[tex]Montrer \\ en \\ posant \\ y= a+b-x , , que [/tex] \\ [tex] \int\limits^b_a {x} f(x)\, dx= \frac{a+b}{2}* \int\limits^b_a f(x) \, dx [/tex] [tex] \

je vais "deviner" l'énoncé complet :
Montrer que [tex] \int\limits^a_b {x.f(x)} \, dx = \frac{a+b}{2} . \int\limits^a_b {f(x)} \, dx [/tex]

Preuve :
[tex] \int\limits^a_b {x.f(x)} \, dx = \int\limits^b_a {(a+b-y).f(a+b-y)} \, (-dy)[/tex]
[tex] \int\limits^a_b {x.f(x)} \, dx= \int\limits^a_b {(a+b-y).f(a+b-y)} \, dy [/tex]
[tex] \int\limits^a_b {x.f(x)} \, dx=(a+b). \int\limits^a_b {f(a+b-y)} \, dy- \int\limits^a_b {y.f(a+b-y)} \, dy [/tex]
[tex] \int\limits^a_b {x.f(x)} \, dx =(a+b). \int\limits^a_b {f(x)} \, dx - \int\limits^a_b {x.f(x)} \, dx [/tex]
[tex] \int\limits^a_b {2x.f(x)} \, dx =(a+b). \int\limits^a_b {f(x)} \, dx [/tex]
[tex] \int\limits^a_b {x.f(x)} \, dx = \frac{a+b}{2}. \int\limits^a_b {f(x)} \, dx [/tex]