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Integral

f' (x)= 2x + 1 dan f(2)= 8
f(x)= ∫ f ' (x) dx = ∫ 2x +1  dx
f(x)=  x² + x + c
f(2)= 8 --> 2² +2 + c = 8
6 + c = 8
c = 2
.
f(x)= x² + x + 2

2)
f'(x)  = ax - 2
f(x)= ∫ ax - 2 dx = 1/2 a x² - 2x + c

f(2)= 6 --> 1/2 a(2²) - 2(2)+ c = 6
2a - 4 + c = 6
2a + c = 10 ....(1)
.
f(-1)= 4 --> 1/2 a (-1)² - 2(-1)+ c = 4
1/2 a + 2 + c = 4
1/2 a + c = 2 ....(2)
.
(1) - (2)
2a + c = 10
1/2 a + c = 2
............(kurangkan)....
3/2 a = 8  --> a = 16/3
.
1/2a + c= 2
c = 2 - 1/2 a
c = 2 - 16/6
c = - 4/6 = - 2/3

f(x)= 1/2 ax² -2x + c  --> f(x)= 1/2 (16/3) x² - 2x - 2/3
f(x) = 8/3 x² - 2x  - 2/3
atau
f(x) = 1/3 ( 8x² -6x -2) = 1/3 (2) (4x² - 3x - 1)
f(x)= 3/2 (4x² -3x -1)