exercicios integrais resolvidos
Anatolie Sochirca ACM DEETC ISEL
Integral definido. Exercícios resolvidos.
a) Calcular os integrais definidos utilizando a fórmula de Barrow.
1
∫
Exercício 1.
1 + x dx .
0
1
1
1
1
(1 + x) 2 +1
1 + x d (1 + x) = ∫ (1 + x) 2 d (1 + x) =
1 +1
0
2
1
∫
1 + x dx = ∫
0
0
3
2
= ⋅ (1 + x ) 2
3
1
0
(
1
∫ (x
−1
∫ (x
−1
x dx
2
)
+1
x dx
2
)
+1
2
1
=
0
.
1 d (x2 )
1
1
1
d (x2 ) d ( x 2 + 1) 1
1
1
=∫ 2
= ⋅∫
= ⋅∫
= ⋅ ∫ x2 +1
2
2
2 −1 x 2 + 1 2
2 −1 x 2 + 1 2
2 −1
−1 x + 1
(
)
(
1
2
0
3
(1 + x) 2
=
3
2
3
3
2 3
2
2
2
2
= ⋅ (1 + 1) − (1 + 0) = ⋅ 2 2 − 1 = ⋅ 2 2 − 1 .
3
3
3
Exercício 2.
1
1
(
)
)
1
− 2 +1
1 x2 +1
= ⋅
2 − 2 +1
−1
(
)
((
)
1
= − ⋅ x2 +1
2
−1
(
)
1
−1
)
1 1
= − ⋅ 2
2 x + 1
1
−1
)
−2
d ( x 2 + 1) =
1 1
1
= − ⋅ 2
1 + 1 − (−1) 2 + 1 = 0
2
(1 + l nx) dx
.
x
1
e
Exercício 3.
∫
(1 + l nx) dx dx =∫ (1 + l nx) ⋅
= ∫ (1 + l nx) ⋅ d (l nx) = ∫ d (l nx) + ∫ l nx d (l nx) =
∫ x x 1
1
1
1
1 e e
= (l nx ) 1
e
e
e
e
e
ln2 x
l n 2e l n 21
1 3
+
2 = (l n e − l n1) + 2 − 2 = 1 + 2 = 2 .
1
1
Matemática 1
Anatolie Sochirca ACM DEETC ISEL
2
dx
∫
Exercício 4.
16 − x 2
0
2
∫
0
2
dx
16 − x
2
dx
=∫
4 −x
2
0
2
.
x
= arcsen
4
2
0
2
0
= arcsen − arcsen =
4
4
π π 1
= arcsen − arcsen(0 ) = − 0 = .
6
6
2
1
∫x
Exercício 5.
0
2
dx
.
+ 4x + 5
O polinómio do denominador não tem raizes reais:
− 4 ± 16 − 20 − 4 ± − 4
=
.
2
2
Portanto é uma fracção elementar