Set I(a, b) = 1 0 x a (1 x) b dx, where a, b
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Set I(a, b) = ∫10xa(1 − x)b dx, where a, b are whole numbers.
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(a) Use substitution to show that I(a, b) = 1(b, a). 1 (b) Show that I(a,0) = I(0, a) = a + 1 (c) Prove that for a ≥ 1 and b≥ 0, (e) Show that I(a, b) = I(a, b) = a b + 1 (d) Use (b) and (c) to calculate I(1, 1) and I(3, 2). a! b! (a + b + 1)! I(a- 1,b + 1)
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a Let u 1 x Then du dx and L u Ia b b 1a0 10 a by part a Further d Ia0 c Using Integration by ...View the full answer
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