Question

integral from 0 to 3 abs(x-1) dx

Answer 1

break it in to two parts, or else use geometry

Answer 2

geometry is probably easiest

Answer 3

First, to get rid of the absolute value signs I'd solve for when it =0 and then you can make one part that's less than 0 negative to that point and then leave the positive part untouched and adjust your limits of integration accordingly.

Answer 4

you have two triangles, left one has base 1 and height 1, right one has base 2 and height 2
use one half base times height to get the area, you can almost do it in your head

Answer 5

ok i understand that but can you explain how i would solve it by breaking it into two parts?

Answer 6

ok

Answer 7

\[f(x) = |x-1| = \left\{\begin{array}{rcc}
1-x & \text{if} & x <1 \\
x-1& \text{if} & x \geq 1
\end{array}
\right.
\]

Answer 8

so
\[\int_0^3 f(x)=\int_0^1(1-x)dx+\int_1^3(x-1)dx\]

Answer 9

i.e. it is a piecewise function, so you have to break the integral in to two pieces
this is a rather long way to do it, but it certainly works

Answer 10

that is why absolute value is such a pain
seems like it is easy when you first learn it means "make it positive" but actually since it is a piecewise function it is quite annoying

Answer 11

much easier to say \(\frac{1}{2}+2=\frac{5}{2}\) and be done with it

Answer 12

Ahh i see. Thank you!

Answer 13

yw