definite integration

2011-03-10 1:25 am
a) Show that ∫f(a-x) =∫f(x) upper limit is a lower limit is 0b) Hence calculate ∫[(cosx)^6]/{[(cosx)^6]+[(sinx)^6]} upper limit is (π/2) lower limit is 0

回答 (1)

2011-03-10 2:26 am
✔ 最佳答案
a) let u = a-x, ∴ du = -dxwhen x = a, u = 0when x = 0, u = a∫f(a-x) dx upper limit is a lower limit is 0 = ∫f(u) (-du) upper limit is 0 lower limit is a= ∫f(u) du upper limit is a lower limit is 0= ∫f(x)dx upper limit is a lower limit is 0 b)by substituting u = π/2 - x∫[(cosx)^6]/{[(cosx)^6]+[(sinx)^6]} dxupper limit is (π/2) lower limit is 0=∫[(sinx)^6]/{[(cosx)^6]+[(sinx)^6]} dxupper limit is (π/2) lower limit is 0=∫[(cosx)^6]/{[(cosx)^6]+[(sinx)^6]} dxupper limit is (π/2) lower limit is 0 = 1/2 ∫{[(cosx)^6]+[(sinx)^6]} /{[(cosx)^6]+[(sinx)^6]} dx upper limit is (π/2) lower limit is 0 = 1/2 ∫dxupper limit is (π/2) lower limit is 0 = π/4


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