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Solution:

∫ (1 + √(9 - (x - 3)^2))dx [0,3]

= ∫ (1 + √(x)*√(6-x))dx [0,3]

Δx = 3/n

Ʃ√(3i/n)*Ʃ√(6 - 3i/n) +Ʃ 1 i = 1, n = n

I am stuck right here... I know that the notation for a constant is 1(n)..

but is there one for a square root? Would is be (n(n+1)/2)^(1/2) ???