求极限lim(t->0+) 1/t^2 ∫(0~t)dx∫(0~t-x)e^(x^2+y^2)dy,高手帮个忙,
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求极限lim(t->0+) 1/t^2 ∫(0~t)dx∫(0~t-x)e^(x^2+y^2)dy,高手帮个忙,
![求极限lim(t->0+) 1/t^2 ∫(0~t)dx∫(0~t-x)e^(x^2+y^2)dy,高手帮个忙,](/uploads/image/z/7999054-70-4.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90lim%28t-%3E0%2B%29+1%2Ft%5E2+%E2%88%AB%280%7Et%29dx%E2%88%AB%280%7Et-x%29e%5E%28x%5E2%2By%5E2%29dy%2C%E9%AB%98%E6%89%8B%E5%B8%AE%E4%B8%AA%E5%BF%99%2C)
f(t²) = ∫[0,t] dx ∫[0,t-x] e^(x²+y²) dy 化为极坐标D:0≤ r ≤ t / (sinθ+cosθ),0≤θ≤ π/2
= ∫[0,π/2] dθ ∫[0,t /(sinθ+cosθ)] e^(r²) r dr
= ∫[0,π/2] (1/2) [ e^ { t² /(sinθ+cosθ)² } ﹣1] dθ
令 u = t²,原式 = lim(u->0+) f (u) / u
f '(u) = ∫[0,π/2] (1/2) e^ {u / (sinθ+cosθ)² } * 1/(sinθ+cosθ)² dθ
原式 = lim(u->0+) f ' (u) 洛必达法则
= lim(u->0) (1/2) ∫[0,π/2] e^ {u / (sinθ+cosθ)² } * 1/(sinθ+cosθ)² dθ
= (1/2) ∫[0,π/2] 1/(sinθ+cosθ)² dθ 第二积分中值定理
= (1/2) (-1/2) cot(θ+π/4) | [0,π/2]
= 1/2
= ∫[0,π/2] dθ ∫[0,t /(sinθ+cosθ)] e^(r²) r dr
= ∫[0,π/2] (1/2) [ e^ { t² /(sinθ+cosθ)² } ﹣1] dθ
令 u = t²,原式 = lim(u->0+) f (u) / u
f '(u) = ∫[0,π/2] (1/2) e^ {u / (sinθ+cosθ)² } * 1/(sinθ+cosθ)² dθ
原式 = lim(u->0+) f ' (u) 洛必达法则
= lim(u->0) (1/2) ∫[0,π/2] e^ {u / (sinθ+cosθ)² } * 1/(sinθ+cosθ)² dθ
= (1/2) ∫[0,π/2] 1/(sinθ+cosθ)² dθ 第二积分中值定理
= (1/2) (-1/2) cot(θ+π/4) | [0,π/2]
= 1/2
求极限lim(t->0+) 1/t^2 ∫(0~t)dx∫(0~t-x)e^(x^2+y^2)dy,高手帮个忙,
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