En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Intégrales 6Intégration en mathématiques/Exercices/Intégrales 6 », n'a pu être restituée correctement ci-dessus.
∫
0
1
x
3
e
x
2
d
x
{\displaystyle \int _{0}^{1}{\frac {x}{3}}\operatorname {e} ^{x^{2}}\,\mathrm {d} x}
Solution
[
e
x
2
6
]
0
1
=
e
−
1
6
{\displaystyle \left[{\frac {\mathrm {e} ^{x^{2}}}{6}}\right]_{0}^{1}={\frac {\mathrm {e} -1}{6}}}
∫
0
π
2
5
sin
t
e
cos
t
d
t
{\displaystyle \int _{0}^{\frac {\pi }{2}}5\sin t\operatorname {e} ^{\cos t}\,\mathrm {d} t}
Solution
[
−
5
e
cos
t
]
0
π
/
2
=
5
(
e
−
1
)
{\displaystyle \left[-5\operatorname {e} ^{\cos t}\right]_{0}^{\pi /2}=5(\mathrm {e} -1)}
∫
−
1
0
(
x
2
+
x
)
e
2
x
d
x
{\displaystyle \int _{-1}^{0}\left(x^{2}+x\right)\operatorname {e} ^{2x}\,\mathrm {d} x}
Solution
[
x
2
e
2
x
2
]
−
1
0
=
−
1
2
e
2
{\displaystyle \left[{\frac {x^{2}\operatorname {e} ^{2x}}{2}}\right]_{-1}^{0}=-{\frac {1}{2\mathrm {e} ^{2}}}}
∫
1
3
1
(
3
t
2
−
4
t
+
1
)
e
t
d
t
{\displaystyle \int _{\frac {1}{3}}^{1}\left(3t^{2}-4t+1\right)\operatorname {e} ^{t}\,\mathrm {d} t}
Solution
[
(
(
3
t
2
−
4
t
+
1
)
−
(
6
t
−
4
)
+
6
)
e
t
]
1
/
3
1
=
[
(
3
t
2
−
10
t
+
11
)
e
t
]
1
/
3
1
=
4
e
−
8
e
3
{\displaystyle \left[\left(\left(3t^{2}-4t+1\right)-\left(6t-4\right)+6\right)\operatorname {e} ^{t}\right]_{1/3}^{1}=\left[\left(3t^{2}-10t+11\right)\operatorname {e} ^{t}\right]_{1/3}^{1}=4\operatorname {e} -8{\sqrt[{3}]{\mathrm {e} }}}
∫
1
2
cos
(
ln
t
)
t
d
t
{\displaystyle \int _{1}^{2}{\frac {\cos(\ln t)}{t}}\,\mathrm {d} t}
Solution
∫
0
ln
2
cos
=
[
sin
]
0
ln
2
=
sin
(
ln
2
)
{\displaystyle \int _{0}^{\ln 2}\cos =\left[\sin \right]_{0}^{\ln 2}=\sin \left(\ln 2\right)}
∫
π
/
3
π
/
2
tan
(
x
/
2
)
ln
(
1
+
cos
x
)
d
x
{\displaystyle \int _{\pi /3}^{\pi /2}\tan(x/2)\ln(1+\cos x)\,\mathrm {d} x}
Solution
[
−
ln
2
(
1
+
cos
)
2
]
π
/
3
π
/
2
=
ln
2
(
3
/
2
)
2
{\displaystyle \left[{\frac {-\ln ^{2}(1+\cos )}{2}}\right]_{\pi /3}^{\pi /2}={\frac {\ln ^{2}(3/2)}{2}}}
∫
0
1
x
2
e
−
x
3
d
x
{\displaystyle \int _{0}^{1}x^{2}\operatorname {e} ^{-x^{3}}\,\mathrm {d} x}
Solution
[
−
e
−
x
3
3
]
0
1
=
1
−
e
−
1
3
{\displaystyle \left[{\frac {-\mathrm {e} ^{-x^{3}}}{3}}\right]_{0}^{1}={\frac {1-\mathrm {e} ^{-1}}{3}}}
∫
0
e
−
1
ln
(
1
+
x
)
d
x
{\displaystyle \int _{0}^{\mathrm {e} -1}\ln(1+x)\,\mathrm {d} x}
Solution
[
t
ln
t
−
t
]
1
e
=
1
{\displaystyle \left[t\ln t-t\right]_{1}^{\mathrm {e} }=1}
∫
1
e
d
t
t
1
−
(
ln
t
)
2
{\displaystyle \int _{1}^{\mathrm {e} }{\frac {\mathrm {d} t}{t{\sqrt {1-(\ln t)^{2}}}}}}
Solution
∫
0
1
d
x
1
−
x
2
=
∫
0
π
/
2
cos
t
d
t
1
−
sin
2
t
=
∫
0
π
/
2
d
t
=
π
2
{\displaystyle \int _{0}^{1}{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\int _{0}^{\pi /2}{\frac {\cos t\,\mathrm {d} t}{\sqrt {1-\sin ^{2}t}}}=\int _{0}^{\pi /2}\mathrm {d} t={\frac {\pi }{2}}}
∫
0
1
(
x
e
x
+
x
2
−
ln
(
1
+
x
)
)
d
x
{\displaystyle \int _{0}^{1}\left(x\operatorname {e} ^{x}+x^{2}-\ln(1+x)\right)\,\mathrm {d} x}
Solution
[
x
e
x
]
0
1
−
∫
0
1
e
x
d
x
+
[
x
3
/
3
]
0
1
−
[
t
ln
t
−
t
]
1
2
=
e
−
(
e
−
1
)
+
1
3
−
2
ln
2
+
1
=
7
3
−
2
ln
2
{\displaystyle [x\operatorname {e} ^{x}]_{0}^{1}-\int _{0}^{1}\operatorname {e} ^{x}\,\mathrm {d} x+[x^{3}/3]_{0}^{1}-[t\ln t-t]_{1}^{2}=\mathrm {e} -(\mathrm {e} -1)+{\frac {1}{3}}-2\ln 2+1={\frac {7}{3}}-2\ln 2}
∫
1
2
ln
(
1
+
x
)
x
2
d
x
{\displaystyle \int _{1}^{2}{\frac {\ln(1+x)}{x^{2}}}\,\mathrm {d} x}
Solution
[
−
ln
(
1
+
x
)
x
]
1
2
+
∫
1
2
d
x
x
(
1
+
x
)
=
−
ln
3
2
+
ln
2
+
∫
1
2
(
1
x
−
1
x
+
1
)
d
x
=
−
ln
3
2
+
ln
2
+
[
ln
x
x
+
1
]
1
2
=
−
3
ln
3
2
+
3
ln
2
{\displaystyle \left[-{\frac {\ln(1+x)}{x}}\right]_{1}^{2}+\int _{1}^{2}{\frac {\mathrm {d} x}{x(1+x)}}=-{\frac {\ln 3}{2}}+\ln 2+\int _{1}^{2}\left({\frac {1}{x}}-{\frac {1}{x+1}}\right)\,\mathrm {d} x=-{\frac {\ln 3}{2}}+\ln 2+\left[\ln {\frac {x}{x+1}}\right]_{1}^{2}=-{\frac {3\ln 3}{2}}+3\ln 2}
![{\displaystyle \left[{\frac {\mathrm {e} ^{x^{2}}}{6}}\right]_{0}^{1}={\frac {\mathrm {e} -1}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb5134eee85cb8a6900a58f3fec90ca5da1e2335)
![{\displaystyle \left[-5\operatorname {e} ^{\cos t}\right]_{0}^{\pi /2}=5(\mathrm {e} -1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5750aa638bcb1b0ca04630af983cfa55404bf493)
![{\displaystyle \left[{\frac {x^{2}\operatorname {e} ^{2x}}{2}}\right]_{-1}^{0}=-{\frac {1}{2\mathrm {e} ^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a91f328ff91a16531094116b79b355835c8e0c7d)
![{\displaystyle \left[\left(\left(3t^{2}-4t+1\right)-\left(6t-4\right)+6\right)\operatorname {e} ^{t}\right]_{1/3}^{1}=\left[\left(3t^{2}-10t+11\right)\operatorname {e} ^{t}\right]_{1/3}^{1}=4\operatorname {e} -8{\sqrt[{3}]{\mathrm {e} }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56069e77965543d35d15cb31a1b8b084fbaff2ad)
![{\displaystyle \int _{0}^{\ln 2}\cos =\left[\sin \right]_{0}^{\ln 2}=\sin \left(\ln 2\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/523b5363ff113e447618b3aa9f3ca86cb53bdf9d)
![{\displaystyle \left[{\frac {-\ln ^{2}(1+\cos )}{2}}\right]_{\pi /3}^{\pi /2}={\frac {\ln ^{2}(3/2)}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fd7835283fe69ac5203598309b4f4a119a6d4db)
![{\displaystyle \left[{\frac {-\mathrm {e} ^{-x^{3}}}{3}}\right]_{0}^{1}={\frac {1-\mathrm {e} ^{-1}}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2ac30496d33a850fbeca9052d1be7723166ac60)
![{\displaystyle \left[t\ln t-t\right]_{1}^{\mathrm {e} }=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9ce6d73f6c5f6dd0358565a69b664af910a0f66)
![{\displaystyle [x\operatorname {e} ^{x}]_{0}^{1}-\int _{0}^{1}\operatorname {e} ^{x}\,\mathrm {d} x+[x^{3}/3]_{0}^{1}-[t\ln t-t]_{1}^{2}=\mathrm {e} -(\mathrm {e} -1)+{\frac {1}{3}}-2\ln 2+1={\frac {7}{3}}-2\ln 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1cd7694f458d82847d1fea8a9b4c081cfc2c349)
![{\displaystyle \left[-{\frac {\ln(1+x)}{x}}\right]_{1}^{2}+\int _{1}^{2}{\frac {\mathrm {d} x}{x(1+x)}}=-{\frac {\ln 3}{2}}+\ln 2+\int _{1}^{2}\left({\frac {1}{x}}-{\frac {1}{x+1}}\right)\,\mathrm {d} x=-{\frac {\ln 3}{2}}+\ln 2+\left[\ln {\frac {x}{x+1}}\right]_{1}^{2}=-{\frac {3\ln 3}{2}}+3\ln 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/def020f4ae87f0a7f1e89094075bad9885180900)
