2道关于反导函数的数学题 f ''(t) = 4e^t + 8 sin t,    f(0) = 0,    f(π) = 0,求ff '''(x) = cos x,    f(0) = 9,    f '(0) = 8,    f ''(0) = 6,求f

逐次积分即可：
1） f ''(t) = 4e^t + 8 sin t,
积分：f'(t)=4e^t-8cost+C1
再积分：f(t)=4e^t-8sint+C1t+C2
代入初始条件:
f(0)=4+C2=0，得C2=-4
f(π)=4e^π+C1π+C2=0,得C1=(4-4e^π)/π
因此f(t)=4e^t-8sint+(4-4e^π)t/π-4
2)f '''(x) = cos x,
积分：f"(x)=sinx+C1,
代入f"(0)=C1=6,
故f"(x)=sinx+6
再积分：f'(x)=-cosx+6x+C2
代入f'(0)=-1+C2=9,得C2=10
故f'(x)=-cosx+6x+10
再积分：f(x)=-sinx+3x^2+10x+C3
代入f(0)=C3=9,
故f(x)=-sinx+3x^2+10x+9

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
当 t = 0时,f(0) = 4*e^0 - 8*sin0 + C1*0 + C2 = 4 + C2 = 0,则 C2 =-4
当 t = π时,f(π) = 4*e^π - 8*sinπ + C1*π + C2 = 4*e^π + C1*π - 4 =0,则 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,当 x = 0 时,f"(0) = 0 + C1=6,则 C1=6,f"(x) = sinx + 6
f'(x) = ∫f"(x)dx = -cosx + 6x + C2,当 x =0 时,f'(0) = -1 + 6*0 + C2 =8,则 C2 = 9
f'(x) = -cosx + 6x + 9
f(x) =∫f'(x)dx = -sinx + 3x^2 + 9x + C3,当 x= 0时,f(0)= -0 + 3*0 + 9*0 + C3 = 9,则 C3=9
f(x) = -sinx + 3x^2 + 9x + 9