# I don't know where to begin for this problem.

• September 26th 2010, 04:02 PM
Zanderist
I don't know where to begin for this problem.
What rules should I be looking to start this problem off?

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You can get full credit for this problem by just answering the last question correctly. The initial questions are meant as hints towards the final answer and also allow you the opportunity to get partial credit. The integral
$$$\displaystyle \int_{-1}^{6}\left|12 x^2 - x^3 - 32 x\right|\, dx$$$ MUST be evaluated by breaking it up into a sum of three integrals:
$$\int_{-1}^a\left|12 x^2 - x^3 - 32 x\right|\, dx +$ $\int_a^c\left|12 x^2 - x^3 - 32 x\right|\, dx +$ $\int_c^{6}\left|12 x^2 - x^3 - 32 x\right|\, dx$$where

$a =$

$c =$

$$$\displaystyle \int_{-1}^a\left|12 x^2 - x^3 - 32 x\right|\, dx$$ =$

$$$\displaystyle \int_a^c\left|12 x^2 - x^3 - 32 x\right|\, dx$$ =$

$$$\displaystyle \int_c^{6}\left|12 x^2 - x^3 - 32 x\right|\, dx$$=$
Thus
$$$\displaystyle \int_{-1}^{6}\left|12 x^2 - x^3 - 32 x\right|\, dx$$=$

Note: You can earn full credit by answering just the last part.

I'm not asking you to solve this for me, I just need to know where to start on this.
• September 26th 2010, 04:19 PM
Traveller
First work out where the function changes sign.
• September 26th 2010, 04:54 PM
Zanderist
I got it now... I just set equal to zero and find the zeros. That gives me my a and me c.
• September 26th 2010, 05:12 PM
Zanderist
The answer to the first sum was 20.25 ( using a calculator program) how do I do this long hand

$$$\displaystyle \int_{-1}^a\left|12 x^2 - x^3 - 32 x\right|\, dx$$ = 20.25$

Is it (1/b-a)(f(b)-f(a))?
• September 26th 2010, 06:01 PM
Zanderist
NOW I HAVE IT!
okay for those reading along, what you need to do after you find your zeros is integrate the function (take the anti-derivative) and just follow this rule.

$f(b)-f(a)\mid_{a}^{b}$