Homework Help: Integration Question

1. Mar 31, 2005

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.

2. Mar 31, 2005

Muzza

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

3. Mar 31, 2005

dextercioby

The integral of any constant A is

[tex] \int A \ dx = Ax+\mathcal{C} [/itex]

,where $\mathcal{C}$ is an integration constant...

Daniel.

4. Mar 31, 2005

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