Integral of x³ - 6x² + 9x + 16 from -1 to 2 gives -3/4, not 47.25

[Briggs]

I am having a little trouble on this problem for calculus..

Curve has equation y = x³ - 6x² + 9x + 16

I am asked to find $$\int_{-1}^{2} (x^3 - 6x^2 + 9x + 16) dx$$
I get the answer to be -3/4 but the book gives the answer as 47.25. I do not know how the book got this. The way I found it was
$$[\frac{x^4}{4} - \frac{6x^3}{3} + \frac{9x^2}{2}]_{-1}^{2}$$ and then putting x=2 and solving it then putting x=-1 and solving it then subtract the answer from x=-1 from x=2 so i found $$(4-16+18)-(\frac{1}{4}--2+\frac{9}{2}) = \frac{-3}{4}$$

It is important that I get this answer correct because It is needed for the next question of finding the area under a curve, could it just be that the book has an incorrect answer as it has known to be in the past or am i going about this completely the wrong way.
Thanks for any help you guys can provide

 

[Muzza]

Your primitive function is wrong. You need an extra term of 16x in there.

 

[dextercioby]

The integral of any constant A is

[tex]\int A \ dx = Ax+\mathcal{C} [/itex]<br /> <br /> ,where $\mathcal{C}$ is an integration constant...<br /> <br /> Daniel.[/tex]

 

[Briggs]

Thanks for the help I see where I was going wrong there, I was a little confused about the constant thing thanks for clearing it up