jennychan12 2 years ago piecewise function. integrals. see below.

1. jennychan12

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

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

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

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

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

6. scarydoor

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

but what about the absolute value?

8. scarydoor

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

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

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

11. jennychan12

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

12. scarydoor

that's right.

13. jennychan12

wait. typo....

14. jennychan12

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

15. jennychan12

wait. but i think it's the same concept

16. scarydoor

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

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

18. scarydoor

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

why from 0 to 2 ?

20. jennychan12

oh wait nvm i got it

21. cinar

|dw:1355978372554:dw|

22. jennychan12

i got -6

23. scarydoor

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

24. scarydoor

wolfram alpha confirms an answer of 10.

25. jennychan12

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

26. jennychan12

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

27. scarydoor
28. jennychan12

thanks.

29. scarydoor

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

30. jennychan12

oh just that ok then yeah i understand

31. scarydoor

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

32. jennychan12

but what about when x = 0?

33. scarydoor

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

34. jennychan12

ok then thanks.

35. scarydoor

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

36. scarydoor

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

37. jennychan12

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

38. scarydoor