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|dw:1358278438909:dw|

you are welcome :)

|dw:1358278984085:dw|

|dw:1358279110805:dw|

|dw:1358279244554:dw|Let's pretend this is what P(x) looks like.

oh, so when you switch the two, you mulply the integration by -1?

P(-x) will be the same function but reflected over the Y-axis.|dw:1358279336129:dw|

So what do you think that area will be? c:

Integral represents the area under the curve. This part was 4.|dw:1358279422383:dw|

Darn they're suppose to be the same, I didn't draw that left curve very well though lol

ya c: good times.

Hmm rave came up with -4, maybe I made a silly mistake somewhere.. I better check again.

Oh i see :o

|dw:1358279740263:dw|

Oh that's a fun one c:

could you please explain how you did that..? i am completely lost...

I'm confused why you're swapping the limits D: hmmm

Oh cause that's what we need to solve for :D I read it incorrectly haha

are

actually we know\[\int\limits_{3}^{5} P(x) dx\] and \[\int\limits_{1}^{5} P(x) dx\]

oh waiat! could you just subtract \[\int\limits_{3}^{5}p(x)dx\] from \[\int\limits_{1}^{5}p(x)dx\] ?

yes

oh yay thank you guys! :)

you are welcome :)