Question

integrate from 0 to 3 |x^3-3x^2+2x|dx

Answer 1

Is that absolute value?

Answer 2

yea

Answer 3

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

Answer 4

why did you inverse the sign in the 2nd summation term??

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yea thats true.. thanx again

Answer 9

please check my other question