我有一個數學問題 想請大家幫一幫忙(微積分面積題)

日本大學的數學題 可能比較難一點 希望數學好的大大可以幫一幫忙

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x的函數f(x)=xe^-x^2 但e是自然數的底

f(x)最大的時候x的值為x=a,即a=√__/__,那個最大值為1/√__e

函數y=f(x)的圖像與x軸,且和直線x=a的所包圍的面積為S

S=1/__-1/__√e

t>0 ,函數y=f(x)的圖像與x軸,且直線x=t所包圍的部分的面積為S(t)

lim_t->∞ S(t)=1/__

請問有大大會算這條題嗎 希望過程可以寫得詳細一點 先謝謝了

1 Answer

Rating
  • linch
    Lv 7
    1 decade ago
    Favorite Answer

    f(x) = xe^(-x^2),

    lim_{x-> - ∞} f(x) = 0, lim_{x->∞} f(x) = 0,

    f'(x) = e^(-x^2) + x*e^(-x^2)*(-2x) = (1 - 2x^2)e^(-x^2)

    f'(x) = 0 ==> 1 - 2x^2 = 0 ==> x = 1/√2, -1/√2 ( √2 / 2), - √2 / 2) )

    f'(x) > 0 if x on (- √2 / 2), √2 / 2) ), f'(x) < 0 if x on (- ∞, -√2 / 2)) and (√2 / 2), ∞ ).

    That is, f is increasing on (- √2 / 2), √2 / 2) ) and drcreasing on

    (- ∞, -√2 / 2)) and (√2 / 2), ∞ ).

    Hence f(√2 / 2)) = √2 / 2) * e^(-1/2) = 1/ √(2e) is absolute maximum value.

    f( - √2/ 2)) = - √2 / 2) * e^(-1/2) = - 1/ √(2e) is absolute minimum value.

    ( a = (√2 / 2) 時有最大值 1/ √(2e) )

    ∫f(x) dx = ∫xe^(-x^2) dx, let u = -x^2, du = -2x dx

    =∫(- 1 / 2) e^u du = (- 1 / 2) e^u + C = ( - 1 / 2) e^(-x^2) + C

    函數y = f(x)的圖像與x軸(上方??), 且和直線x = a的所包圍的面積為S

    S =∫_[0,√2 / 2] f(x) dx (下限 0, 上限 √2 / 2)

    = ( - 1 / 2) e^(-x^2) |_[0,√2 / 2]

    = - (1/2) e^(-1/2) + 1/2

    = 1/2 - 1/(2√e)

    S(t) =∫_[0, t] f(x) dx (下限 0, 上限 t)

    = ( - 1 / 2) e^(-x^2) |_[0, t]

    = - (1/2) e^(-t^2) + 1/2

    = 1/2 - (1/2) e^(-t^2)

    lim_{t -> ∞} S(t) = 1/2 (Hint: lim_{t -> ∞} e^(-t^2) = 0 )

    O = P = Q = R = S = T = 2 ( 所有空格答案都是 2 )

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