Question

integral SOS!!!

Answer 1

\[\int\limits \left| 1-x \right|dx\]

Answer 2

@perl?

Answer 3

limits of integration ?

Answer 4

no limits

Answer 5

oh right between 2 to zero

Answer 6

\[\int\limits_{2}^{0}\left| 1-x \right|dx\]

Answer 7

tnx

Answer 8

the opposite!

Answer 9

I need to divide it somhow

Answer 10

right?

Answer 11

the graph of this looks like

Answer 12

you can do this:
\(\large\color{slate}{\displaystyle\int\limits_{0}^{2}|1-x|~dx}\)
you can brake it down

\(\large\color{slate}{\displaystyle\int\limits_{1}^{2}(x-1)~dx+\int\limits_{0}^{1}(1-x)dx}\)

Answer 13

because from 1 to 2, the graph is a line y=x-1
and from 0 to 1, the graph is a line y=-x+1

Answer 14

how do you know which intgral need to be multiplied by -1?

Answer 15

lets look at the function together.
https://www.desmos.com/calculator/vqveqvcocx

Answer 16

ok

Answer 17

you can see that from x=1 to x=2 the graph is a line y=x-1
and
that from x=0 to x=1 the graph is a line y=-x+1

Answer 18

and so, we are going to find each area under the curve separately, and then add the areas.

Answer 19

you can also find what to use without drawing it if you prefer
|f(x)|=f(x) if f(x)>0
|f(x)|=-f(x) if f(x)<0

so you have
|1-x|=1-x if 1-x>0
|1-x|=-(1-x) if 1-x<0


giving the same thing with the inequalities solved
|1-x|=1-x if 1>x
|1-x|=-(1-x) if 1<x

Answer 20

that is just using the piecewise definition of |u|

Answer 21

if you prefer,