Question

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Answer 1

Find the interval within [-5, 5] where the expression within the absolute bars, namely, x^2+x-12, is positive and negative. Do this by factoring x^2+x-12 first.

Answer 2

So x=-4, x=3?

Answer 3

Yes. Those are the points where the graph will cross the x-axis.
So the original interval [-5,5] has to be split into the following intervals:
[-5, -4), (-4, 3), (3, 5]
Find out in which interval x^2+x-12 = (x+4)(x-3) is positive and in which interval it is negative. In the interval where it is positive, you can get rid of the absolute bars: |x^2+x-12| = x^2+x-12. In the interval where it is negative, |x^2+x-12| = -( x^2+x-12 )

Answer 4

A quick way to find the intervals where x^2+x-12 is positive or negative is to realize this is a parabola that open upward (because the leading coefficient is positive). This implies the graph has a minimum between the roots -4 and 3. So the graph will be negative in the interval (-4,3) and positive everywhere else.

Answer 5

Awesome. So correct me if I'm wrong here...to find the area of the region would the integrals look like \[\int\limits_{-5}^{-4}x^2+x-12 dx + \int\limits_{-4}^{3}-(x^2+x-12)dx +\int\limits_{3}^{5}x^2+x-12dx\]

Answer 6

Correct!

Answer 7

Great thank you so much!