## anonymous 5 years ago integrate from 0 to 3 |x^3-3x^2+2x|dx

1. watchmath

Is that absolute value?

2. anonymous

yea

3. watchmath

$$x^3-3x^2+2x=x(x^2-3x+2)=x(x-1)(x-2)$$ So the integral is equal to $\int_0^1 x^3-3x^2+2x\,dx+\int_1^2-x^3+3x^2-2x\,dx+\int_2^3x^3-3x^2+2x\,dx$

4. anonymous

5. watchmath

Because over $$[1,2]$$ the function $$f(x)=x(x-1)(x-2)$$ is negative. So $$|f(x)|=-f(x)$$ on $$[1,2]$$.

6. anonymous

i see... do we have to include 1 and 2?? because you used square brackets to show them

7. watchmath

Well, in the integration it won't make any difference whether we have open interval or closed interval.

8. anonymous

yea thats true.. thanx again

9. anonymous