Integrating Curve y=x³ - 6x² + 9x + 16: Help Needed

[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

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

 

[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