Question

How to solve? Help please!
Integral problem below(:
----The integral must be evaluated by breaking it up into a sum of three integrals.

Answer 1

The integral \[\int\limits_{-1}^{3} |7x^2-x^3-6x| dx\]
is split into three separate sums:
\[\int\limits_{-1}^{0} |7x^2-x^3-6x| dx + \int\limits_{0}^{c} |7x^2-x^3-6x| dx +\int\limits_{c}^{3} |7x^2-x^3-6x| dx \]

--->Find c. The values of each integral and the sum of all three.

Answer 2

I figured out c. haha it's 1.

Answer 3

well \(-x^3+7x^2-6x=-x(x^2-7x+6)=-x(x-1)(x-6)\)

Answer 4

hence our discontinuities are at \(x=0,x=1,x=6\) as our absolute value changes "direction"

Answer 5

yes, \(c=1\)

Answer 6

so now i just evaluate each integral right?

Answer 7

you gotta figure out over what intervals it is positive, and over what intervals it is negative
if it is positive, then \(|f(x)|=f(x)\) and if it is negative, then \(|f(x)|=-f(x)\)

Answer 8

but because in with in |abs| doesn't the function always come out positive? @satellite73

Answer 9

@pdd21 you don't know a way of integrating arbitrary absolute values, though, so you need to determine (like @satellite73 said) where the function is positive and negative so you know how to simplify each integrand over its respective interval

Answer 10

$$-x^3+7x^2-6x=-x(x^2-7x+6)=-x(x-1)(x-6)$$so we know how it factorizes... observe since the leading coefficient is negative and in the long run it dominates our behavior is a lot like \(-x^3\) i.e. its shape looks like this: