is my answer correct?

Question

whpalmer4 · Answer

Yes.  No.  Maybe. :-)

anonymous · Answer

$$\int\limits_{1}^3 (x^2-x+1) dx$$
= 9-5/6?

anonymous · Answer

lol

anonymous · Answer

@Limitless  and @satellite73  :)

whpalmer4 · Answer

Mmm...doesn't look like it is correct.

anonymous · Answer

why?

whpalmer4 · Answer

Because I get a different result :-)

whpalmer4 · Answer

Can you show your indefinite integral?  we can decide if it is an integration mistake or an arithmetic mistake

anonymous · Answer

my antiderivative was: $$\frac{ x^3 }{ 3 }-\frac{ x^2 }{ 2 }+x$$

anonymous · Answer

Yup....ur Integration is Correct

anonymous · Answer

ok good

anonymous · Answer

but the another guy got as total answer 6.67.. so what arithmetic mistake and i making?

whpalmer4 · Answer

My guess is your subtraction of the -x^2/2 for x=1 might have gone awry

anonymous · Answer

1/3-1/2?

anonymous · Answer

@Yahoo!

abb0t · Answer

Keep your eye on the negative sign and similarly, know that it also follows the distributive property. I am not sure if that's where you made a mistake, but that's a common error that people often make:
$$\left[ (\frac{ x^3 }{ 3 }) - (\frac{ x^2 }{ 2 }) + x \right]$$

whpalmer4 · Answer

evaluate at x = 3:
$$3^3/3-3^2/2+3=9-9/2+3=12-9/2=7\frac{1}{2}$$
evaluate at x = 1:
$$1^3/3-1^2/2+1=5/6$$
$$7\frac{1}{2}-\frac{1}{6} = 7\frac{3}{6}-\frac{5}{6}=?$$

anonymous · Answer

ah thx! i said that 6=3^2 =.= sorry thank you!

whpalmer4 · Answer

I make the mistake that abb0t refers to all the time, it's the first thing I check :-)