Analisi Matematica 3 Esercizi

∑ l^x+3 (2m+x) / m+1 = ∑ (2m‑ x) l^2m l^x ∑ y^x+3 / m+1

l^x+3 / l = l^x+3

∑ y^m = ∑

l ∑ (2m+x) / 2l^x+3

∑y^m+1m = ∑ (2m+x)

l^x+3 ∑ l^inx 2m_m

l^{x3 (-ln(1‑ ex1))}

SERIE DI FOURIER

f(x) = x³

z₀ = 1/π ∫ π³ dx = 1/π [ x⁴_π ‑ π^π/4] = 0

c_n = 2/π ∫ π x³ cos(n_nx) dx = ∫ x³sin(nx) ‑ 3x sin(nx) dx x 1/π =

= 1/π { x³ sin(nx)_n

∫ y^x+3 y^m

= 1/π { x³ sin(nx) ‑ 3x² sin(nx)_n² ‑ 6x sin(nx)_n² dx } =

= 1/π { x³ sin(nx) ‑ 3x² sin(x)_n² + [ 6x cos(x)ⁿ_m³ + 6 cos(x)n^m3 dx }

= ‑ 1/π { x³ sin(x)_m ‑ 3x² sin(x)_m² ‑ ( 6x cos(x)ⁿ³ + ∫ [ 6 cos(x)^m3 dx }₀

ψ(t) = (t, cos(t), cos(2t))

Dψ(t) = (1, −sin(t), −2 sin(2t))

∫₀^2π ||Dψ(t)|| dt = ∫₀^2π √1 + (−sin(t))² + (−2 sin(2t))² =

∫₀^2π √1 + sin²(t) + 4 sin²(2t) = ∫₀^2π √1 + 1−cos²(t) + 4 − 4cos²(2t)

-₇∕₇₀[225 ln(15)] - [^x²∕₂]₀¹⁵ = -₇∕₇₀[225 ln(15)] - [^225×5∕₂]= -7,09

Svolgendo

∫₀³ -3 = -3 ∫ [v]₀³ = -3 × 3 = -9

∫ v dv = [^v²∕₂]₀³ = ⁹∕₂

∫₀³ 3 ln(3) dv = 3 ln(3) ∫ v dv = ⁹∕₂ ln(3)

∫₀³ v ln(3) = ⁹∕₂ ln(3)

92 -7,09 -9 + ⁹∕₂ ln(3) - ⁹∕₂ ln(3) = ⁵⁴⁶∕₃ 1,82

ln(2x+y)

T= { 0, (1, 3), (2, 2) }

y-3/3 = x-1/-1 - y + 3 = - 3x + 3 = 3x - y = 0 ✔

y - 2x + 2y + x - z + x - 2 = -2x + 2y + x - z = x+z + 2 - 2

- y = -3v - 2u/4

J =

1/4 ∫₀¹ ln(4u + 2v/2u + v + 3v/4)

= (1/6) ∫₀⁶ 6 ln (6u + 5v) - ln(x)

[(6u+5v) ln(6u + 5v)|_u⁰

4∫₀²(2x+v) ln(2x-v)

l(x)=ln(x)

l'(x)=1/x

g'(x)=xg(x)=x²/2

∫ [600 ln (40) - 256 ln (8)] - 1600/12 + 256/12

∫ ∫ v ln (5v) dv f(x) = ln (5v) f'(x) = 1/5v g'(x) = 5v g(x) = 5v²/2

5/2 ∫ v ln (5v) - 5/5 v² dv = ∫ 160 ln (40) u - 5/2 [v²/2] 0

160 ln (40) - 80

[600 ln (40) - 256 ln (16) - 1600/12 + 256/12 - 260 ln (40) + 80 - 64 + 32 ⋅ ln (8)]

+ 32 ln (8) [7/16] - 9.04