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anonymous
 3 years ago
evaluate the following:
( calculus )
anonymous
 3 years ago
evaluate the following: ( calculus )

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358278438909:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u can evaluate this easily by using the properties of definite integral \[\int\limits_{3}^{5} \left( 2P(x)+Q(x) \right)dx=\int\limits_{3}^{5} 2P(x) dx+\int\limits_{3}^{5}Q(x)dx\]\[=2\int\limits_{3}^{5}P(x) dx + \int\limits_{3}^{5}Q(x) dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohh thank you so much! @exraven !

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358278984085:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358279110805:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for the last one, you can swap the limits of integration and multiply the whole thing by 1\[\int\limits_{5}^{1} P(x) dx = \int\limits_{1}^{5} P(x) dx\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358279244554:dwLet's pretend this is what P(x) looks like.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, so when you switch the two, you mulply the integration by 1?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0P(x) will be the same function but reflected over the Yaxis.dw:1358279336129:dw

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0So what do you think that area will be? c:

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Integral represents the area under the curve. This part was 4.dw:1358279422383:dw

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

anonymous
 3 years ago
Best 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 !

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for 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\[\int\limits_{3}^{5} Q(u) du\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but thank you @exraven . and yup thanks as always @zepdrix

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358279740263:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{5} P(x) dx = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]\[\int\limits_{1}^{5} P(x) dx = \int\limits_{3}^{1} P(x) dx + \int\limits_{3}^{5} P(x) dx\]solve for \[\int\limits_{3}^{1}P(x) dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could you please explain how you did that..? i am completely lost...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay, 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\[\int\limits_{1}^{5} P(x) dx = A_1 + A_2 = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then, just swap the limit of integration for\[\int\limits_{1}^{3} P(x) dx\]it becomes\[\int\limits_{3}^{1} P(x) dx\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0I'm confused why you're swapping the limits D: hmmm

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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 \[\int\limits_{3}^{5}p(x)dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually we know\[\int\limits_{3}^{5} P(x) dx\] and \[\int\limits_{1}^{5} P(x) dx\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh yay thank you guys! :)
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