Esercizi con svolgimento di Analisi matematica I

∫ x cos x dx = x sin x - ∫ sin x ⋅ sin x + cos x + c

f: x f': 1

g': cos x g: sin x

∫ x e^x dx = x e^x - ∫ e^x e^x + c

f: x f': 1

g': e^x g: e^x

∫ x² e^2x dx = x² e^x - ∫ x e^x ⋅ 2

:: x² e^x + x² e^x - 2

f: x² f': 1

g': e^x g: e^x x²

∫ x ln x dx = x² ln x = ∫ v² ln x - 1/4 x²

:: ∫ dx x² ln x - 1/4 x²

f: ln x f': 1

g': x g: x²/2

∫ cos ⋅ sin(sin x) dx = ∫ sin y dy g = - cos y

:: c = - cos (sin x) + c

y: sin x

dy: cos dx

∫ sin(e^t)e^t dx = ∫ sin y dy = - cos y

c = - cos(e^t) + c

g: e(x)

dg: e^x dx

∫ x e^t dx = ∫ x e^t dx

:: ∫ 1/1+2

dy: dx

y: e^x

dy: e^x dx

∫ x cos(x) dx = ∫ y cos(y) dy = ∫ cos g dg = 1/2 [sin s] - 1/2 [0 - 0] = 0

y = x²

dy = 2x dx [with: x =]

∫ 1/(x ln x)² dx = ∫ 1/(x ln x) dx = ∫ 1/y dy = ln|y| + c = ln|ln x| + c

y = ln x

dy = 1/x dx

∫ √1 - x² dx = ∫ √1 - sin² y cos 2y dy = ∫ cos g cos g dg = ∫ cos² g dg

x = sin y

dx = cos y dy

cos² r + sin² i = 1

[∫ sin⁴ x cos³ x dx - ∫ sin² x cos² x cos x dx = ∫ y⁴ (1 - y²) dy = ∫ y⁴ dy - ∫ y⁶ dy = 0 - 0]

y = sin x

dy = cos x dx

1 - y²

∫ x - 1/(1 + √x) dx = ∫ y² - 1/1 + y dy = ∫ (y - 1)(1/y + 1) dy = ∫ y² dy - ∫ 2y dy - ∫ 2 dy

y = √x → x = y²

dy = 1/2x^-1/2 x dx = 2y dy

z = ∫ y³ dy - ∫ y² dy - 2∫ y dy - ∫ y³ dy = 2/3 y^{3 - y2 = 2/3 (x)3/2 x - 2√x + c}

oppure

∫ x - 1/(1 + √x) dx = ∫ [√x - 1/(√x + 1)]/1 + √x dx = ∫ √x - 1 dx = ∫ x^1/2 dx - ∫ dx = ∫ x^1/2 dx - ∫ dx

∫ dx = 3/2 x^3/2 - x + c = 2/3 √x³ + c

∫√3x-1 dx →∫ t ² dt = ∫(t ²) dt = t ³/3 + c

2/3 t ³ dt . 2/3 + c

t = √3x - 1

dt = 3/2√3x - 1

x = t ² + 1/3

∫2/3(3x -1)√3x -1/9

2/3(3x-1) ³ / 9

(√3x -1) ³

2/(√3x -1) ³ + c

(3x - x) ²(3x - 4)

(3x - x)√3x -1

∫sen(1 - √x) / √x dx = ∫sin(1-t)/ t = 2 ∫sin1x -1 dt = 2 sin1x -1dt - 2cos(1-t) + c

t = √x

t ² = x

dx/dt = 2t → dx = 2t dt

∫4x ³ √x ⁴ + 4 dx = ∫u(4u ³ x ³) t ² / √t ² x ⁴ dx = ∫ t ^3/2 dt = 4.2 .t ⁴ + c = x ⁴ + c

t = √x ⁴ + 4

t ³ = √x ⁴ + 4

x = 3/2 √x ⁴ + 4

dx/dt = 4t/3t ⁴ + 4.tdx

→ dx = 3t/√x ³ (t ³ + 4)

(x ³ + 4) ^3/3 + c

∫₀^∞ e^x/(x² - 1) dx = ∫₄⁹ 1/(y - 4)(9 - y) ∫₄⁹ 1/(y - 2i) (y - 1) dy

e^-2x

da e^-2 · e^-2iθ

A = B

₄1 A = B (y - 4) + By + B

(y - 4)(y - 1)

g5 + A = (B - A)(B)

(y - 2i)(y - 1)

A + B = 0; A = -B

A + B = i

0 + B + 2

B = 1/2

∫₀¹ 1/(y - 4) dy + ∫₂⁴ 1/(y - 2i) dy + 1/2 [ln|y² - 1| = -1/2 ln|e^{iθ x}| + 1/2 ln|e^{i x}|]

lim_{s →∞} ∫₀^s e^{x2 - 1} dx = lim_{s →∞} [1/2 ln|y²| + 1/2 ln |s - 1| + 1/2 ln |s-1| + 1/2 ln|e^{i x} - 1| - 1/2]

-

lim_{s →∞} [1/2 ln(e^x|) + 1/2 ln(e^{x + t}) - 1/2 ln(e^t - 1)] =

lim_{s →∞}

∫ ₁^{1+cos x} (1/sin⁴x) dx = ln|x+sinx| + c

∫ [ (3x+1)/(x²+1) dx + ∫ (3x)/(x²+1) dx + 2 ∫ x dx + 3 ∫ (2x)/(x²+1) dx + 2 ∫ (1)/(x²+1) dx ] =

= (3/2) ln|x²+1| + 2 arctan(x) + c

∫ [ 1/(sin²x+cos²x) dx = ∫ sin²x/(sin²x+cos²x) dx + ∫ cos²x/(sin²x+cos²x) dx ] = ∫ (1)/cos²x dx + ∫ (1)/sin²x dx

= tan(x) - cot(x) + c

∫ arcsin x = x arcsin x - ∫ (x/√(1-x²)) dx = x arcsin x + √(1-x²) + c

f = arcsin x; f' = 1

g' = 1; g = x

( x²y )( z/2 ) + x² - 2 = 3 ∫ ( y z )

∑_n=2 (^(√a-2)n/_logn)

lim _n→∞ |(√a-2)ⁿ⁺¹|/|log_n+1|

|(√a-1)| < 1

1 < √a-1 < 1

lim _n→∞ |(log_n|

{ 0 < √a < 2 → 0 < a < 4}

Per a>4 diverge

Per a<0 non esiste

Per a=0 per Leibnitz converge ∑_n1/log_n=lim 1/log_n=0

Per a<4 converge diverge

c. a. 4

div.

1/log_n > 1/n sapendo che 1/4 diverge anche 1/log_n diverge

Teorema di Torricelli-Barrow

Se \( f:[a,b] \) una funzione limitata e integrabile, che ammette primitiva, allora

\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]

Dim. Sia \( D = \{x_1, x_2, \ldots, x_n \} \) una divisione dell'intervallo \([a,b]\)

\[\inf f(c_i) = \sup f \quad \forall c \in [a,b]\]

\[I(f,D) = \sum_{j=1}^{n} f(x_i)(x_i - x_{i-1}) \leq I(f)\]

applico il teorema di Lagrange

\[\frac{F(x_i) - F(x_{i-1})}{(x_i-x_{i-1})} = f(c_i) \implies F(x_i) - F(x_{i-1}) = f(c_i)(x_i - x_{i-1})\]

\[\sum_{i}^{i=k}F(x_i - F(a_i))f(x_i - x_{i-1})= F(b) - F(a)\]

I(S_x) \leq F(b) - F(a) \leq I(S_y)

Dato che la funzione a gradini per difetto e quella per eccesso sono due insiemi

l'assero è così dimostrato

\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]

anno

2001    2002    2003    2004    2005    2006    2007    2008    2009    2010    2011

• 2,66     3,12     3,83     4,26     5,37     4,80     4,51     5,14
• ⁴ (4,31)   (5,41)

q = ¹/₂    r = ⁶/_n    q = a y = 0,37 x + 1,35

• lim ^{| a - 2 |}/_n²

[ ^{(-1)m (a - 3)m}/_m² ]

Per a >> 3 → diverg

Per a → 1 → segni alterni → per lambdaj residuciuouni

Per a = 3 → converge

Per a = 1 → per lambdaj converge

lim_h→0 h sin(1/h)

lim_x→0 x sin(1/x) = 0

lim_x→0 (1 + 1/x)^x = [1 + (1/(x [ln (1+1/x)]])]^x = e^1/x e^-1 = e^1/x (1/e)

∫₀¹ x - 9/(x² + 3x + 2) dx : ∫₀¹ x - 9/[(x+5) (x-2)]

A/(x+6) + B/(x-2)

Ax - 2A + Bx + 5B = 0

{ A+B=1 } { A - 2B = 1 } { -2A + 5B = 9 }

A = 2 B = 1

∫₀¹ 1/(x+5) +∫₀¹ 1/(x-2) = 2 [ln|x+5|]₀¹ - [ln|x-2|]₀

= 2[ln 6/5 ] - [ln 1/2 ] = ln 36/25 - ln 1/2 = ln 36/25/1/2

= ln 72/25

∫₀^∞ 1/(1+x^2t) dt = ∫ 1/(1+x) + 1/2 log₂ e^x dx = log₂ ∫ 1/(1+2 log₂ x²) dx

x_t= x

t = log₂ x

dt/dx = 1/2

∫ dt = x