Integrals Containing Absolute Values

[CashNash88]

Hey I was wondering what the special rule is for integrals including absolute values is. Any help would be much appreciated. Here is the problem just incase...

P.S. I don't know how to do the integral sign so... for now here it is.

0~4 |X^2 - 1| dx

Thanks again.[/img][/list]

 

[camlax38]

Hey I was wondering what the special rule is for integrals including absolute values is. Any help would be much appreciated. Here is the problem just incase...

P.S. I don't know how to do the integral sign so... for now here it is.

0~4 |X^2 - 1| dx

Thanks again.[/img][/list]

Im gonna assume that 0~4 is what your integral is over.
First you need to find if your function is continuous over the whole interval, so set it equal to 0. You would get x=1,-1. -1 is arbitrary to you b/c it isnt on the interval 0-4.

No split your interval.
0~1(x^2 -1) + 1~4(x^2 -1)

If you draw a number line and plug in values btwn 0 and 1 you notice your function is negative so for the first part it should be

0~1 -(x^2 -1) +...

On your number line if you plug in values btwn 1 and 4 the function is positive so you can leave that part alone.

then just evaluate
0~1 -(x^2 -1) + 1~4(x^2 -1)

 

[soroban]

Hello,l CashNash88!

\(\displaystyle \L\int^{\;\;\;4}_0|x^2\,-\,1|\,dx\)

I recommend sketching or visualizing the graph.

We have the parabola, \(\displaystyle y\:=\:x^2\,-\,1\)
It opens upward, has x-intercepts \(\displaystyle \pm1\), and y-intercept -\(\displaystyle 1.\)

In the graph of \(\displaystyle y \:=\:|x^2\,-\,1|\), anything below the x-axis is reflected upward.

         *            |            *
                      |           .:
          *           |           *:
                      |           ::
           *          |          *::
            *        ***        *:::
              *   *   |:::*  .*:::::
      -----------*----+----*-------+---
                      0    1       4

And now you can see the required area . . .

. . \(\displaystyle \L\int^{\;\;\;1}_0(1\,-\,x^2)\,dx \:+\:\int^{\;\;\;4}_1(x^2\,-\,1)\,dx\)

 

[CashNash88]

Thanks!