Question

find the area inbetween x^2+6 and 12-x^2

Answer 1

We could scale these two functions down so the intersection points lie on the x-axis.

Answer 2

Since I'm quite lazy, let's just put them in Wolfram|Alpha and http://puu.sh/1qVS8
Intersection points are \((\sqrt3, 9)\) and \((-\sqrt3, 9)\)

Answer 3

So let's move these two functions downwards, and we get y=x^2-3 and y=3-x^2

Answer 4

Then since our two functions look like this, we can take the integral of both functions from -sqrt3 to sqrt3 and add them up

Answer 5

thx i got that one..im on another now that i used wolfram aplia and when i plug it into my homework website it marks it as wrong...i have 1 chance left

Answer 6

\[\large \int_{-\sqrt{3}}^{\sqrt{3}}x^2-3dx+\int_{-\sqrt{3}}^{\sqrt{3}}3-x^2dx\]

Answer 7

its the area between 1/x and 1/(x^2) between .5 and 16

Answer 8

Umm, let's ask @TuringTest for this one :s

Answer 9

if im correct they integrate @ 1 so i put in
integrate (1/(x^2) - (1/x) from .5 to 1 and got .0306853
integrate (1/x) - (1/(x^2) from 1 to 16 and got log 16 - (15/6)
i added them together
for the approximate answer i put 1.8657740222
and for the exact i put (ln 16 - (15/6)) + .0306853

Answer 10

do i go to his page and ask him?

Answer 11

He's probably busy atm, hold on a second

Answer 12

Sorry, you'll have to wait a little bit, I don't know how to do that one :/

Answer 13

its ok.....but im waiting for you to talk to him right?

Answer 14

Split the atom.

Answer 15

@AccessDenied will answer your question, while he's typing out the question, lemme post a plot of this situation: In blue: 1/x, in red: 1/x^2

Answer 16

I think your approach is correct (I was originally going to type that explanation), although I seem to get a different value in calculator than you specified for the first integral.

I get about the same answer, except one decimal place to the right (0.306...).