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En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Intégrales 1Intégration en mathématiques/Exercices/Intégrales 1 », n'a pu être restituée correctement ci-dessus.
∫
0
1
x
(
x
2
+
1
)
5
d
x
{\displaystyle \int _{0}^{1}x(x^{2}+1)^{5}\,\mathrm {d} x}
Solution
∫
0
1
x
(
x
2
+
1
)
5
d
x
=
1
2
∫
1
2
y
5
d
y
=
2
6
−
1
6
12
=
21
4
{\displaystyle \int _{0}^{1}x(x^{2}+1)^{5}\,\mathrm {d} x={\frac {1}{2}}\int _{1}^{2}y^{5}\,\mathrm {d} y={\frac {2^{6}-1^{6}}{12}}={\frac {21}{4}}}
∫
2
7
(
x
3
−
7
x
+
15
)
d
x
{\displaystyle \int _{2}^{7}(x^{3}-7x+15)\,\mathrm {d} x}
Solution
∫
2
7
(
x
3
−
7
x
+
15
)
d
x
=
7
4
−
2
4
4
−
7
7
2
−
2
2
2
+
15
(
7
−
2
)
=
2385
4
−
315
2
+
75
=
2055
4
{\displaystyle \int _{2}^{7}(x^{3}-7x+15)\,\mathrm {d} x={\frac {7^{4}-2^{4}}{4}}-7{\frac {7^{2}-2^{2}}{2}}+15(7-2)={\frac {2385}{4}}-{\frac {315}{2}}+75={\frac {2055}{4}}}
∫
0
1
(
4
x
+
2
)
(
x
2
+
x
−
2
)
2
d
x
{\displaystyle \int _{0}^{1}(4x+2)(x^{2}+x-2)^{2}\,\mathrm {d} x}
Solution
∫
0
1
(
4
x
+
2
)
(
x
2
+
x
−
2
)
2
d
x
=
2
∫
−
2
0
y
2
d
y
=
16
3
{\displaystyle \int _{0}^{1}(4x+2)(x^{2}+x-2)^{2}\,\mathrm {d} x=2\int _{-2}^{0}y^{2}\,\mathrm {d} y={\frac {16}{3}}}
∫
0
2
x
2
(
x
3
−
8
)
d
x
{\displaystyle \int _{0}^{2}x^{2}(x^{3}-8)\,\mathrm {d} x}
Solution
∫
0
2
x
2
(
x
3
−
8
)
d
x
=
1
3
∫
−
8
0
y
d
y
=
−
32
3
{\displaystyle \int _{0}^{2}x^{2}(x^{3}-8)\,\mathrm {d} x={\frac {1}{3}}\int _{-8}^{0}y\,\mathrm {d} y=-{\frac {32}{3}}}
∫
0
1
d
t
t
2
+
2
t
+
1
{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} t}{t^{2}+2t+1}}}
Solution
∫
0
1
d
t
(
t
+
1
)
2
=
[
−
1
t
+
1
]
0
1
=
1
2
{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} t}{(t+1)^{2}}}=\left[{\frac {-1}{t+1}}\right]_{0}^{1}={\frac {1}{2}}}
∫
0
1
x
d
x
(
x
2
+
1
)
3
{\displaystyle \int _{0}^{1}{\frac {x\,\mathrm {d} x}{\sqrt {(x^{2}+1)^{3}}}}}
Solution
∫
0
1
x
d
x
(
x
2
+
1
)
3
=
1
2
∫
1
2
y
−
3
/
2
d
y
=
1
−
2
−
1
/
2
{\displaystyle \int _{0}^{1}{\frac {x\,\mathrm {d} x}{\sqrt {(x^{2}+1)^{3}}}}={\frac {1}{2}}\int _{1}^{2}y^{-3/2}\,\mathrm {d} y=1-2^{-1/2}}
∫
−
1
0
2
x
+
3
(
x
2
+
3
x
−
1
)
2
d
x
{\displaystyle \int _{-1}^{0}{\frac {2x+3}{(x^{2}+3x-1)^{2}}}\,\mathrm {d} x}
Solution
∫
−
1
0
2
x
+
3
(
x
2
+
3
x
−
1
)
2
d
x
=
∫
−
3
−
1
d
y
y
2
=
1
−
1
3
=
2
3
{\displaystyle \int _{-1}^{0}{\frac {2x+3}{(x^{2}+3x-1)^{2}}}\,\mathrm {d} x=\int _{-3}^{-1}{\frac {\mathrm {d} y}{y^{2}}}=1-{\frac {1}{3}}={\frac {2}{3}}}
∫
−
4
4
d
x
(
x
+
5
)
3
{\displaystyle \int _{-4}^{4}{\frac {\mathrm {d} x}{\sqrt {(x+5)^{3}}}}}
Solution
∫
−
4
4
(
x
+
5
)
−
3
/
2
d
x
=
−
2
[
(
x
+
5
)
−
1
/
2
]
−
4
4
=
2
(
1
−
9
−
1
/
2
)
=
4
3
{\displaystyle \int _{-4}^{4}(x+5)^{-3/2}\,\mathrm {d} x=-2\left[(x+5)^{-1/2}\right]_{-4}^{4}=2(1-9^{-1/2})={\frac {4}{3}}}
∫
0
1
3
x
+
1
x
2
+
1
d
x
{\displaystyle \int _{0}^{1}{\frac {3x+1}{\sqrt {x^{2}+1}}}\;\mathrm {d} x}
Solution
∫
0
1
3
x
+
1
x
2
+
1
d
x
=
[
3
2
ln
(
x
2
+
1
)
+
arctan
x
]
0
1
=
3
2
ln
2
+
π
4
{\displaystyle \int _{0}^{1}{\frac {3x+1}{\sqrt {x^{2}+1}}}\;\mathrm {d} x=\left[{\frac {3}{2}}\ln(x^{2}+1)+\arctan x\right]_{0}^{1}={\frac {3}{2}}\ln 2+{\frac {\pi }{4}}}
∫
0
1
x
−
2
(
2
x
−
3
)
2
d
x
{\displaystyle \int _{0}^{1}{\frac {x-2}{(2x-3)^{2}}}\;\mathrm {d} x}
∫
−
1
0
x
x
2
+
2
x
−
3
d
x
{\displaystyle \int _{-1}^{0}{\frac {x}{x^{2}+2x-3}}\;\mathrm {d} x}
![{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} t}{(t+1)^{2}}}=\left[{\frac {-1}{t+1}}\right]_{0}^{1}={\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d9301872a5a90a7a67144a593add616d39427cf)
![{\displaystyle \int _{-4}^{4}(x+5)^{-3/2}\,\mathrm {d} x=-2\left[(x+5)^{-1/2}\right]_{-4}^{4}=2(1-9^{-1/2})={\frac {4}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47b415131281a2cc8e674028934a7f68ca5f843c)
![{\displaystyle \int _{0}^{1}{\frac {3x+1}{\sqrt {x^{2}+1}}}\;\mathrm {d} x=\left[{\frac {3}{2}}\ln(x^{2}+1)+\arctan x\right]_{0}^{1}={\frac {3}{2}}\ln 2+{\frac {\pi }{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2307de634fed626bf8092bfef23ce1cc8d248080)
![{\displaystyle \int _{0}^{1}\left({\frac {1/4}{x-3/2}}-{\frac {1/8}{(x-3/2)^{2}}}\right)\;\mathrm {d} x=\left[{\frac {1}{4}}\ln |x-3/2|+{\frac {1}{8(x-3/2)}}\right]_{0}^{1}=-{\frac {\ln 3}{4}}-{\frac {1}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f702a184ac6528e2e0b7a6b039cdde7e0c89134)
![{\displaystyle \int _{-1}^{0}\left({\frac {1/4}{x-1}}+{\frac {3/4}{x+3}}\right)\;\mathrm {d} x={\frac {1}{4}}\left[3\ln |x+3|+\ln |x-1|\right]_{-1}^{0}={\frac {3\ln 3}{4}}-\ln 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b8ff8b69cf8ae2d78a544779cc546762bf8537b)
