Question

Question attached

Answer 1

x(x-1)(x-2) => x = 0,1,2

|x³ - 3x² + 2x| = -x³ + 3x² - 2x or x³ - 3x² + 2x

I have to solve that but I can't draw the graph

Answer 2

\[\int\limits_{-2}^{1} |x(x-1)(x-2)| dx\]

Answer 3

We'll begin by changing the limits to -2 to 1
\[-(\int _{-2}^{1} |x(x-1)(x-2)| dx)\]
=>\[-(\int _{-2}^{1} |x||(x-1)|(x-2)| dx)\]
when -2<x<-1
|x|=-x
|x-1|=-(x-1)
|x-2|=-(x-2)
when -1<x<0
|x|=-x
|x-1|=-(x-1)
|x-2|=-(x-2)
when 0<x<1
|x|=x
|x-1|=-(x-1)
|x-2|=-(x-2)
so
\[-(\int _{-2}^{-1} -x\times -(x-1)\times -(x-2) dx+\int _{-1}^{0} -x\times -(x-1)\times -(x-2) dx+\int _{0}^{1} -x\times -(x-1)\times -(x-2) dx)\]
So
\[-(\int _{-2}^{0} -x\times -(x-1)\times -(x-2) dx+\int _{0}^{1} x\times -(x-1)\times -(x-2) dx)\]
Can you do it now melinda?

Answer 4

Mmmm thanks Ash, but I can't. I guess I've learned that in a different way. The example says to not change the limits. I just try to use some properties, then.

Answer 5

We are not changing limits. You can now change the limits back and remove the minus sign

Answer 6

Is that so? How can I solve the last part?

Answer 7

Should I use this? \[\int\limits_{a}^{c} f(x)dx = \int\limits_{a}^{b} f(x)dx + \int\limits_{b}^{c} f(x)dx\]

Answer 8

I've changed the limits back
\[\int _{0}^{-2} x\times (x-1)\times (x-2) dx+ \int _{1}^{0} -x\times (x-1)\times (x-2) dx\]
They can't be combined as there is - sign in the second one
\[\int _{0}^{-2}(x^3-3x^2+2x) dx +\int _{1}^{0}-(x^3-3x^2+2x) dx\]

Answer 9

Ok, so can I do this?
[-8 -12 -4] - [ - (1 - 3 + 2)]

Answer 10

No you have to individually evaluate the limits
\[([(-8-12-4)]-[0])-([0]-[1-3+2])\]

Answer 11

Isn't that what I just did?

Answer 12

Yeah sorry:(

Answer 13

It's cool. Thank you so much for your help !

Answer 14

Any ideas on how drawing the graph?

Answer 15

Let me think

Answer 16

@ Melinda plot (x)(x-1)(x-2) and the portion which is below x-axis. Flip it upwards

Answer 17

Thanks!

Answer 18

@MelindaR did you understand?