En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : ComparaisonIntégration en mathématiques/Exercices/Comparaison », n'a pu être restituée correctement ci-dessus.
Soit
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} }
|
∫
a
b
f
(
x
)
d
x
|
=
∫
a
b
|
f
(
x
)
|
d
x
{\displaystyle \left|\int _{a}^{b}f(x)\,\mathrm {d} x\right|=\int _{a}^{b}|f(x)|\,\mathrm {d} x}
alors la fonction
f
{\displaystyle f}
Soient
f
,
g
:
[
a
,
b
]
→
R
{\displaystyle f,g:[a,b]\to \mathbb {R} }
a
≤
b
{\displaystyle a\leq b}
g
{\displaystyle g}
|
f
|
{\displaystyle |f|}
A
{\displaystyle A}
|
∫
a
b
f
(
x
)
g
(
x
)
d
x
|
⩽
A
∫
a
b
g
(
x
)
d
x
{\displaystyle \left|\int _{a}^{b}f(x)g(x)\,\mathrm {d} x\right|\leqslant A\int _{a}^{b}g(x)\,\mathrm {d} x}
Solution
−
A
g
≤
f
g
≤
A
g
⇒
−
A
∫
a
b
g
(
x
)
d
x
≤
∫
a
b
f
(
x
)
g
(
x
)
d
x
≤
A
∫
a
b
g
(
x
)
d
x
{\displaystyle -Ag\leq fg\leq Ag\Rightarrow -A\int _{a}^{b}g(x)\,\mathrm {d} x\leq \int _{a}^{b}f(x)g(x)\,\mathrm {d} x\leq A\int _{a}^{b}g(x)\,\mathrm {d} x}
Soient
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
0
<
a
<
b
{\displaystyle 0<a<b}
F
{\displaystyle F}
f
{\displaystyle f}
F
(
b
)
−
F
(
a
)
2
b
⩽
∫
a
b
f
(
t
2
)
d
t
⩽
F
(
b
)
−
F
(
a
)
2
a
{\displaystyle {\frac {F(b)-F(a)}{2{\sqrt {b}}}}\leqslant \int _{\sqrt {a}}^{\sqrt {b}}f(t^{2})\,\mathrm {d} t\leqslant {\frac {F(b)-F(a)}{2{\sqrt {a}}}}}
Solution
Par changement de variable,
∫
a
b
f
(
t
2
)
d
t
=
∫
a
b
f
(
s
)
2
s
d
s
{\displaystyle \int _{\sqrt {a}}^{\sqrt {b}}f(t^{2})\,\mathrm {d} t=\int _{a}^{b}{\frac {f(s)}{2{\sqrt {s}}}}\,\mathrm {d} s}
Par positivité de
f
{\displaystyle f}
s
↦
1
s
{\displaystyle s\mapsto {\frac {1}{\sqrt {s}}}}
∫
a
b
f
(
s
)
2
b
d
s
≤
∫
a
b
f
(
s
)
2
s
d
s
≤
∫
a
b
f
(
s
)
2
a
d
s
{\displaystyle \int _{a}^{b}{\frac {f(s)}{2{\sqrt {b}}}}\,\mathrm {d} s\leq \int _{a}^{b}{\frac {f(s)}{2{\sqrt {s}}}}\,\mathrm {d} s\leq \int _{a}^{b}{\frac {f(s)}{2{\sqrt {a}}}}\,\mathrm {d} s}
c'est-à-dire
F
(
b
)
−
F
(
a
)
2
b
≤
∫
a
b
f
(
s
)
2
s
d
s
≤
F
(
b
)
−
F
(
a
)
2
a
{\displaystyle {\frac {F(b)-F(a)}{2{\sqrt {b}}}}\leq \int _{a}^{b}{\frac {f(s)}{2{\sqrt {s}}}}\,\mathrm {d} s\leq {\frac {F(b)-F(a)}{2{\sqrt {a}}}}}
Trouver un encadrement de chacune des intégrales suivantes :
∫
0
1
2
d
t
(
3
+
2
cos
t
)
2
{\displaystyle \int _{0}^{\frac {1}{2}}{\frac {\mathrm {d} t}{(3+2\cos t)^{2}}}}
∫
0
1
2
d
t
1
−
t
5
{\displaystyle \int _{0}^{\frac {1}{2}}{\frac {\mathrm {d} t}{\sqrt {1-t^{5}}}}}
Soit
α
>
0
{\displaystyle \alpha >0}
Prouver que pour
x
∈
[
0
,
1
]
{\displaystyle x\in [0,1]}
1
−
x
α
≤
1
−
x
α
≤
1
−
x
α
2
{\displaystyle 1-x^{\alpha }\leq {\sqrt {1-x^{\alpha }}}\leq 1-{\frac {x^{\alpha }}{2}}}
En déduire que :
α
α
+
1
<
∫
0
1
1
−
x
α
d
x
<
α
+
1
2
α
+
1
{\displaystyle {\frac {\alpha }{\alpha +1}}<\int _{0}^{1}{\sqrt {1-x^{\alpha }}}\,\mathrm {d} x<{\frac {\alpha +{\frac {1}{2}}}{\alpha +1}}}
Solution
y
:=
x
α
∈
[
0
,
1
]
{\displaystyle y:=x^{\alpha }\in [0,1]}
z
:=
1
−
y
∈
[
0
,
1
]
{\displaystyle z:=1-y\in [0,1]}
z
≤
z
{\displaystyle z\leq {\sqrt {z}}}
1
−
y
≤
1
−
y
/
2
{\displaystyle {\sqrt {1-y}}\leq 1-y/2}
(
1
−
y
/
2
)
2
=
1
−
y
+
y
2
/
4
≥
1
−
y
{\displaystyle (1-y/2)^{2}=1-y+y^{2}/4\geq 1-y}
∫
0
1
(
1
−
x
α
)
d
x
=
1
−
1
α
+
1
=
α
α
+
1
{\displaystyle \int _{0}^{1}(1-x^{\alpha })\,\mathrm {d} x=1-{\frac {1}{\alpha +1}}={\frac {\alpha }{\alpha +1}}}
∫
0
1
(
1
−
x
α
2
)
d
x
=
1
−
1
/
2
α
+
1
=
α
+
1
2
α
+
1
{\displaystyle \int _{0}^{1}\left(1-{\frac {x^{\alpha }}{2}}\right)\,\mathrm {d} x=1-{\frac {1/2}{\alpha +1}}={\frac {\alpha +{\frac {1}{2}}}{\alpha +1}}}
∫
0
1
1
−
x
α
d
x
{\displaystyle \int _{0}^{1}{\sqrt {1-x^{\alpha }}}\,\mathrm {d} x}
]
0
,
1
[
{\displaystyle ]0,1[}
Soit
f
:
[
0
,
1
]
→
R
{\displaystyle f:[0,1]\to \mathbb {R} }
x
{\displaystyle x}
[
0
,
1
]
{\displaystyle [0,1]}
∫
x
1
f
(
t
)
d
t
≥
1
−
x
2
2
{\displaystyle \int _{x}^{1}f(t)\,\mathrm {d} t\geq {\frac {1-x^{2}}{2}}}
Soit
F
{\displaystyle F}
f
{\displaystyle f}
Prouver que :
F
(
1
)
=
∫
0
1
x
f
(
x
)
d
x
+
∫
0
1
F
(
x
)
d
x
{\displaystyle F(1)=\int _{0}^{1}xf(x)\,\mathrm {d} x+\int _{0}^{1}F(x)\,\mathrm {d} x}
En déduire que :
∫
0
1
x
f
(
x
)
d
x
≥
1
3
{\displaystyle \int _{0}^{1}xf(x)\,\mathrm {d} x\geq {\frac {1}{3}}}
En déduire que :
∫
0
1
[
f
(
x
)
]
2
d
x
≥
1
3
{\displaystyle \int _{0}^{1}[f(x)]^{2}\,\mathrm {d} x\geq {\frac {1}{3}}}
Solution
x
f
+
F
=
(
x
F
)
′
{\displaystyle xf+F=(xF)'}
∫
0
1
x
f
(
x
)
d
x
+
∫
0
1
F
(
x
)
d
x
=
[
x
F
]
0
1
=
F
(
1
)
{\displaystyle \int _{0}^{1}xf(x)\,\mathrm {d} x+\int _{0}^{1}F(x)\,\mathrm {d} x=[xF]_{0}^{1}=F(1)}
F
(
1
)
−
∫
0
1
F
(
x
)
d
x
=
∫
0
1
(
F
(
1
)
−
F
(
x
)
)
d
x
≥
∫
0
1
1
−
x
2
2
d
x
=
1
−
1
/
3
2
=
1
3
{\displaystyle F(1)-\int _{0}^{1}F(x)\,\mathrm {d} x=\int _{0}^{1}\left(F(1)-F(x)\right)\,\mathrm {d} x\geq \int _{0}^{1}{\frac {1-x^{2}}{2}}\,\mathrm {d} x={\frac {1-1/3}{2}}={\frac {1}{3}}}
Par Cauchy-Schwarz,
1
3
2
≤
(
∫
0
1
x
f
(
x
)
d
x
)
2
≤
∫
0
1
x
2
d
x
∫
0
1
[
f
(
x
)
]
2
d
x
=
1
3
∫
0
1
[
f
(
x
)
]
2
d
x
{\displaystyle {\frac {1}{3^{2}}}\leq \left(\int _{0}^{1}xf(x)\,\mathrm {d} x\right)^{2}\leq \int _{0}^{1}x^{2}\,\mathrm {d} x\int _{0}^{1}[f(x)]^{2}\,\mathrm {d} x={\frac {1}{3}}\int _{0}^{1}[f(x)]^{2}\,\mathrm {d} x}
Démontrer que pour tout entier naturel
n
{\displaystyle n}
1
2
(
n
+
1
)
≤
∫
0
1
x
n
1
+
x
2
d
x
≤
1
n
+
1
{\displaystyle {\frac {1}{2(n+1)}}\leq \int _{0}^{1}{\frac {x^{n}}{1+x^{2}}}\,\mathrm {d} x\leq {\frac {1}{n+1}}}
Solution
∀
x
∈
[
0
,
1
]
1
≤
1
+
x
2
≤
2
{\displaystyle \forall x\in [0,1]\quad 1\leq 1+x^{2}\leq 2}
∫
0
1
x
n
d
x
=
1
n
+
1
{\displaystyle \int _{0}^{1}x^{n}\,\mathrm {d} x={\frac {1}{n+1}}}
Étudier les variations de la fonction
f
{\displaystyle f}
f
(
x
)
=
e
x
2
−
(
e
−
1
)
x
2
−
1
{\displaystyle f(x)=\mathrm {e} ^{x^{2}}-(\mathrm {e} -1)x^{2}-1}
Prouver que :
4
3
<
∫
0
1
e
x
2
d
x
<
e
+
2
3
{\displaystyle {\frac {4}{3}}<\int _{0}^{1}\mathrm {e} ^{x^{2}}\,\mathrm {d} x<{\frac {\mathrm {e} +2}{3}}}
Solution
f
{\displaystyle f}
R
+
{\displaystyle \mathbb {R} _{+}}
f
′
(
x
)
=
2
x
(
e
x
2
−
e
+
1
)
{\displaystyle f'(x)=2x(\mathrm {e} ^{x^{2}}-\mathrm {e} +1)}
a
:=
ln
(
e
−
1
)
{\displaystyle a:={\sqrt {\ln(\mathrm {e} -1)}}}
f
{\displaystyle f}
[
0
,
a
]
{\displaystyle [0,a]}
[
a
,
+
∞
[
{\displaystyle [a,+\infty [}
0
{\displaystyle 0}
1
{\displaystyle 1}
>
a
{\displaystyle >a}
ln
(
e
−
1
)
<
ln
e
{\displaystyle \ln(\mathrm {e} -1)<\ln \mathrm {e} }
f
(
a
)
=
e
a
2
−
(
e
−
1
)
a
2
−
1
=
(
e
−
1
)
(
1
−
ln
(
e
−
1
)
)
−
1
=
e
−
2
−
(
e
−
1
)
ln
(
e
−
1
)
{\displaystyle f(a)=\mathrm {e} ^{a^{2}}-(\mathrm {e} -1)a^{2}-1=(\mathrm {e} -1)\left(1-\ln(\mathrm {e} -1)\right)-1=\mathrm {e} -2-(\mathrm {e} -1)\ln(\mathrm {e} -1)}
lim
+
∞
f
=
+
∞
{\displaystyle \lim _{+\infty }f=+\infty }
∀
x
∈
]
0
,
1
[
f
(
x
)
<
0
{\displaystyle \forall x\in ]0,1[\quad f(x)<0}
∫
0
1
e
x
2
d
x
<
∫
0
1
(
(
e
−
1
)
x
2
+
1
)
d
x
=
e
−
1
3
+
1
=
e
+
2
3
{\displaystyle \int _{0}^{1}\mathrm {e} ^{x^{2}}\,\mathrm {d} x<\int _{0}^{1}\left((\mathrm {e} -1)x^{2}+1\right)\,\mathrm {d} x={\frac {\mathrm {e} -1}{3}}+1={\frac {\mathrm {e} +2}{3}}}
∀
x
∈
[
0
,
1
]
f
(
x
)
≥
f
(
a
)
{\displaystyle \forall x\in [0,1]\quad f(x)\geq f(a)}
∫
0
1
e
x
2
d
x
≥
e
+
2
3
+
f
(
a
)
=
(
e
−
1
)
(
4
3
−
ln
(
e
−
1
)
)
≃
1
,
36
>
4
3
{\displaystyle \int _{0}^{1}\mathrm {e} ^{x^{2}}\,\mathrm {d} x\geq {\frac {\mathrm {e} +2}{3}}+f(a)=(\mathrm {e} -1)\left({\frac {4}{3}}-\ln(\mathrm {e} -1)\right)\simeq 1{,}36>{\frac {4}{3}}}
Démontrer que :
1
,
13
<
∫
0
π
4
4
−
cos
2
t
3
d
t
<
1
,
20
{\displaystyle 1{,}13<\int _{0}^{\frac {\pi }{4}}{\sqrt[{3}]{4-\cos ^{2}t}}\,\mathrm {d} t<1{,}20}
![{\displaystyle f:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b592d102ccd1ba134d401c5b3ea177baaba3ffac)
![{\displaystyle f,g:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e569de0fc730b7a75b46b6fd57598e0c94712f91)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle x\in [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64a15936df283add394ab909aa7a5e24e7fb6bb2)
![{\displaystyle y:=x^{\alpha }\in [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51a7a302b201d0c3a9aaa33ad2dc7dbfeb4c39c9)
![{\displaystyle z:=1-y\in [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b71cc459749b17608761ea611ffbf05f7853828)
![{\displaystyle ]0,1[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f6a83a50a400fb17f0c9abe6e674c6526a7b0e1)
![{\displaystyle f:[0,1]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2de6d0d4c98d4ca7ad937c772dc3e3e914b062f5)
![{\displaystyle [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/738f7d23bb2d9642bab520020873cccbef49768d)
![{\displaystyle \int _{0}^{1}[f(x)]^{2}\,\mathrm {d} x\geq {\frac {1}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9bb2e2b0cf329fc67dd2aa97fbd9a84445ff78c)
![{\displaystyle \int _{0}^{1}xf(x)\,\mathrm {d} x+\int _{0}^{1}F(x)\,\mathrm {d} x=[xF]_{0}^{1}=F(1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7099a5b8a33f06b6a4a2b6235ea453b56374d2c)
![{\displaystyle {\frac {1}{3^{2}}}\leq \left(\int _{0}^{1}xf(x)\,\mathrm {d} x\right)^{2}\leq \int _{0}^{1}x^{2}\,\mathrm {d} x\int _{0}^{1}[f(x)]^{2}\,\mathrm {d} x={\frac {1}{3}}\int _{0}^{1}[f(x)]^{2}\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed00f0133914756c35ad9f41dec14ade677a8a2d)
![{\displaystyle \forall x\in [0,1]\quad 1\leq 1+x^{2}\leq 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0315c418714720d03622c7363139b725a73f0712)
![{\displaystyle [0,a]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63d050b0ffe6cc6f635808b9a013366a60e6d0c0)
![{\displaystyle \forall x\in ]0,1[\quad f(x)<0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d09758a8da24f00b9123864f8a58e0b815b6ce81)
![{\displaystyle \forall x\in [0,1]\quad f(x)\geq f(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32363495c5986f590fe92fa0914ca9512b311dd4)
![{\displaystyle 1{,}13<\int _{0}^{\frac {\pi }{4}}{\sqrt[{3}]{4-\cos ^{2}t}}\,\mathrm {d} t<1{,}20}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96b38f98f7a5751e3c740779db87990741ec5f93)
![{\displaystyle \left[0,{\frac {\pi }{4}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c416907085576976c05d6a7a6716f23c8e448583)
![{\displaystyle t\mapsto {\sqrt[{3}]{4-\cos ^{2}t}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab41aad15f196c41a0d5854b531f0ff69b5520a8)
![{\displaystyle {\sqrt[{3}]{4-\cos ^{2}0}}={\sqrt[{3}]{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5da80b9fe90a74d162ffad7fc0f68b5087187110)
![{\displaystyle {\sqrt[{3}]{4-\cos ^{2}{\frac {\pi }{4}}}}={\sqrt[{3}]{\frac {7}{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b56f4cfceabbca33ff515f430ae1d2c6a6fe836)
![{\displaystyle {\frac {\pi }{4}}{\sqrt[{3}]{3}}\approx 1{,}13}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb9b6701a4d84c07c9dbdeec4f744b9d80237bb8)
![{\displaystyle {\frac {\pi }{4}}{\sqrt[{3}]{\frac {7}{2}}}\approx 1{,}20}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cf61408a8816abac5e0a381d3d863eab3984eed)
