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Question Number 70818 by oyemi kemewari last updated on 08/Oct/19
what the prove that  ∫_a ^b f(x) dx =∫_a ^b f(a+b−x) dx
whattheprovethatabf(x)dx=abf(a+bx)dx
Commented by kaivan.ahmadi last updated on 08/Oct/19
u=a+b−x⇒du=−dx   { ((x=a⇒u=b)),((x=b⇒u=a)) :}  ⇒∫_a ^b f(a+b−x)dx=∫_b ^a −f(u)du=∫_a ^b f(u)du=∫_a ^b f(x)dx
u=a+bxdu=dx{x=au=bx=bu=aabf(a+bx)dx=baf(u)du=abf(u)du=abf(x)dx
Answered by MJS last updated on 08/Oct/19
∫_a ^b f(x)dx=[F(x)]_a ^b =F(b)−F(a)  ∫_a ^b f(a+b−x)dx=       [t=a+b−x ⇒ dx=−dt]  =−∫_b ^a f(t)dt=−[F(t)]_b ^a =[F(t)]_a ^b =F(b)−F(a)
baf(x)dx=[F(x)]ab=F(b)F(a)baf(a+bx)dx=[t=a+bxdx=dt]=abf(t)dt=[F(t)]ba=[F(t)]ab=F(b)F(a)

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