## jennychan12 Group Title piecewise function. integrals. see below. one year ago one year ago

1. jennychan12 Group Title

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?

2. scarydoor Group Title

Riemann integrals are not affected by singular points. So you can do the integral as though the point x=0 doesn't exist.

3. scarydoor Group Title

Plus, the point x=0 doesn't even come up in the integral, since you're taking the integral from 1 to 4.

4. scarydoor Group Title

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)

5. jennychan12 Group Title

so basically $\int\limits_{1}^{4} \frac{ x^2 }{ \left| x \right| } dx$ ?

6. scarydoor Group Title

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|

7. jennychan12 Group Title

but what about the absolute value?

8. scarydoor Group Title

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.

9. scarydoor Group Title

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 $x = | x |$

10. scarydoor Group Title

And so where you see $|x |$ you can just write $x$

11. jennychan12 Group Title

wait. the abs. value doesnt matter cuz it's always positive from 1-4 right?

12. scarydoor Group Title

that's right.

13. jennychan12 Group Title

wait. typo....

14. jennychan12 Group Title

$\int\limits_{-4}^{2} f(x)dx$

15. jennychan12 Group Title

wait. but i think it's the same concept

16. scarydoor Group Title

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: $\int\limits_{-4}^{2}f(x) dx = \int\limits_{-4}^{0} f(x) dx + \int\limits_{0}^{2} f(x) dx$

17. jennychan12 Group Title

but how would you integrate that? just assume it's x^2/x ??

18. scarydoor Group Title

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 $|x| = x$

19. jennychan12 Group Title

why from 0 to 2 ?

20. jennychan12 Group Title

oh wait nvm i got it

21. cinar Group Title

|dw:1355978372554:dw|

22. jennychan12 Group Title

i got -6

23. scarydoor Group Title

I got 10. I might have been a little rough in calculation though.

24. scarydoor Group Title

wolfram alpha confirms an answer of 10.

25. jennychan12 Group Title

stupid negative sign... either way, the answer's 2

26. jennychan12 Group Title

ohh can i see the wolfram alpha one? do u have a link?

27. scarydoor Group Title
28. jennychan12 Group Title

thanks.

29. scarydoor Group Title

$\int\limits_{-4}^{2} \frac{x^2}{|x|} dx = \int\limits_{-4}^{0} \frac{x^2}{|x|} dx + \int\limits_{0}^{2} \frac{x^2}{|x|} dx$

30. jennychan12 Group Title

oh just that ok then yeah i understand

31. scarydoor Group Title

$= \int\limits_{-4}^{0} - x dx + \int\limits_{0}^{2} x dx$

32. jennychan12 Group Title

but what about when x = 0?

33. scarydoor Group Title

Oh yeah just use limits like the other person did. And it'll come out to what I wrote up there.

34. jennychan12 Group Title

ok then thanks.

35. scarydoor Group Title

$\lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2} {|x|}dx = \lim_{b \rightarrow 0^{-}}\int\limits_{-4}^{b} \frac{x^2}{-x} dx$

36. scarydoor Group Title

You'd probably write it like that if you wanted to be heaps formal.

37. jennychan12 Group Title

just kidding my teacher printed all the answers wrong -_-

38. scarydoor Group Title