Jika Integral f(x) dx=ax^2+bx+c,dan a tidak sama dengan 0 . Jika a, f(a) ,2b membentuk barisan aritmetika,dan f(b)=6 maka integral batas atas 1 dan batas bawah

Integral

∫ f(x)  dx =  ax²  + bx + c
f(x)= d(ax² +bx+c)/df
f(x) = 2ax + b

f(a) = 2a(a) + b
f(a) = 2a² + b

f(b) = 6
2ab + b = 6
b(2a+1) = 6
b = 6/(2a +1)...(1)

a , f(a), 2b  membentuk barisan aritmetika
maka  a + 2b = 2 . f(a)
a + 2b = 2 (2a² + b)
a + 2b = 4a² + 2b
a = 4a²
a - 4a² = 0
a(1- 4a) = 0 --> a = 0  atau 1 - 4a = 0
a = 0  (tidak memenuhi)
1 - 4a = 0
4a = 1
a = 1/4 --> b = 6/(2+1)
b = 6/(2(1/4) + 1)  = 6/(1/2 + 1) = 6/(3/2)
b = 4

f(x) = 2ax + b
f(x)= 2(1/4) x + (4)
f(x)= 1/2 x +4

₀¹∫ f(x) dx = ₀¹∫ 1/2 x + 4 dx
= [1/2 (1/2) x² + 4x ]¹₀
= 1/4 (1) + 4(1)
= 1/4 + 4
= 17/4