百度作业网2道关于反导函数的数学题
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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f
f '''(x) = cos x, f(0) = 9, f '(0) = 8, f ''(0) = 6,求f
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
/>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
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