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mlddmlnogBest ResponseYou've already chosen the best response.0
dw:1358278438909:dw
 one year ago

exravenBest ResponseYou've already chosen the best response.1
u can evaluate this easily by using the properties of definite integral \[\int\limits_{3}^{5} \left( 2P(x)+Q(x) \right)dx=\int\limits_{3}^{5} 2P(x) dx+\int\limits_{3}^{5}Q(x)dx\]\[=2\int\limits_{3}^{5}P(x) dx + \int\limits_{3}^{5}Q(x) dx\]
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
ohh thank you so much! @exraven !
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
dw:1358278984085:dw
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
dw:1358279110805:dw
 one year ago

exravenBest ResponseYou've already chosen the best response.1
for the last one, you can swap the limits of integration and multiply the whole thing by 1\[\int\limits_{5}^{1} P(x) dx = \int\limits_{1}^{5} P(x) dx\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
dw:1358279244554:dwLet's pretend this is what P(x) looks like.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
oh, so when you switch the two, you mulply the integration by 1?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
P(x) will be the same function but reflected over the Yaxis.dw:1358279336129:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
So what do you think that area will be? c:
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Integral represents the area under the curve. This part was 4.dw:1358279422383:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Darn they're suppose to be the same, I didn't draw that left curve very well though lol
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
@zepdrix  ohh the area stays the same since it's only changed to negative but the mount of space under curve? stays the same !
 one year ago

exravenBest ResponseYou've already chosen the best response.1
for this one\[\int\limits_{3}^{5} Q(x) dx\] use the substitution rule, let\[u = x\]\[du = dx\]\[x = 3, u = 3\]\[x = 5, u = 5\]the definite integral becomes\[\int\limits_{3}^{5} Q(u) du\]
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
but thank you @exraven . and yup thanks as always @zepdrix
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Hmm rave came up with 4, maybe I made a silly mistake somewhere.. I better check again.
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
no the right answer is 4 :) she gave us the answer on the study guide. but just the answers, and no process. :p
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
dw:1358279740263:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Oh that's a fun one c:
 one year ago

exravenBest ResponseYou've already chosen the best response.1
\[\int\limits_{1}^{5} P(x) dx = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]\[\int\limits_{1}^{5} P(x) dx = \int\limits_{3}^{1} P(x) dx + \int\limits_{3}^{5} P(x) dx\]solve for \[\int\limits_{3}^{1}P(x) dx\]
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
could you please explain how you did that..? i am completely lost...
 one year ago

exravenBest ResponseYou've already chosen the best response.1
okay, first of all suppose P(x) is like thisdw:1358280041569:dwnotice that I divide the area from x = 1 to x = 5 into two parts, and that's why I can write it like this\[\int\limits_{1}^{5} P(x) dx = A_1 + A_2 = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]
 one year ago

exravenBest ResponseYou've already chosen the best response.1
and then, just swap the limit of integration for\[\int\limits_{1}^{3} P(x) dx\]it becomes\[\int\limits_{3}^{1} P(x) dx\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
I'm confused why you're swapping the limits D: hmmm
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Oh cause that's what we need to solve for :D I read it incorrectly haha
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
oh i understand that part now, but i still dont understand how to solve for the value when you do not know what \[\int\limits_{1}^{3}p(x)dx\] and \[\int\limits_{3}^{5}p(x)dx\]
 one year ago

exravenBest ResponseYou've already chosen the best response.1
actually we know\[\int\limits_{3}^{5} P(x) dx\] and \[\int\limits_{1}^{5} P(x) dx\]
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
oh waiat! could you just subtract \[\int\limits_{3}^{5}p(x)dx\] from \[\int\limits_{1}^{5}p(x)dx\] ?
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
so like.. \[\int\limits_{1}^{5}p(x)dx  \int\limits_{3}^{5}p(x)dx = 13 = 4\] and then, since you have to switch the limits to 3 to 1, you multiply the 4 by 1 to get 4 as our answer?
 one year ago

mlddmlnogBest ResponseYou've already chosen the best response.0
oh yay thank you guys! :)
 one year ago
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