < 2π
Dγ(t) = (-5sin(t), 5cos(t), -5cos(5t))
|Dγ(t)| = ∫₀^2π √25sin²(t) + 25cos²(t) + 25cos²(5t) dt == ∫₀^2π √25 - 25cos²(t) + 25cos²(t) + 25cos²(5t) = ∫₀^2π √25 + 25cos²(5t) ≈ 38,2

Equation and transformation

∫_τ ln (3x + y) dx dy
OA: y - 2 = x - 1 → y = 2/3x - 2x = 0 2x - y = 0
OB: y - 1 = x - 2 → -2y + 2x - x = 0 x - 2y = 0 ∆ x = 4 - 1 = 31 - 4 = -3 → -x + 2y → 2x + 3y = 3

• u = 2x - y
• v = -x + 2y
• 2x = u + y
• y = (u + 2v)-v + 2y = u + y
• y = (u + 2v)/3

J = |1 2/3| |2/3 1/3|= 1/3
∫³₂ ∫^3-v₀ ln ((6u + 3v + u + 2v)/3) du dv - ∫³₀ ∫ ln (7u + 5v) dx - ln (3) du dv

l'(x) = ln (7u + 5v)
g'(x) = 7g(x) = 7u + 5v
∫(7u + 5v) ln (7u + 5v) - ∫(7u + 5v)/(7u + 5v) - [∫(7u + 5v) ln (7u + 5v)]³₀ - 1 [ (7u + 5v) ln(7u + 5v) ]^7-v - [U]^7-v [ (7u + 5v) ln(7u + 5v) - 5v ln(5v) ]₃ [ (7 - 7v + 5v) ln(7 - 7v + 5v) ] - 5v ln(5v) ] - 3 + v [ (7 - 2v) ln(7 - 2v) - 5v ln(5v) ] - 3 + v - 3 ln(3) + v ln(3) g(x) ln(7 - 2v)
g'(x) = x - 2 - 2v f(x) ln(x) x^-1 - 2x f(x) = ln(x)

• g'(x) = xf'(x) = ¹⁄_x
• g⁸(x) = ^x⁄₂¹⁄₁₄

∫₁²⁷ x ln(x)[ x ln(x) ]₁⁷⁵ x dx 1/14 [ x ln(x) ]₂₁²⁷ x dx [ x ln(x) ]₂₇²¹ x dx [ 27 ln(27) - 15 ln 75 ] - 27 x 75¹⁄₁₄ ln(27) - 15 ln(75) - 27 + 5 = ¹⁄₂₄¹⁄₂₈ [ x² ln(x) ]₁²² x dx 1/28 [ 441 ln(27) - 725 ln(75) ] - ⁷⁴⁷¹⁄₂ ²⁸⁵⁵⁄₂ x 2 x 23/6 5v ln(5v) 5v = x f(y) = ln(x) g'(x) = xf'(x) = ¹⁄_x g'(x) = ^x⁄₂

∫₀⁷⁵ x² ln(x) ∫₀⁷⁵ x dx -₁/₇₀ [225 ln (15)] - [^x/₂] ¹⁵₀ = -₁/₇₀[225 ln (15)] - [^{225 5}/₂]) = -7,09-∫³₀ 3 = - 3 []³₀ = -3 . 3 = -9

∫vdv = ∫³₀ ^v2/₂ dv = [^v2/₂] ³₀ = ⁹/₂
∫₀³ 3 ln (3) dv = 3 ln (3) ∫₀³ v dv = ¹/₃[2 ln (3)∫₀³ vln (3) = ⁹/₂ ln (3)92 - 7,09 - 9 + ⁹/₂ - 9 ln (3) + ⁹/₂ ln (3) = -⁵⁴⁶/₃, 182

∫ ln (3x+4y) dx
T = {0, A(4, -1), B (3, 2)}
OA: y + 1 = _x/⁴→ 4y = x - 4→ x + 4y = 0
OB: y - 2 = _{x - 3}/³→ 3y - 6 = 2x - 6→ 2x - 3y = 0x + 4y = 3 + 8 = 11 √2x - 3y = 8 + 3 = 11 √u = x + 4y → {x = u - 4y}v = 2x - 3y → {x = _{v + 3y}/²}v + 3y = 2u - 8y4y = 20 - v{y = 2u - v}{y = _{2u - v}/¹¹}(y = _{2u - v}/¹¹ 3 →) {x = _{u + 3v}/²}y = _{2u - v}/¹¹x = _{4v + 3u}/¹¹

J = |₂/¹¹ - ₁/¹¹| |₈/¹¹ + ₃/¹¹ - ₁/¹¹||₃/¹¹ 4 ||₁₁ |∫ ₀^11-v ∫ ₀^11-u ln (_{3u + 12 v + 2u - v}/¹¹) dudv = ₁/¹¹ ∫ ₀^11-u ln (14u + 11v) - ln (11) dudv

∫ ₀^11-u ln (14u + 11v) dvf(x) = ln(x)g'(x) = 14f'(x) = ₁/^{14u + 11v}g*(x) = 14u + 11v∫₀^11-u [(11v + 11u)(ln (14u + 11v))]∫₀^11-u(14u + 11v)(ln (14u + 11v)) dv - ∫₀∫_v dv - {|(121 - 33u + 33u)(ln(321 - 33u + 11v))|}= {*|∫₀¹¹ ∫₀ 2u ln (121) - 11 + v dv|}∫ ln(t+1) du = -ln(t+1)∣₀^11-v u du = -ln(t+1)(11-v) = -11ln(t+1) + vln(t+1) ∫₀¹¹ 11 ln(t+1) - 11 + v + 11 ln(t+1) + v ln(t+1) du ∫₀¹¹ v du - ∫₀¹¹ v²/₂ = ∫₀¹¹ v²/₂ - ∫₀¹¹ 11²/₂ g'(x) = v g(x) = v²/₂

∫ 11 ln(t+1) dv v ln(t+1) dv f(x) = ln(t+1) f'(x) = ¹/_t+1 ∫_a¹¹ [v²/₂ ln(t+1)] - ∫₀¹¹ -v²/₂ dv = [11²/₂ ln(t+1)] - [-11/₆ v²/₂]¹¹₀ = -¹¹/₆ + 11²/₂ ln(t+1) - 11/₂∫ ln(3x+y) dx dyOA: OB: x + 4y3x + y = 02x - 3y = 0u = x + 4yu = 2x - 3yJ = ₂/^{11 4}/₁₁V + 3y = 2U - 8y1/11 ∫ ₁¹¹ ln(8u + 12v + 2u - v)/11g'(x) = 1l'(x) = 1/u + vg(x) = u + v[-x + v]∫ ¹¹[(₀) - 17] ln(vt)]12 ln(t_^t) = t 21 ln(x)- 17t - 11∫₀^v ln (v)f'(x) = ln (v)g'(x) = vf(x) = ¹/_vg(x) = ^v/₂[^v2/₂ ln(v)]^v₀ - ∫₀^{v v}/₂ = ^v2/₂ ln (v) - ^v2/₂¹/₁₀ [1² ln (v) - 1²/₂ ln (v) + ¹¹/₂ 11 + ¹¹/₂ ] = 11 ln (v) - 11 ln (v+1)= ¹¹/₂ ln (v-1) + ¹¹/₄ + ¹¹/₂ - v∫₁⁴ ln(2x+3y)T = { 0,^A (1,3),^B (2,2) }O^A y/3=x-1/x -y+3= -3x+3=3x-y=0 4√O^B y/2=x-2/2 -2y+x=2x+x=2x-4y=0 -2x+2y=04√U=-3x-yV=-2x+2yx=u+y/3x=u+y/3x=-u+4y2u+4y = 3v+6y4y=3v-2uy=-3v-2u/4x=4u-3v-2u/122u-3v/12U=3x-yV=-2x+2yy=3x-uy=v+2x/2x=v+2uy=3v+2u/4J =| -2 4 | 1/4| 3 4 | 6/16 = 4/16 = 1/4∫₁⁰ ln(4u+2v+2u+3v/4)= (1/6)∫₀⁴ 6ln(6u+5v)-ln(x)[ (6u+5v) ln(6u+5v) ]^4-v_-v - [(24-6v+5v) ln(24-v) - ( 5v) ln(5v) -4+v](1/24)∫₀⁴ (2x-v) ln(2x-v) dv = ∫₂₀²⁴ f(x) ln(x) dx = -∫₂₄²⁰ (x) ln(x) dvl(x)=ln (x)l'(x) = 1/xg⁻¹(x) = xg⁻¹(x) = x²/2[₃/² ln(x)]₂₀²⁴ - ∫₂₀²⁴ x/₂ dx576/₂ ln(24) - 280 ln(20) - [_x²/⁴]₂₀²⁴576/₂ ln(24) - 200 ln(20) - 576/₄ + 200228 ln(24) - 200 ln(20) - 200 + 100

∫₀⁵v ln(5v)f(u) = ln(5v)g"_(u) = 5g(x) = _5v²/²∫₀^5v/₂ 5u/₂(∫₀⁵v ln(5v))∫₀⁴[40 ln(20) - ₅/ln(^v2/₂)]⁴-40 ln(20) - - 201/24 [228 ln(24) - 200 ln(20) - 4x - 40 ln(20) + 20 - 16 + 8 - 16 ln(x) + 8 ln(x)]19/₂ ln(x) - 85/₃ ln(20) - 4/₃ - 5/₃ ln(20) - ₇/³ ln(1) + 1/₃ ln(2)-1,56

ϕ(t) = (t cos(t), t sin(t)) t ∈ [0, 10π]

Dϕ(t) = (-tsin(t)+cos(t), tcos(t)+sin(t))
∫ ||Dϕ(t)|| = 0^10π ∫ √

∫₀^∞ t² + 1 dt ≈ 495
γ(t) = (t²cos(t), t²sin(t)) t ∈ ]0, 2π]
Dγ(t) = (2t cos(t) - t²sin(t) , 2t sin(t) + t²cos(t))

||Dγ(t)|| = ∫₀^2π √ 4t² cos²(t) + t⁴ sin²(t) * 2t³ sin(t) + 4t² sin²(t) + t⁴ cos²(t) + t⁴
= ∫₀^2π √ 4t² cos²(t) + t⁴ sin²(t) + 4t² sin²(t) + t⁴ cos²(t)
= ∫₀^2π √ 4t² + t⁴

∫ ln (x - y) dx dy
OA: y = -1/2 x + 4: x = -6/6 → -6y + 8 = -x + 8 → x - 6y = 0 → x + 6y = 8
OB: y = - x/2 + 4 → -4y + 8 = -2x + 8 → 2x - 4y = 0 = 8

• u = -x + 6y
• {x = -u + 6y
• {x = -u + 6y
• v = 2x - 4y
• {x = v + 4y/2
• {-2u + 12y = v + 4y
• x = 40 + 6v/8y = v + 20/8

J = | 4/8 6/8 | | 2/8 1 |= 4/64 + 12/64 - 8/64 = 1/81/8
∫₀^8-v ∫₀ ln (4v + 6v - v + 2v/8) du dv - 1/8 ∫₀^8-v ln (2u + 5v) -ln (8) du dv

1/2∫₀^8-v z ln (2u + 5v) du
f (x) = ln (2u + 5v)
f' (x) = 2/2u + 5v
g' (x) = 2g (x) = 8u + 5v

• {[(2u + 5v) ln (2u + 5v)]₀^8-v -∫₀^8-v v u du
• { [ (8 - 2u + 5v) ln (16 + 3v) ] 0 -∫₀^8-v (5v) ln (5v)] - 8 + v-8 ln (8) + v ln (8)
• { (16 + 3v) ln (16 + 3v) dv = 1/36∫₀¹⁶ (x) ln (x) dx f (x) = ln (x) f ' (x) = 1/x

{ - ∫₁₆⁴⁰ ln (x){ 1/6 [1600 ln (40)-256 ln (16)] - 1/6 [x²/2]⁴⁰₁₆x²[1600 ln (40) - 256 ln (8)] - ¹⁶⁰⁰/₁₂ + ²⁵⁶/₁₂

∫₀⁸ 5v ln (5v) dv = F(x) = ln (5v)   F'(x) = ¹/₅v   g'(x) = 5v g(x) = ^v2/₂⁵/₂ [∫₀^v2/₂ (ln (5v)^v)⁸ dv = 160 ln (40) - ⁵/₂ [∫₀^v2/₂]⁸160 ln (40) - 80[¹⁶⁰⁰/₆ ln (40) - ²⁵⁶/₆ ln (16) - ¹⁶⁰⁰/₁₂ + ²⁵⁶/₁₂ = 260 ln (40) + 80 - 64 + 32 ln (8)]

+ 32 ln (8) ] ⁷/₁₆ = 9,04
φ(t) = (5 cos(t), sin(3t), 5 sin(t) + cos(3t)) t ∈ [0, 2π]

D_Y(t) = (-5 sin(t), 3 cos(3t), 5 cos(t) - 3 sin(3t))
|D_Y(t)| = ∫₀^2π √(25 sin²(t) + 9 cos²(3t) + 30 sin(t) cos(3t) + 25 cos²(t) + 9 - 36 cos(t) sin(3t)) dt
= ∫₀^2π √(25 - 25 cos²(t) + 15 sin²(t) + 45 sin²(2t) - 9 - 9 - 25 cos²(t) - 15 sin²(4t)) dt
= ∫₀^2π √(25 + 15 sin²(t) + 3 sin²(3t) - 9 - 15 sin²(t)) dt
= ∫₀^2π √(34 - 30 sin²(t)) - √(36 = ∫₀^2π √(1 - 30 sin²(t)) dt = 34,31