anonymous
  • anonymous
CALCULUS.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1358986487783:dw|
anonymous
  • anonymous
so you need to find the antiderivative correct?
anonymous
  • anonymous
yes :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
hmm are u guys doing Integration rite now? haha xD
anonymous
  • anonymous
cause im taking Calculus atm :)
anonymous
  • anonymous
oh, no our exam is tom.. so i'm doing the review sheet and I have no idea how to do this... :/
anonymous
  • anonymous
@UnkleRhaukus help? :'(
anonymous
  • anonymous
@zepdrix help??? :'(
anonymous
  • anonymous
hm sorry it looked easy for a minute but idk what to do with the |x|
anonymous
  • anonymous
sorry.. :\
zepdrix
  • zepdrix
Hmmmm I'm not familiar with seeing absolutes in integrals either. I think this is how we would deal with it though. Think about the function \(y=|x|\). It's a V-shape, formed by 2 lines, \(y=-x\) when \(x \lt0\) and \(y=x\) when \(x \gt0\). So I think we can replace the abs(x) with -x when our integral is dealing with values of x which are less than 0. And then also replace abs(x) with +x when we're dealing with positive values of x. Let's start by splitting up the integral.\[\large \int\limits_{-1}^3 \frac{|x|}{x}dx \qquad \rightarrow \qquad \int\limits_{-1}^0 \frac{|x|}{x}dx+\int\limits_{0}^3 \frac{|x|}{x}dx\]
zepdrix
  • zepdrix
Notice that we're dealing with NEGATIVE x's in the boundaries of the first integral, and POSITIVE x's in the boundaries of the second. \[\large \int\limits\limits_{-1}^0 \frac{-x}{x}dx+\int\limits\limits_{0}^3 \frac{x}{x}dx\]
zepdrix
  • zepdrix
I think that's what you're suppose to do.... I'm not positive though :3
anonymous
  • anonymous
ah thank you zepdrix. haha even though i just gave up on this one. it's too confusing for me.. i will post the next quesiton though. exam's tom :( and open study has been acting really really slowww. i couldnt get on for like an hour.

Looking for something else?

Not the answer you are looking for? Search for more explanations.