Matematica Aplicada - Aula 01

Prévia do material em texto

Matema´tica Aplicada – Aula 1
Prof. Erne´e Kozyreff Filho
1 Revisa˜o de derivadas
• d
dt
a = 0
• d
dt
tn = ntn−1
• d
dt
et = et
• d
dt
eat = aeat
• d
dt
ln t =
1
t
• d
dt
sen t = cos t
• d
dt
sen at = a cos at
• d
dt
cos t = − sen t
• d
dt
cos at = −a sen at
2 Revisa˜o de integrais
2.1 Integrais indefinidas (diretas)
•
∫
t dt =
t2
2
+ C
•
∫
t2 dt =
t3
3
+ C
•
∫
t3 dt =
t4
4
+ C
•
∫
tn dt =
tn+1
n + 1
+ C, n 6= −1
•
∫ √
t dt =
1
2
√
t
+ C
•
∫
1
t3
dt = − 1
2t2
+ C
•
∫
1
t2
dt = −1
t
+ C
•
∫
1
t
dt = ln |t|+ C
•
∫
et dt = et + C
•
∫
e2t dt =
1
2
e2t + C
•
∫
eat dt =
1
a
eat + C, a 6= 0
•
∫
e−t dt = −e−t + C
•
∫
e(b−a)t dt =
1
b− ae
(b−a)t + C, a 6= b
•
∫
sen t dt = − cos t + C
•
∫
sen at dt = −1
a
cos at + C, a 6= 0
•
∫
cos t dt = sen t + C
•
∫
cos at dt =
1
a
sen at + C, a 6= 0
1
User
Realce
2.2 Me´todo da substituic¸a˜o
•
∫
3(3t + 4)5 dt =
(3t + 4)6
6
+ C •
∫
t sen(2t2) dt = −1
4
cos(2t2)
2.3 Integrac¸a˜o por partes
∫
uv′ = uv +
∫
u′v
•
∫
tet dt = tet − et + C
•
∫
t2et dt = et(t2 − 2t + 2) + C
•
∫
t3 cos at dt =
t3 sen at
a
+
3t2 cos at
a2
− 6t sen at
a3
− 6 cos at
a4
+ C
•
∫
et sen t dt =
et(sen t− cos t)
2
+ C
2.4 Integrais definidas e impro´prias
•
∫ 4
0
2t + 3 dt = 28
•
∫ 2
0
et dt = e2 − 1 ≈ 6,39
•
∫ 2
0
e−t dt = 1− e−2 ≈ 0,86
•
∫ ∞
0
e−t dt = 1
•
∫ ∞
0
te−t dt = 1
•
∫ ∞
0
t dt =∞
2
	Sem título