Here's the question you clicked on:
mlddmlnog
CALCULUS.
|dw:1358986487783:dw|
so you need to find the antiderivative correct?
hmm are u guys doing Integration rite now? haha xD
cause im taking Calculus atm :)
oh, no our exam is tom.. so i'm doing the review sheet and I have no idea how to do this... :/
@UnkleRhaukus help? :'(
hm sorry it looked easy for a minute but idk what to do with the |x|
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.\[\large \int\limits_{-1}^3 \frac{|x|}{x}dx \qquad \rightarrow \qquad \int\limits_{-1}^0 \frac{|x|}{x}dx+\int\limits_{0}^3 \frac{|x|}{x}dx\]
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. \[\large \int\limits\limits_{-1}^0 \frac{-x}{x}dx+\int\limits\limits_{0}^3 \frac{x}{x}dx\]
I think that's what you're suppose to do.... I'm not positive though :3
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.