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It is proved that if the function f (x, y) is continuous on the bounded closed domain D, the function g (x, y) is integrable on D, and G (x, y) ≥ 0, (x, y) belongs to D, then at least one point (a, b) belongs to D, such that ∫ (domain D) f (x, y) g (x, y) d Δ = f (a, b) ∫ (domain D) g (x, y) d Δ = f (a, b) ∫ (domain D) g (x, y) d Δ
Because f (x, y) is continuous over a bounded closed domain D, there exists a minimum m and a maximum m for F;
Then m * ∫ (region D) g (x, y) d Δ=
Logx (x + 1), how to derive its monotonicity
Y = logx (x + 1) = ln (x + 1) / LNX, (x > 0 and X ≠ 1) y '= {(LNX) / (x + 1) - [ln (x + 1)] / X} / ln & # 178; X = [xlnx - (x + 1) ln (x + 1)] / [x (x + 1) ln & # 178; x] Let G (x) = xlnx, then G' (x) = LNX + 1, let g '(x) > 0, the solution is x > 1 / E, so G (x) is an increasing function at (1 / E, + ∞). (1) if
Finding the derivative of F (x) = logx (LNX)
If y = log x (LNX) = > log x (x ^ y) = log x (LNX) = > x ^ y = LNX, y ^ (Y-1) * y '= 1 / x = > y' = x ^ (- 1-y + 1) * y = YX ^ y, y = log x (LNX) is brought into y '= [log x (LNX)] * x ^ (log x (LNX))
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