Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

piecewise function. integrals. see below.

See more answers at
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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

the piecewise function f(x) = \[\frac{ x^2 }{ \left| x \right| }, x \ne 0\] \[0, x=0\] find \[\int\limits_{1}^{4} f(x)dx\] how would u do that?
Riemann integrals are not affected by singular points. So you can do the integral as though the point x=0 doesn't exist.
Plus, the point x=0 doesn't even come up in the integral, since you're taking the integral from 1 to 4.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

and so then you know also that x > 0, so on the integral (1,4) f(x) = x^2 / x. (explicitly: you can take away the absolute value sign)
so basically \[\int\limits_{1}^{4} \frac{ x^2 }{ \left| x \right| } dx\] ?
Yeah that's right. This is intuitively why singular points don't really matter: In this picture, there's a singular point where the function is different at just one point. But, you remember when you do the Riemann approximation of the integral (the area under the curve) how you multiply the value of the function by the width? Well Singular points basically have no width... so the area underneath them, is effectively nothing. So you can ignore them. |dw:1355969105016:dw|
but what about the absolute value?
What you can do for the absolute value, at least in this case, is figure out whether x, over the interval of integration, is positive or negative, and then use that knowledge to remove the absolute value sign.
For example, you are integrating from x=1 to x=4. You know that x is positive for all of those values. So you know that across that whole integral \[ x = | x | \]
And so where you see \[ |x | \] you can just write \[ x \]
wait. the abs. value doesnt matter cuz it's always positive from 1-4 right?
that's right.
wait. typo....
\[\int\limits_{-4}^{2} f(x)dx\]
wait. but i think it's the same concept
Ah. okay. Well, you have something like this: |dw:1355970713671:dw| Where the integral is summing up that "shaded" area. So you can write it like this: \[\int\limits_{-4}^{2}f(x) dx = \int\limits_{-4}^{0} f(x) dx + \int\limits_{0}^{2} f(x) dx\]
but how would you integrate that? just assume it's x^2/x ??
That, in effect, is just breaking up the integral into two more. Now you know that from x=-4 to x=0 that x <= 0, so that \[|x| = -x \]. And on x=0 to x=2 \[ |x| = x \]
why from 0 to 2 ?
oh wait nvm i got it
i got -6
I got 10. I might have been a little rough in calculation though.
wolfram alpha confirms an answer of 10.
stupid negative sign... either way, the answer's 2
ohh can i see the wolfram alpha one? do u have a link?^2+%2F+abs%28x%29+dx
\[\int\limits_{-4}^{2} \frac{x^2}{|x|} dx = \int\limits_{-4}^{0} \frac{x^2}{|x|} dx + \int\limits_{0}^{2} \frac{x^2}{|x|} dx\]
oh just that ok then yeah i understand
\[= \int\limits_{-4}^{0} - x dx + \int\limits_{0}^{2} x dx\]
but what about when x = 0?
Oh yeah just use limits like the other person did. And it'll come out to what I wrote up there.
ok then thanks.
\[\lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2} {|x|}dx = \lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2}{-x} dx\]
You'd probably write it like that if you wanted to be heaps formal.
just kidding my teacher printed all the answers wrong -_-
huh? Kidding about what?

Not the answer you are looking for?

Search for more explanations.

Ask your own question