تمارين - 2BACSEF
التانية باكالوريا العلوم التجريبية – خيار فرنسي
درس :
Calcul intégral
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Exercice 1
Calculer les intégrales suivantes :
∫
1
2
2
x
d
x
;
;
∫
e
e
2
1
t
d
t
;
;
∫
0
π
6
c
o
s
(
2
θ
)
d
θ
\int_{1}^{2}2xdx~~;; ~~\int_{e}^{e^2}\dfrac1t dt~~;; ~~\int_{0}^{\dfrac{\pi}{6}}cos(2\theta)d\theta
∫
1
2
2
x
d
x
;;
∫
e
e
2
t
1
d
t
;;
∫
0
6
π
cos
(
2
θ
)
d
θ
Correction
Calculons l’intégrale
∫
1
2
2
x
d
x
\int_{1}^{2} 2x \, dx
∫
1
2
2
x
d
x
:
∫
1
2
2
x
d
x
=
[
x
2
]
1
2
=
2
2
−
1
2
=
4
−
1
=
3
\begin{align*} \int_{1}^{2} 2x \, dx &= \left[ x^2 \right]_{1}^{2} = 2^2 - 1^2 \\&= 4 - 1 = 3 \end{align*}
∫
1
2
2
x
d
x
=
[
x
2
]
1
2
=
2
2
−
1
2
=
4
−
1
=
3
Donc,
∫
1
2
2
x
d
x
=
3
\boxed{\int_{1}^{2} 2x \, dx = 3}
∫
1
2
2
x
d
x
=
3
Calculons l’intégrale
∫
e
e
2
1
t
d
t
\int_{e}^{e^2} \dfrac{1}{t} \, dt
∫
e
e
2
t
1
d
t
:
∫
e
e
2
1
t
d
t
=
[
ln
∣
t
∣
]
e
e
2
=
ln
∣
e
2
∣
−
ln
∣
e
∣
=
2
−
1
=
1
\begin{align*} \int_{e}^{e^2} \dfrac{1}{t} \, dt &= \left[ \ln |t| \right]_{e}^{e^2} = \ln |e^2| - \ln |e| \\ &= 2 - 1 = 1 \end{align*}
∫
e
e
2
t
1
d
t
=
[
ln
∣
t
∣
]
e
e
2
=
ln
∣
e
2
∣
−
ln
∣
e
∣
=
2
−
1
=
1
Donc,
∫
e
e
2
1
t
d
t
=
1
\boxed{\int_{e}^{e^2} \dfrac{1}{t} \, dt = 1}
∫
e
e
2
t
1
d
t
=
1
Calculons l’intégrale
∫
0
π
6
cos
(
2
θ
)
d
θ
\int_{0}^{\dfrac{\pi}{6}} \cos(2\theta) \, d\theta
∫
0
6
π
cos
(
2
θ
)
d
θ
:
∫
0
π
6
cos
(
2
θ
)
d
θ
=
[
sin
(
2
θ
)
2
]
0
π
6
=
sin
(
π
3
)
2
−
sin
(
0
)
2
=
3
2
2
−
0
=
3
4
\begin{align*} \int_{0}^{\dfrac{\pi}{6}} \cos(2\theta) \, d\theta &= \left[ \dfrac{\sin(2\theta)}{2} \right]_{0}^{\dfrac{\pi}{6}} \\ &= \dfrac{\sin(\dfrac{\pi}{3})}{2} - \dfrac{\sin(0)}{2} \\&= \dfrac{\dfrac{\sqrt{3}}{2}}{2} - 0 = \dfrac{\sqrt{3}}{4} \end{align*}
∫
0
6
π
cos
(
2
θ
)
d
θ
=
[
2
sin
(
2
θ
)
]
0
6
π
=
2
sin
(
3
π
)
−
2
sin
(
0
)
=
2
2
3
−
0
=
4
3
Donc,
∫
0
π
6
cos
(
2
θ
)
d
θ
=
3
4
\boxed{ \int_{0}^{\dfrac{\pi}{6}} \cos(2\theta) \, d\theta = \dfrac{\sqrt{3}}{4}}
∫
0
6
π
cos
(
2
θ
)
d
θ
=
4
3