Subiectul al III-lea, 2., toate variantele

2. a) ∫₀² f(x)·√(x+4) dx = ∫₀² 4x dx = [2x²]₀² = - 3p = 8 − 0 = 8 - 2p

2. b) ∫₅¹² (1/(2x))·f(x) dx = ∫₅¹² 4/(2√(x+4)) dx = 4∫₅¹² (x+4)′/(2√(x+4)) dx = [4√(x+4)]₅¹² = - 3p = 16 − 12 = 4 - 2p

2. c) g(x) = √x/√(x+4), x ∈ [a,a+8], deci V = π∫ₐᵃ⁺⁸ (g(x))² dx = π∫ₐᵃ⁺⁸ (1 − 4/(x+4)) dx = = π[x − 4ln(x+4)]ₐᵃ⁺⁸ = π(8 − 4ln((a+12)/(a+4))) - 3p 4π(2 − ln((a+12)/(a+4))) = 4π(2 − ln 2), de unde obținem a = 4, care convine - 2p

∫₀⁴ (f(x) − xeˣ) dx = ∫₀⁴ (x + 1) dx = [x²/2]₀⁴ + [x]₀⁴ = - 3p = 8 + 4 = 12 - 2p

∫₀¹ (f(x) − x − 1) dx = ∫₀¹ x(eˣ)′ dx = [xeˣ]₀¹ − [eˣ]₀¹ = - 3p. = e − e + 1 = 1 - 2p.

2. c) V = π∫₀¹(g(x))² dx = π∫₀¹[f′(x)/(f(x))²] dx = π · [−1/f(x)]₀¹ = - 3p = π(−1/(e + 2) + 1) = π(e + 1)/(e + 2) - 2p

2.a) ∫₁²(f(x) − √(x + 3)) dx = ∫₁² 2x dx = [x²]₁² = - 3p = 4 − 1 = 3 - 2p

b) ∫₁⁶ 1/(f(x) − 2x) dx = ∫₁⁶ 1/√(x + 3) dx = 2∫₁⁶ (x + 3)′/(2√(x + 3)) dx = [2√(x + 3)]₁⁶ = - 3p = 6 − 4 = 2 - 2p

c) g(x) = x/√(x + 3), x ∈ [0,6], deci V = π∫₀⁶(g(x))² dx = π∫₀⁶ x²/(x + 3) dx = - 2p = π∫₀⁶(x − 3 + 9/(x + 3)) dx = π([x²/2]₀⁶ − [3x]₀⁶ + [9 ln(x + 3)]₀⁶) = 9π ln 3 - 3p

2.a) ∫₀¹ (f(x) − x² − 8) dx = ∫₀¹ 4x dx = [2x²]₀¹ = - 3p = 2 − 0 = 2 - 2p

b) ∫₀⁸ x/(f(x) − 4x) dx = ∫₀⁸ x/(x² + 8) dx = (1/2)∫₀⁸ (x² + 8)′/(x² + 8) dx = [(1/2)ln(x² + 8)]₀⁸ = - 3p = (ln 72)/2 − (ln 8)/2 = ln 3 - 2p

c) ∫₀ᵃ 1/(f(x) − 4) dx = ∫₀ᵃ 1/(x + 2)² dx = [−1/(x + 2)]₀ᵃ = −1/(a + 2) + 1/2, pentru orice a ∈ (0,+∞) - 3p −1/(a + 2) + 1/2 = 1/4, de unde obținem a = 2 - 2p

Reguli generale:

2.a) ∫₁³ (f(x) − 2ln x) dx = ∫₁³ (x − 1) dx = [x²/2]₁³ − [x]₁³ = - 3p = 4 − 2 = 2 - 2p

Reguli generale:

2.b) ∫₁ᵉ ((f(x) − x + 1)/x) dx = ∫₁ᵉ (2ln x/x) dx = ∫₁ᵉ 2ln x · (ln x)′ dx = [ln² x]₁ᵉ = - 3p = ln² e − ln² 1 = 1 - 2p

Reguli generale:

2.c) g(x) = 3x(x + 2ln x), x ∈ (0, +∞), deci 𝒜 = ∫₁³ |g(x)| dx = [x³]₁³ + ∫₁³ (3x²)′ · ln x dx = - 3p = 26 + [3x² ln x]₁³ − [3x²/2]₁³ = 14 + 27ln 3, deci a + 27ln 3 = 14 + 27ln 3, de unde obținem a = 14 - 2p

2. a) ∫₀¹ f(x)/eˣ dx = ∫₀¹ ((x² + 1)eˣ)/eˣ dx = ∫₀¹ (x² + 1) dx = [x³/3 + x]₀¹ = - 3p = 1/3 + 1 − 0 − 0 = 4/3 - 2p

2. b) Funcția F este derivabilă și F′(x) = (2x − 2)eˣ + (x² − 2x + 3)eˣ = (x² + 1)eˣ - 3p F′(x) = f(x), pentru orice număr real x, deci funcția F este o primitivă a funcției f - 2p

2. c) A = ∫₀² f(x) dx = ∫₀² (x² + 1)eˣ dx = [F(x)]₀² = F(2) − F(0) = - 3p = 3e² − 3 = 3(e² − 1) - 2p

2.a) ∫₀¹ (x² + x + 1)f(x) dx = ∫₀¹ (x² + x + 1) · (2x + 1)/(x² + x + 1) dx = ∫₀¹ (2x + 1) dx = (x² + x)|₀¹ = - 3p = 1 + 1 − 0 − 0 = 2 - 2p

b) ∫₀¹ f(x) dx = ∫₀¹ (2x + 1)/(x² + x + 1) dx = ∫₀¹ (x² + x + 1)′/(x² + x + 1) dx = ln(x² + x + 1)|₀¹ = - 3p = ln 3 − ln 1 = ln 3 - 2p

c) g(x) = eˣ(2x + 1), x ∈ ℝ, deci 𝒜 = ∫₀¹ |g(x)| dx = ∫₀¹ eˣ(2x + 1) dx = eˣ(2x + 1)|₀¹ − ∫₀¹ 2eˣ dx = - 3p = eˣ(2x + 1)|₀¹ − 2eˣ|₀¹ = 3e − 1 − 2e + 2 = e + 1 - 2p

2.a) ∫₀¹ f(x)/(x + 1) dx = ∫₀¹ eˣ dx = eˣ|₀¹ = - 3p = e − 1 - 2p

b) ∫₀¹ f(x) dx = ∫₀¹ eˣ(x + 1) dx = eˣ(x + 1)|₀¹ − ∫₀¹ eˣ dx = eˣ(x + 1)|₀¹ − eˣ|₀¹ = - 3p = 2e − 1 − e + 1 = e - 2p

c) ∫₁ᵃ [2x f(x²)/(x² + 1)] dx = ∫₁ᵃ 2x e^(x²) dx = ∫₁ᵃ e^(x²)(x²)′ dx = e^(x²)|₁ᵃ = e^(a²) − e - 3p e^(a²) − e = e(e³ − 1) ⇒ e^(a²) = e⁴ și, cum a > 1, obținem a = 2 - 2p

∫₁³ (f(x) − 6/(2x + 3)) dx = ∫₁³ eˣ dx = [eˣ]₁³ = - 3p = e³ − e = e(e² − 1) - 2p

∫₋₁⁰ (f(x) − eˣ) dx = ∫₋₁⁰ 6/(2x + 3) dx = 3∫₋₁⁰ (2x + 3)′/(2x + 3) dx = [3 ln(2x + 3)]₋₁⁰ = - 3p = 3(ln 3 − ln 1) = 3 ln 3 - 2p

g(x) = (2x² + 3x)eˣ + 6x, x ∈ (−3/2, +∞), deci 𝒜 = ∫₀¹ |g(x)| dx = ∫₀¹ ((2x² + 3x)eˣ + 6x) dx = = [(2x² + 3x)eˣ]₀¹ − [(4x + 3)eˣ]₀¹ + [4eˣ]₀¹ + [3x²]₀¹ = - 3p = 5e − 7e + 3 + 4e − 4 + 3 = 2e + 2 = 2(e + 1) - 2p

2.a) ∫₀¹ (x + 9) · f(x) dx = ∫₀¹ 8x dx = [4x²]₀¹ = - 3p = 4 − 0 = 4 - 2p

b) ∫₁⁶ (1/(8x)) · f(x) dx = ∫₁⁶ 1/(x + 9) dx = ∫₁⁶ (x + 9)′/(x + 9) dx = [ln(x + 9)]₁⁶ = - 3p = ln 15 − ln 10 = ln(3/2) - 2p

c) ∫₀³ f(x²) dx = ∫₀³ 8x²/(x² + 9) dx = 8∫₀³ (1 − 9/(x² + 9)) dx = [8x]₀³ − 8 · (9/3)[arctg(x/3)]₀³ = 24 − 6π - 3p 24 − 6π = 6(4 + aπ), de unde obținem a = −1 - 2p

Reguli generale:

∫₁² (f(x) − 3x) dx = ∫₁² 4x³ dx = [x⁴]₁² = - 3p = 16 − 1 = 15 - 2p

Reguli generale:

∫₂⁵ 1/(f(x) − 4x³ + 3) dx = (1/3)∫₂⁵ 1/(x + 1) dx = [(1/3) ln(x + 1)]₂⁵ = - 3p = (1/3)(ln 6 − ln 3) = (1/3) ln 2 - 2p

Reguli generale:

g(x) = 5x² + 3, x ∈ [1,2], deci V = π∫₁² (25x⁴ + 30x² + 9) dx = π[5x⁵ + 10x³ + 9x]₁² = 234π - 3p f(3) = 4·3³ + 3·3 = 117, deci V = 2πf(3) - 2p

∫₂³ (f(x)−3eˣ) dx=∫₂³ 2x dx=[x²]₂³= - 3p =9−4=5 - 2p

∫₀¹ x(f(x)−2x) dx=∫₀¹ 3xeˣ dx=[3xeˣ]₀¹−[3eˣ]₀¹= - 3p =3e−0−3e+3=3 - 2p

∫₀¹ [(f′(x)−x)/(2f(x)−x²)] dx=(1/2)∫₀¹ [(2f(x)−x²)′/(2f(x)−x²)] dx=(1/2)[ln|2f(x)−x²|]₀¹=(1/2)ln(e+1/2) - 3p (1/2)ln(e+1/2)=a ln(e+1/2), de unde obținem a=1/2 - 2p

2.a) ∫₀² (f(x) − 2/(x + 1)) dx = ∫₀² 6x dx = 6 · [x²/2]₀² = - 3p = 12 − 0 = 12 - 2p

b) ∫₀¹ (f(x) − 6x) dx = ∫₀¹ 2/(x + 1) dx = [2 ln(x + 1)]₀¹ = - 3p = 2 ln 2 − 2 ln 1 = 2 ln 2 - 2p

c) ∫₁ᵉ (f(x) − 2/(x + 1)) · ln² x dx = ∫₁ᵉ 6x ln² x dx = ∫₁ᵉ (3x²)′ ln² x dx = [3x² ln² x]₁ᵉ − ∫₁ᵉ 6x ln x dx = = 3e² − [3x² ln x]₁ᵉ + [3x²/2]₁ᵉ = 3(e² − 1)/2 - 3p 3(e² − 1)/2 = a(e² − 1)/2, de unde obținem a = 3 - 2p

Reguli generale:

2. a) ∫₀² [f(x)/eˣ] dx = ∫₀² (x − 1) dx = (x²/2 − x)|₀² = - 3p = 2 − 2 = 0 - 2p

Reguli generale:

2. b) ∫₀¹ f(x) dx = ∫₀¹ (x − 1)eˣ dx = (x − 1)eˣ|₀¹ − eˣ|₀¹ = - 3p = 1 − (e − 1) = 2 − e - 2p

Reguli generale:

2. c) ∫₂ⁿ x/[f(x)·f(−x)] dx = ∫₂ⁿ x/(1 − x²) dx = −(1/2)∫₂ⁿ (x² − 1)′/(x² − 1) dx = −(1/2)ln|x² − 1| |₂ⁿ = (1/2)ln[3/(n² − 1)] - 3p (1/2)ln[3/(n² − 1)] = (1/2)ln(3/8) și, cum n este număr natural, n > 2, obținem n = 3 - 2p

∫₁² (f(x) − eˣ − 3) dx = ∫₁² 3x² dx = [x³]₁² = - 3p = 8 − 1 = 7 - 2p

∫₀¹ x(f(x) − 3x²) dx = ∫₀¹ (xeˣ + 3x) dx = [xeˣ − eˣ]₀¹ + [3x²/2]₀¹ = - 3p = 1 + 3/2 = 5/2 - 2p

f′(x) = eˣ + 6x ⇒ f(x) − f′(x) = 3(x − 1)², deci ∫₀ᵃ 1/(f(x) − f′(x)) dx = (1/3)∫₀ᵃ 1/(x − 1)² dx = = −(1/3)·[1/(x − 1)]₀ᵃ = −(1/3)·(1/(a − 1) + 1), pentru orice a ∈ (0,1) - 3p −(1/3)·(1/(a − 1) + 1) = 1/6, de unde obținem a = 1/3, care convine - 2p

Reguli generale:

SUBIECTUL al III-lea, 2. a) ∫₁⁴ (f(x) − ln x + 1) dx = ∫₁⁴ x² dx = [x³/3]₁⁴ = - 3p = 64/3 − 1/3 = 21 - 2p

Reguli generale:

SUBIECTUL al III-lea, 2. b) ∫₂⁴ x/(f(x) − ln x) dx = ∫₂⁴ x/(x² − 1) dx = (1/2)∫₂⁴ (x² − 1)'/(x² − 1) dx = [(1/2)ln(x² − 1)]₂⁴ = - 3p = (1/2)ln 15 − (1/2)ln 3 = (1/2)ln 5 - 2p

Reguli generale:

SUBIECTUL al III-lea, 2. c) ∫₁ᵃ f(x)/x² dx = ∫₁ᵃ (1 + (ln x)/x² − 1/x²) dx = ∫₁ᵃ 1 dx + ∫₁ᵃ (−1/x)' ln x dx − ∫₁ᵃ 1/x² dx = = [x]₁ᵃ − [(1/x)ln x]₁ᵃ + ∫₁ᵃ 1/x² dx − ∫₁ᵃ 1/x² dx = a − 1 − (ln a)/a, pentru orice a ∈ (1, +∞) - 3p a − 1 − (ln a)/a = (a − ln a)/a, de unde obținem a = 2, care convine - 2p

2.a) ∫₀¹ (f(x) − 2x) dx = ∫₀¹ (5x² + 1) dx = (5x³/3)|₀¹ + x|₀¹ = - 3p = 5/3 + 1 = 8/3 - 2p

b) ∫₀² 1/(f(x) − 5x²) dx = ∫₀² 1/(2x + 1) dx = (1/2)∫₀² (2x + 1)′/(2x + 1) dx = (1/2) ln(2x + 1)|₀² = - 3p = (ln 5)/2 − (ln 1)/2 = (ln 5)/2 - 2p

c) g(x) = 5x√x + 1/√x, x ∈ (0,+∞); ∫ g(x) dx = ∫ (5x√x + 1/√x) dx = 2x²√x + 2√x + C, deci G(x) = 2x²√x + 2√x + c, unde c este număr real - 3p G(1) = 5 ⇒ c = 1, deci G(x) = 2x²√x + 2√x + 1, x ∈ (0,+∞) - 2p

Reguli generale:

2.a) ∫₀⁴ (f(x) − 2/√(x + 3)) dx = ∫₀⁴ (x + 2) dx = [x²/2]₀⁴ + [2x]₀⁴ = - 3p = 8 + 8 = 16 - 2p

Reguli generale:

2.b) ∫₁⁶ (f(x) − x − 2) dx = ∫₁⁶ 2/√(x + 3) dx = 4∫₁⁶ (x + 3)′/(2√(x + 3)) dx = [4√(x + 3)]₁⁶ = - 3p = 12 − 8 = 4 - 2p

Reguli generale:

2.c) ∫₋₂¹ f′(x)(1 + 1/f(x)) dx = ∫₋₂¹ f′(x) dx + ∫₋₂¹ f′(x)/f(x) dx = [f(x)]₋₂¹ + [ln|f(x)|]₋₂¹ = 2 + ln 2 - 3p Cum ln(2eᵃ) = a + ln 2, obținem a = 2 - 2p

2. a) ∫₁²(f(x)−3+6√x) dx = ∫₁²4x dx = [2x²]₁² = - 3p = 8−2 = 6 - 2p

2. b) ∫₁⁴((4x−f(x))/√x) dx = ∫₁⁴((−3+6√x)/√x) dx = ∫₁⁴(−3/√x+6) dx = [−6√x]₁⁴+[6x]₁⁴ = - 3p = −6+18 = 12 - 2p

2. c) G′(x) = g(x), x ∈ (0,+∞); G″(x) = g′(x) = 4√x+(4x+3)·1/(2√x)−6 = - 3p = (12x−12√x+3)/(2√x) = 3(2√x−1)²/(2√x) ≥ 0, pentru orice x ∈ (0,+∞), de unde rezultă că G este convexă - 2p

∫₀¹ (f(x) − 3) dx = ∫₀¹ 2x dx = [x²]₀¹ = - 3p = 1 − 0 = 1 - 2p

∫₀¹ eˣf(x) dx = ∫₀¹ eˣ(2x + 3) dx = [eˣ(2x + 3)]₀¹ − ∫₀¹ 2eˣ dx = 5e − 3 − [2eˣ]₀¹ = - 3p = 5e − 3 − 2e + 2 = 3e − 1 - 2p

∫₁² f(x)/(x(x + 3)) dx = ∫₁² (2x + 3)/(x² + 3x) dx = ∫₁² (x² + 3x)′/(x² + 3x) dx = [ln(x² + 3x)]₁² = ln(5/2) - 3p ln(5/2) = ln(a/2), de unde obținem a = 5, care convine - 2p

∫₀¹ (f(x) − 2x) dx = ∫₀¹ (eˣ + 2) dx = [eˣ + 2x]₀¹ = - 3p = e + 2 − 1 = e + 1 - 2p

∫₀³ 1/(f(x) − eˣ) dx = ∫₀³ 1/(2x + 2) dx = (1/2)∫₀³ 1/(x + 1) dx = [(1/2)ln(x + 1)]₀³ = - 3p = (ln 4)/2 − (ln 1)/2 = ln 2 - 2p

∫₀¹ f(x)/eˣ dx = ∫₀¹ (1 + (2x + 2)e⁻ˣ) dx = [x]₀¹ + ∫₀¹ (2x + 2)(−e⁻ˣ)′ dx = - 3p = 1 − [(2x + 4)e⁻ˣ]₀¹ = 5 − 6/e, deci 5 + a/e = 5 − 6/e, de unde obținem a = −6 - 2p

2.a) ∫₀² (f(x) − (x + 3)²) dx = ∫₀² x dx = [x²/2]₀² = - 3p = 4/2 − 0 = 2 - 2p

b) ∫₋₂⁰ 1/(f(x) − x) dx = ∫₋₂⁰ 1/(x + 3)² dx = ∫₋₂⁰ (x + 3)′/(x + 3)² dx = [−1/(x + 3)]₋₂⁰ = - 3p = −1/3 + 1 = 2/3 - 2p

c) ∫₀⁶ f(x)/(x + 3) dx = ∫₀⁶ [x + (x + 3)²]/(x + 3) dx = ∫₀⁶ (x + 4 − 3/(x + 3)) dx = [x²/2]₀⁶ + [4x]₀⁶ − [3ln(x + 3)]₀⁶ = 42 − 3ln 3 - 3p 42 − 3ln 3 = 3(a − ln 3), de unde obținem a = 14 - 2p

2. a) ∫₂⁴ (f(x) − ln x) dx = ∫₂⁴ (x + ln x − ln x) dx = ∫₂⁴ x dx = [x²/2]₂⁴ = - 3p = 16/2 − 4/2 = 6 - 2p

2. b) ∫₁ᵉ ((f(x) − x)/x) dx = ∫₁ᵉ ((x + ln x − x)/x) dx = ∫₁ᵉ (ln x/x) dx = ∫₁ᵉ ln x · (ln x)′ dx = [(ln x)²/2]₁ᵉ = - 3p = (ln e)²/2 − (ln 1)²/2 = 1/2 - 2p

2. c) ∫₁ⁿ f(x) dx = ∫₁ⁿ (x + ln x) dx = ∫₁ⁿ (x + x′ ln x) dx = [x²/2 + x ln x − x]₁ⁿ = (n² − 2n + 1)/2 + n ln n, pentru orice număr natural nenul n - 3p (n − 1)²/2 + n ln n = 2 + 3 ln n ⇔ (n − 3)(n + 1 + 2 ln n) = 0 și, cum n este număr natural nenul, obținem n = 3 - 2p

2.a) ∫₀³ (1/f(x)) dx = ∫₀³ (x² + 1) dx = [x³/3 + x]₀³ = - 3p = 27/3 + 3 = 12 - 2p

b) ∫₂³ 2x f(x) dx = ∫₂³ (2x/(x² + 1)) dx = ∫₂³ ((x² + 1)′/(x² + 1)) dx = [ln(x² + 1)]₂³ = - 3p = ln 10 − ln 5 = ln 2 - 2p

c) ∫₁² (eˣ/f(x)) dx = ∫₁² eˣ(x² + 1) dx = [eˣ(x² + 1)]₁² − ∫₁² 2xeˣ dx = 5e² − 2e − [2xeˣ − 2eˣ]₁² = 3e² − 2e - 3p e(3e − 2) = e(ae − a + 1), de unde obținem a = 3 - 2p

2. a) ∫₁³ (f(x) − 1/√(x + 1)) dx = ∫₁³ (x + 1) dx = [x²/2 + x]₁³ = - 3p = 9/2 + 3 − 1/2 − 1 = 6 - 2p

2. b) ∫₀⁸ (f(x) − x − 1) dx = ∫₀⁸ 1/√(x + 1) dx = 2∫₀⁸ (x + 1)′ · 1/(2√(x + 1)) dx = [2√(x + 1)]₀⁸ = - 3p = 2 · 3 − 2 · 1 = 4 - 2p

2. c) V = π∫₀³ g²(x) dx = π∫₀³ ((x + 1)² + 2√(x + 1) + 1/(x + 1)) dx = - 2p = π∫₀³ (x + 1)′((x + 1)² + 2√(x + 1) + 1/(x + 1)) dx = π[(x + 1)³/3 + 4 · (x + 1)√(x + 1)/3 + ln(x + 1)]₀³ = π(64/3 + 32/3 + ln 4 − 1/3 − 4/3) = π(91/3 + ln 4) - 3p

2.a) ∫₀² (f(x) − 4x) dx = ∫₀² (3x² + 2) dx = [x³ + 2x]₀² = - 3p = 2³ + 2 · 2 = 12 - 2p

2.b) ∫₀¹ (f(x) − 3x² − 2)eˣ dx = ∫₀¹ 4xeˣ dx = [4(x − 1)eˣ]₀¹ = - 3p = 0 + 4 = 4 - 2p

2.c) ∫₋₁⁰ a · f′(x) · (f(x))^(a−1) dx = [(f(x))ᵃ]₋₁⁰ = 2ᵃ − 1, pentru orice a ∈ (0, +∞) - 3p 2ᵃ − 1 = 63, de unde obținem a = 6, care convine - 2p

∫₀¹ (f(x) − 1) dx = ∫₀¹ 2x² dx = [2x³/3]₀¹ = - 3p = 2·1³/3 − 2·0³/3 = 2/3 - 2p

∫₀² (4x/f(x)) dx = ∫₀² (4x/(2x² + 1)) dx = ∫₀² ((2x² + 1)′/(2x² + 1)) dx = [ln(2x² + 1)]₀² = - 3p = ln9 − ln1 = 2ln3 - 2p

∫₁ᵉ f(1/x)·ln x dx = ∫₁ᵉ (2/x² + 1)·ln x dx = ∫₁ᵉ (−2/x + x)′·ln x dx = [(−2/x + x)·ln x]₁ᵉ − [2/x + x]₁ᵉ = 3 − 4/e - 3p 3 − 4/e = 2n² + 1 − 4/e și, cum n este număr natural, obținem n = 1 - 2p

∫₁² (x + 4)f(x) dx = ∫₁² 4x dx = [2x²]₁² = - 3p = 8 − 2 = 6 - 2p

∫₁⁴ (1/x) · f(x²) dx = 2∫₁⁴ 2x/(x² + 4) dx = 2∫₁⁴ (x² + 4)′ · 1/(x² + 4) dx = [2 ln(x² + 4)]₁⁴ = - 3p = 2 ln 20 − 2 ln 5 = 4 ln 2 - 2p

Dacă F : (−4, +∞) → ℝ este o primitivă a funcției f, atunci F′(x) = f(x), pentru orice x ∈ (−4, +∞) - 2p F″(x) = f′(x) = 16/(x + 4)² > 0, pentru orice x ∈ (−4, +∞), deci orice primitivă a funcției f este convexă - 3p

2. a) ∫₀² (f(x) − eˣ/2) dx = ∫₀² (x + 1) dx = [x²/2 + x]₀² = - 3p = 2²/2 + 2 = 4. - 2p

2. b) ∫₀¹ 2x(f(x) − 1) dx = ∫₀¹ (2x² + xeˣ) dx = [2x³/3]₀¹ + [(x − 1)eˣ]₀¹ = - 3p = 2/3 + 1 = 5/3. - 2p

2. c) Cum f′(x) = 1 + eˣ/2 = f(x) − x, x ∈ ℝ, obținem ∫₋₁⁰ (f(x) − x)·f(x) dx = ∫₋₁⁰ f′(x)·f(x) dx = = [f²(x)/2]₋₁⁰ = (9e² − 1)/(8e²). - 3p (9e² − 1)/(8e²) = ((3e + 1)(3e + a))/(8e²), de unde obținem a = −1. - 2p

2.a) ∫₁² (f(x)/√x) dx = ∫₁² (x+1) dx = [x²/2+x]₁² = - 3p = (2+2)−(1/2+1) = 5/2 - 2p

2.b) V = π∫₀¹ g²(x) dx = π∫₀¹ x(x+1)² dx = π∫₀¹ (x³+2x²+x) dx = π[x⁴/4+2x³/3+x²/2]₀¹ = - 3p = π(1/4+2/3+1/2) = 17π/12 - 2p

2.c) ∫₁ᵉ (f(x)√x ln x)/(x+1) dx = ∫₁ᵉ x ln x dx = ∫₁ᵉ (x²/2)′ ln x dx = [(x²/2) ln x]₁ᵉ − ∫₁ᵉ (x²/2)·(1/x) dx = e²/2 − (1/2)∫₁ᵉ x dx = - 3p = e²/2 − (1/2)[x²/2]₁ᵉ = e²/2 − e²/4 + 1/4 = (e²+1)/4, deci a = 1 - 2p

∫₁³ f(x) dx = ∫₁³ (x − 1/x + 1/x) dx = ∫₁³ x dx = [x²/2]₁³ = - 3p = 9/2 − 1/2 = 4 - 2p

∫₁² (f(x) + 1/x) ln x dx = ∫₁² x ln x dx = ∫₁² (x²/2)′ ln x dx = [(x²/2) ln x]₁² − ∫₁² (x²/2) · (1/x) dx = - 3p = (4/2) ln 2 − (1/2) ln 1 − [x²/4]₁² = 2 ln 2 − 3/4 - 2p

∫₁^{√2} xⁿ⁺¹ fⁿ(x) dx = ∫₁^{√2} x(x² − 1)ⁿ dx = (1/2) ∫₁^{√2} (x² − 1)′(x² − 1)ⁿ dx = [(x² − 1)ⁿ⁺¹/(2(n + 1))]₁^{√2} = 1/(2(n + 1)) - 3p Cum 2(n + 1) ≤ 2021 ⇔ n ≤ 2019/2, obținem că 1009 este cel mai mare număr natural nenul pentru care ∫₁^{√2} xⁿ⁺¹ fⁿ(x) dx ≥ 1/2021 - 2p

∫₁² (f(x) − x³ − 1) dx = ∫₁² 6x dx = [3x²]₁² = - 3p = 12 − 3 = 9 - 2p

∫₀¹ [x²/(f(x) − 6x)] dx = ∫₀¹ [x²/(x³ + 1)] dx = (1/3)∫₀¹ [(x³ + 1)′/(x³ + 1)] dx = [(1/3) ln(x³ + 1)]₀¹ = - 3p = (1/3) ln 2 − (1/3) ln 1 = (1/3) ln 2 - 2p

∫₀¹ f(√x) dx = ∫₀¹ (x√x + 6√x + 1) dx = [2x²√x/5 + 6·(2x√x/3) + x]₀¹ = 27/5 - 3p a³/5 = 27/5, de unde obținem a = 3 - 2p