Intégration par Parties

Intégration par Parties (IPP)

Théorème

Soient $u$ et $v$ deux fonctions de classe $\mathcal{C}^1$ sur $[a ; b]$. On a :

$$\int_{a}^{b} u(x)\,v'(x)\,dx = \Big[u(x)\,v(x)\Big]_{a}^{b} - \int_{a}^{b} u'(x)\,v(x)\,dx$$

Origine

C'est la transposition de la formule de dérivation d'un produit :

$$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$

En intégrant les deux membres sur $[a ; b]$, on obtient la formule d'IPP.

Méthode

1. Choisir $u$ et $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$0 (on dérive $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$1, on primitive $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$2)
2. Calculer $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$3 et $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$4
3. Appliquer la formule

Règle de choix (LIATE)

Ordre de priorité pour le choix de $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$5 :
- Logarithme, Inverse trigo, Algébrique, Trigo, Exponentielle

Exemples

Exemple 1 : $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$6

On pose $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$7, $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$8, donc $$(uv)' = u'v + uv' \implies uv' = (uv)' - u'v$$9, $$\int_{0}^{1} x\,e^x\,dx = \Big[x\,e^x\Big]_{0}^{1} - \int_{0}^{1} e^x\,dx = e - \Big[e^x\Big]_{0}^{1} = e - (e - 1) = 1$$0.

$$\int_{0}^{1} x\,e^x\,dx = \Big[x\,e^x\Big]_{0}^{1} - \int_{0}^{1} e^x\,dx = e - \Big[e^x\Big]_{0}^{1} = e - (e - 1) = 1$$

Exemple 2 : $$\int_{0}^{1} x\,e^x\,dx = \Big[x\,e^x\Big]_{0}^{1} - \int_{0}^{1} e^x\,dx = e - \Big[e^x\Big]_{0}^{1} = e - (e - 1) = 1$$1

On pose $$\int_{0}^{1} x\,e^x\,dx = \Big[x\,e^x\Big]_{0}^{1} - \int_{0}^{1} e^x\,dx = e - \Big[e^x\Big]_{0}^{1} = e - (e - 1) = 1$$2, $$\int_{0}^{1} x\,e^x\,dx = \Big[x\,e^x\Big]_{0}^{1} - \int_{0}^{1} e^x\,dx = e - \Big[e^x\Big]_{0}^{1} = e - (e - 1) = 1$$3, donc $$\int_{0}^{1} x\,e^x\,dx = \Big[x\,e^x\Big]_{0}^{1} - \int_{0}^{1} e^x\,dx = e - \Big[e^x\Big]_{0}^{1} = e - (e - 1) = 1$$4, $$\int_{0}^{1} x\,e^x\,dx = \Big[x\,e^x\Big]_{0}^{1} - \int_{0}^{1} e^x\,dx = e - \Big[e^x\Big]_{0}^{1} = e - (e - 1) = 1$$5.

$$\int_{1}^{e} \ln(x)\,dx = \Big[x\ln(x)\Big]_{1}^{e} - \int_{1}^{e} 1\,dx = e - (e - 1) = 1$$