Question

Jika f'(t) = 5 - 2kt dengan k sembarang konstanta, f(0) = 0, dan f(1) = 1, tentukan nilai k dan formula fungsi f (integral)
plis bantu ya :'(

Answer 1

F'(t) = 5 - 2kt
f(t) = ∫(5 - 2kt) dt
f(t) = 5t - kt^2 + c

f(0) = 0
5(0) - k(0)^2 + c = 0
c = 0

f(t) = 5t - kt^2
f(1) = 1
5(1) - k(1)^2 = 1
5 - k = 1
k = 5 - 1
k = 4

f(t) = 5t - kt^2
f(t) = 5t - 4t^2
f(t) = -4t^2 + 5t

jadi,

k = 4 dan f(t) = -4t^2 + 5t