Loser66
  • Loser66
\[\int_{-3}^{3} |x+1|dx\] Please, help
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jtvatsim
  • jtvatsim
Integrate separately on parts of the domain that are convenient is the quick tip. :)
Loser66
  • Loser66
Show me, please.
jtvatsim
  • jtvatsim
Or, just draw it and find the area. :)

Looking for something else?

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

More answers

jtvatsim
  • jtvatsim
Will do, I'll do the drawing method since it's more clever.
Loser66
  • Loser66
I show you my confusion.
jtvatsim
  • jtvatsim
|dw:1433638297626:dw|
Loser66
  • Loser66
no confuse, hehehe.. I am sorry. I posted the wrong question. But I would like to know how to solve it traditionally
jtvatsim
  • jtvatsim
No worries. :) You should be able to see from the graph that the answer is the area of two triangles. But I will do the traditional (algebraic nastiness) approach as well.
Loser66
  • Loser66
I meant integrate separately..........
jtvatsim
  • jtvatsim
First, we need the definition of what |x + 1| means: |x + 1| = { x + 1, for x >= -1; -(x + 1), for x < -1. This is the usual, "make the number positive" rule we remember in our heads expressed algebraically.
jtvatsim
  • jtvatsim
It may take a bit of pondering to fully see that. :)
jtvatsim
  • jtvatsim
Then, this naturally gives us two convenient domains. (-infinity, -1) and (-1, +infinity). The split point is at x = -1.
jtvatsim
  • jtvatsim
We want (-3, 3) so we break this into the domains (-3, -1), and (-1, 3).
jtvatsim
  • jtvatsim
Then we integrate using the definition of absolute value.
jtvatsim
  • jtvatsim
\[\int\limits_{-3}^3 |x + 1| \ dx = \int\limits_{-3}^{-1} -(x+1) \ dx + \int\limits_{-1}^3 x+1 \ dx\]
jtvatsim
  • jtvatsim
The integration can then be carried out as usual.
Loser66
  • Loser66
Got you. Thanks a lot
jtvatsim
  • jtvatsim
No problems! :)

Looking for something else?

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