討論串[考古] 95成大
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1. ∫ (x^(-2/3))(lnx) dx. 0. 1. = lim ∫ (x^(-2/3))(lnx) dx. R→0+ R. |1 1 1. = lim (3)(x^(1/3))(lnx) | - ∫ (3)(x^(1/3))(---) dx. R→0+ |R R x. (令 u = lnx
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令 x = (r)(cosθ) , y = (r)(sinθ) , 則 |J| = r. 0 ≦ r ≦ 1 , 0 ≦ θ ≦ π/2. 1 √(1 - x^2). ∫ ∫ sin(x^2 + y^2) dydx. 0 0. π/2 1. = ∫ ∫ (sin(r^2))(r) drdθ. 0 0
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令 g(x,y,z) = x^2 + y^2 + z^2 - 14. F(x,y,z) = f(x,y,z) + (λ)(g(x,y,z)). = x + 3y - 2z + (λ)(x^2 + y^2 + z^2 - 14). Fx = 1 + (2λ)(x) = 0 => λ = -1/(2x)
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3x^(1/3)lnx - S 3x^(-2/3)dx. = 0 - 0 - [9] = -9. P.S. lim lnx/x^(-2/3) = lim 1/x * (-3/2) * x^(5/3) = 0 Pi/2 1. = S S sin(r^2)rdrdt. 0 0. = Pi/4 * (-c
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1.Compute the following limit:. １ ｎ. (a) lim － Σ √[(3i/n)^7+(3i/n)^4]. n→∞ ｎ i=1. X 1. (b) lim [ (-----)-(-----)].. n→∞ lnX X-1. x. e ln(x+e) 1. 2.Let
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