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

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jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0the 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Riemann integrals are not affected by singular points. So you can do the integral as though the point x=0 doesn't exist.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Plus, the point x=0 doesn't even come up in the integral, since you're taking the integral from 1 to 4.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and 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)

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0so basically \[\int\limits_{1}^{4} \frac{ x^2 }{ \left x \right } dx\] ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah 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

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0but what about the absolute value?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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 \[ x =  x  \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And so where you see \[ x  \] you can just write \[ x \]

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0wait. the abs. value doesnt matter cuz it's always positive from 14 right?

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{4}^{2} f(x)dx\]

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0wait. but i think it's the same concept

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah. 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: \[\int\limits_{4}^{2}f(x) dx = \int\limits_{4}^{0} f(x) dx + \int\limits_{0}^{2} f(x) dx\]

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0but how would you integrate that? just assume it's x^2/x ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That, 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 \[ x = x \]

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait nvm i got it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355978372554:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got 10. I might have been a little rough in calculation though.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wolfram alpha confirms an answer of 10.

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0stupid negative sign... either way, the answer's 2

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0ohh can i see the wolfram alpha one? do u have a link?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integral+from+x+%3D+4+to+x%3D2+of+x^2+%2F+abs%28x%29+dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{4}^{2} \frac{x^2}{x} dx = \int\limits_{4}^{0} \frac{x^2}{x} dx + \int\limits_{0}^{2} \frac{x^2}{x} dx\]

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0oh just that ok then yeah i understand

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[= \int\limits_{4}^{0}  x dx + \int\limits_{0}^{2} x dx\]

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0but what about when x = 0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yeah just use limits like the other person did. And it'll come out to what I wrote up there.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{b \rightarrow 0^{}}\int\limits_{4}^{b} \frac{x^2} {x}dx = \lim_{b \rightarrow 0^{}}\int\limits_{4}^{b} \frac{x^2}{x} dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You'd probably write it like that if you wanted to be heaps formal.

jennychan12
 4 years ago
Best ResponseYou've already chosen the best response.0just kidding my teacher printed all the answers wrong _

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0huh? Kidding about what?
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