# Integration of a polynomial problem

1. Nov 7, 2014

### MartinJH

Hi,
I'm using KA Stroud 6th edition (for anyone with the same book, P407) and there is a example question where I just can't seem to get the answer they have suggested:

1. The problem statement, all variables and given/known data

Question:
Determine the value of I = ∫(4x3-6x2-16x+4) dx
when x = -2, given that at x = 3, I = -13
Their answer is when x = -2, I = 12.

3. The attempt at a solution

I found the integral: I = ∫(4x3-6x2-16x+4) dx = x4-2x3-8x2+4x + C
and then substituted x for 3 and getting:
-13 = -33 + C
thus:
C = 20
Now when I replace x with -2, plus the constant, I get:
-24-2(-2)3-8(-2)2+4(-2) + 20 = -20

I'm a few days into Integrals so I feel I may be doing something daft?

Many thanks.

2. Nov 7, 2014

### Staff: Mentor

Your work is fine except for one minor thing. At the end you wrote -24 instead of (-2)4. In the first, 2 is raised to the 4th power, and then you take the negative, resulting in -16. In the latter, -2 is raised to the 4th power, resulting in +16.

3. Nov 7, 2014

### MartinJH

That was an honest slip. I appreciate there is a difference between them both.
I finally got the answer, it was a case of not respecting the brackets and powers... I need a break.

Thanks for pointing that out and explaining! :)

4. Nov 7, 2014

### LCKurtz

HEY!! I thought micromass's avatar was retired.

5. Nov 8, 2014

### MartinJH

I assume that is for me? :). I use this logo for most online things. EEVBlog, my Steam account etc etc. I was thinking about using Floyds new album cover.

6. Nov 8, 2014

### LCKurtz

Yes, but it was really directed at the old timers, sort of tongue-in-cheek. Turns out one of our previous highly regarded members used to use that logo. Nothing to worry about though.

7. Nov 9, 2014

### MartinJH

Yeah, that's cool. I Understand :).