EASY: I= ∫(4x^3-9x^2+8x-2)dx given that I=11/16 when x=1/2

I= ∫(4x^3-9x^2+8x-2)dx= x^4-x^3+4x^2- 2x+C

11/16=(1/2)^4-3(1/2)^3+4(1/2)^2- 2(1/2)+C

16/16=C

so I=x^4-x^3+4x^2-2x+1

But C supposedly equals -2.

So is C=1 or is C=-2 ?

Thanks.

C is 1 only. {How you obtained C = -2? Your solution that 11/16=(1/2)^4-3(1/2)^3+4(1/2)^2- 2(1/2)+C is correct; which solves for C = 1 only]

I = x^4 - 3x^3 + 4x^2 - 2x + C
11/16 = (1/2)^4 - 3(1/2)^3 + 4(1/2)^2 - 2(1/2) + C
11/16 = 1/16 + 3/8 - 1 + 1 + C
C= 11/16 - 1/16 + 6/16
C = (11-1+6)/16 = 16/16 = 1

C=1

It's easy ,i'm not sure but i think C = 1