Question

Hey, I need help with some math.

Answer 1

All help is appreciated

Answer 2

If you graph them, you'll see that you have to split this into two parts because the upper curve switches at \(x=1\).

Answer 3

So your limits will be from -1 to 1 and 1 to 3.

Answer 4

\(2x^3 - 6x^2 - 2x + 6\) is the upper curve from -1 to 1.

Answer 5

and \(-x^3 + 3x^2 + x - 3\) from 1 to 3

Answer 6

k

Answer 7

i think they want you to factor it to find the integral bounds.

nobody expects you to graph that from scratch (but you'll want to know how to make a rough sketch to get an idea of what it looks like once you find the bounds [lmk if you need help w that], its calculus 1, however)

do you know any methods to factor 3rd order polynomials?

Answer 8

and its always

function above minus function below

Answer 9

thnxs

Answer 10

You might find wolfram alpha helpful in the future to check your work on questions like this, e.g. http://www.wolframalpha.com/input/?i=area+between+2x%5E3+-+6x%5E2+-+2x+%2B+6+and+-x%5E3+%2B+3x%5E2+%2B+x+-+3

Answer 11

k

Answer 12

they get the answer by doing.

\[\int\limits_{-1}^{1}[f(x) - g(x)]dx + \int\limits_{1}^{3}[g(x) - f(x)]dx\]

to find intersection point(s), which in this case there are 3 (bounds + 1 in between), you can solve for x equation f(x) = g(x)

Answer 13

@Euler271 U r a such a badasa Euler..lolz

Answer 14

badass*

Answer 15

lol thanks

Answer 16

k