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  • 2道關于反導函數(shù)的數(shù)學題

    2道關于反導函數(shù)的數(shù)學題

     f ''(t) = 4e^t + 8 sin t,    f(0) = 0,    f(π) = 0,求f



    f '''(x) = cos x,    f(0) = 9,    f '(0) = 8,    f ''(0) = 6,求f


    數(shù)學人氣:706 ℃時間:2020-04-26 17:15:32
    優(yōu)質解答

    f'(t) = ∫f"(t)*dt = ∫(4*e^t + 8*sint)dt = 4*e^t - 8*cost + C1
    f(t) = ∫f'(t)*dt = ∫(4*e^t - 8*cost + C1)*dt = 4*e^t - 8*sint + C1*t + C2
    當 t = 0時,f(0) = 4*e^0 - 8*sin0 + C1*0 + C2 = 4 + C2 = 0,則 C2 =-4
    當 t = π時,f(π) = 4*e^π - 8*sinπ + C1*π + C2 = 4*e^π + C1*π - 4 =0,則 C1= (4 - 4*e^π)/π
    所以,f(t) = 4*e^t - 8*sint + (4-4*e^π)*t/π - 4

    f"(x) =∫f'"(x)dx = sinx + C1,當 x = 0 時,f"(0) = 0 + C1=6,則 C1=6,f"(x) = sinx + 6
    f'(x) = ∫f"(x)dx = -cosx + 6x + C2,當 x =0 時,f'(0) = -1 + 6*0 + C2 =8,則 C2 = 9
    f'(x) = -cosx + 6x + 9
    f(x) =∫f'(x)dx = -sinx + 3x^2 + 9x + C3,當 x= 0時,f(0)= -0 + 3*0 + 9*0 + C3 = 9,則 C3=9
    f(x) = -sinx + 3x^2 + 9x + 9
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