Evaluate the integral
\[\int\limits\limits_{-1}^{4}||x^2+x-6|-6| dx\]
The double absolute values are tripping me up.
you may be able to change the integration if theres symmetry. sec
I can factor it but I don't see symmetry.
I mean like how would I even do the cases?
if you plug in -1 and 4, its always going to be negative, so we can simplify it to be -x^2+x+36
Are you sure about that?
CUz don't we have to break it into cases?
you might be able to do that
No. I get the wrong answer.
what did you get
Should be 71/6.
got to split up the intervals and apply appropriate sign so that the function is positive
Yes I got that. I factored the innermost quadratic but not sure how to deal with the outermost one.
when is x^2 + x - 6 > 6 ?
you may solve below to get the numbers where the intgrand changes sign \[||x^2+x-6|-6| =0\]
Did you mean greater than 0 pg? That's true for x<3 and x>2 .
x<-3 sorry.
if you're allowed, you may simply graph it |dw:1448715042781:dw|
No I cannot graph this haha.
I propose you go with ganeshie and integrate according to the graph
is it hard to solve : \[||x^2+x-6|-6| =0\] ? @Dido525
||x^2+x-6|-6| =0 |x^2+x-6|-6 =0 |x^2+x-6| =6
you can eyeball the solutions : x = -1, 3
What about the negative case?
Also don't we have to consider ++,+-, -+, +- ?
definitely, we need to consider all that and sketch the graph
||x^2+x-6|-6| =0 |x^2+x-6|-6 =0 |x^2+x-6| =6 x^2+x-6 = 6 or x^2+x-6 = -6 you can solve right ?
Yes I can solve that. I think I have it. Hol don.
take ur time
Okay I am pretty far. I got to: \[\int\limits_{-1}^4||x^2+x-6|-6|dx=\int\limits_{-1}^2|x^2+x|dx+\int\limits_{2}^4|(x+4)(x-3)|dx\]
Aha! I got it :) .
THanks everyone!
Awesome ! that integral looks perfect!
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