Any ideas as to go about doing this particular integral?
$$\int\limits\limits_{-1}^{4}||x^2+x-6|-6| dx$$
I'm a bit confused as to how to consider the cases into account. My idea was to consider 4 cases and the split the limits as needed but I'm not sure if that's exactly the best approach.
Any suggestions?
While $x\in[-1,2]$ so $|x^2+x-6|=-x^2-x+6$ and so $$||x^2+x-6|-6|\to|-x^2-x|=|x^2+x|$$ and when $x\in[2,4]$ so $|x^2+x-6|=+x^2+x-6$ and so $$||x^2+x-6|-6|\to|x^2+x-12|=|(x+4)(x-3)|$$
As the integrand is a integrable function on $[-1,4]$ so we get $$\int_{-1}^4||x^2+x-6|-6|dx=\int_{-1}^2|x^2+x|dx+\int_{2}^4|(x+4)(x-3)|dx$$
Now consider the same way for each of integrals above.