Esercizi di Analisi matematica 1

Esercizi di integrali e studio di funzione

∫(xⁿ+1)(xⁿ⁻¹)dx∫√(x⁴−x)dx∫(tan(x))(sec⁴(x))dx∫(2+x)/[(x²+4)(x²+1)]dx∫4/(x²−5^½)dx = 2∫dx/√(x⁴−x) + x/√(x⁴−x) + C

∫√(2x-ln(x))dx∫[1/(x²−5x+6)]dx = ∫[1/(x-2)]−[1/(x-3)]dxx³+x⁺²ⁿ∫x⁻²(3x+1+x)dx =∫[√x+4]/[x²(x+8)]dxx=x±^½∫3log√x = ∫ 3log√[x-1]dx + C

∫(exm+dx(f(x)=ln(x), p(x)=ex)x(e^ln(x))dx(x+t²)/4−√3cos(x)dxe^xsen(3x)dxf(x)=e^x g(n)=sen(3x) f'(x)=e^x g'(x)=cos(3x)⋅3= e^x⋅sen(3x)−∫e^xcos(3x)⋅3dx

f(n)=e^xg(x)=−cos(3x)= e^x⋅sen(3x)−3[e^xcos(3x)−∫e^xsen(3x)⋅3]dx

f(n)=e^x g(x)=cos(3x)= e^x⋅cos(3x)−3∫e^xsen(3x)dx∫e^xsen(3x)dx =sen(3x)⋅3f(n)=e^xsin(3x)−e^xcos(3x)−∫e^xsen(3x)dx

f(x)=arctom(cx) * np(n):=arctom(cx)=∫arctom(cx)*n dx∫^c_aarctom(x)→∫^b_aarctom(x)=n2+arctom(x)⋅n2+∫arctom(x)⋅1dx+∫arctom(x)∗1dx=n2arctom(x)−n(e)n2arctom(x)+Clog(n)(1+sen(cx))+C.1+e^xdx1ⁿ+t+e^x++1ⁿ+t−e^xdx

∫=1 to e^x(13)ⁿ_n−t^t−1∫1 to e^xdxarctom(e^x)+e^xlog(e^x+C)+Carctom(e^x)+1+log(e^x) + C= t: e^u(x)1+log(e^xt)+Ce^u(x) − ln(21e^u(x)) + t +C.

g(u)=∫1⁴e + dte^u1⁴u = √1+√1+4 + C.x + arctomlog(1+sen(x))+cx/sinx cosxtg(x)·cot(x).ds = ∫sen(x)dx= sin(x) = (sen(n)−15u=∫1+u⋅udx =

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Studio di funzione

f(x) = x³ - 3x + 2

Dominio: x ∈ R

Shame tema

f(x) = (x³ - 3x) + 2 ⇒ x = x³ + 3x + 2

Intersezioni: asse xy = 0x = 3x² + 0 = 20x - 2 = x = -2y = 0(-2,0)(1,0)

Sezione F(x) ≥ 0x³ - 3x + 2 > 0(x + 2) (x - 1) (x - 1) > 0

lim x → ∞f(x)x³ - 3x + →x → -∞diinci!

f'(x) ≥ 0⇒ 3x² ≥ 3xx > 0

f'(x) ≥ 6x∀ x > 0(x + 2) (x - 1) (x - 1)(-2) -2 = +4(1, 0) 0 = 0

f(x) = e^-1/x3 / x+4

Dominio: x+3 ≠ 0 → x ≠ -3 x+4 ≠ 0 → x ≠ -4 D: ∀x ∈ ℝ / {-3, -4} ∀x ∈ (-∞, -4) ∪ (-4, -3) ∪ (-3, +∞)

Intersez: y=0 → e^-1/x3 = 0 → 2^1/x3 = 0 x=0 → 0, 3√(2/4)

Asintoti: lim_x→-3⁺ e^-1/x3 / x+4 = lim_x→-3⁻ e^-1/x3 / x+4 = 0 lim_x→-4⁺ e^-1/x3 / x+4 = lim_x→-4⁻ e^-1/x3 / x+4 = 0 lim_x→-∞ e^-1/x3 / x+4 = lim_x→+∞ e^-1/x3 / x+4 = 0 lim_x→-4 √(e^-1/x3) / (x+3) = 1

f'(x) = [e^-1/x3(x+4) - e^-1/x3(1)] / (x+3)² = e^-1/x3 / (x+3)²

f'(x) > 0 → (x+1)² > 0 ∀x ∈ D e^-1/x3 > ∀x ∈ ℝ x+4 > 0 per x ≠ -4 x+3 > 0 ∀x ∈ Dx² + 8x +16

Conclusions: - x₁, x₂ = -7 ± √49

f(x)= x² - log(x)

Polinomio: x > 0 => x ∈ (0,+∞)

Asintoto: f(x) => e^x2-log(x)log x

Segnof(x) => x > 0 f(x) => maif(x) => x - log(x)x² + log xx / x² + 1 > 0 ∈ D

f(x)= x > 0∫ tan x sec²xdx= ∫ tan²x + tanx sec xdx∫ (x² + 1)^-2 = cos x2. ∫ dx3. ∫x²-1dv =4. ∫ x /(x+1)u =arctan u + cf(x)=2x => f’(x)= xg’(x)= xe^xf(x)= f’(x) = x