## mlddmlnog 2 years ago CALCULUS.

1. mlddmlnog

|dw:1358986487783:dw|

2. lilfayfay

so you need to find the antiderivative correct?

3. mlddmlnog

yes :)

4. lilfayfay

hmm are u guys doing Integration rite now? haha xD

5. lilfayfay

cause im taking Calculus atm :)

6. mlddmlnog

oh, no our exam is tom.. so i'm doing the review sheet and I have no idea how to do this... :/

7. mlddmlnog

@UnkleRhaukus help? :'(

8. mlddmlnog

@zepdrix help??? :'(

9. lilfayfay

hm sorry it looked easy for a minute but idk what to do with the |x|

10. lilfayfay

sorry.. :\

11. 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.$\large \int\limits_{-1}^3 \frac{|x|}{x}dx \qquad \rightarrow \qquad \int\limits_{-1}^0 \frac{|x|}{x}dx+\int\limits_{0}^3 \frac{|x|}{x}dx$

12. 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. $\large \int\limits\limits_{-1}^0 \frac{-x}{x}dx+\int\limits\limits_{0}^3 \frac{x}{x}dx$

13. zepdrix

I think that's what you're suppose to do.... I'm not positive though :3

14. mlddmlnog

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.