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evaluate the following: ( calculus )

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u can evaluate this easily by using the properties of definite integral \[\int\limits_{3}^{5} \left( 2P(x)+Q(x) \right)dx=\int\limits_{3}^{5} 2P(x) dx+\int\limits_{3}^{5}Q(x)dx\]\[=2\int\limits_{3}^{5}P(x) dx + \int\limits_{3}^{5}Q(x) dx\]

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Other answers:

ohh thank you so much! @exraven !
you are welcome :)
for the last one, you can swap the limits of integration and multiply the whole thing by -1\[\int\limits_{5}^{1} P(x) dx = -\int\limits_{1}^{5} P(x) dx\]
|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?
ok :) @zepdrix
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
@zepdrix - ohh the area stays the same since it's only changed to negative but the mount of space under curve? stays the same !
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\[-\int\limits_{3}^{5} Q(u) du\]
ya c: good times.
but thank you @exraven . and yup thanks as always @zepdrix
Hmm rave came up with -4, maybe I made a silly mistake somewhere.. I better check again.
no the right answer is 4 :) she gave us the answer on the study guide. but just the answers, and no process. :p
Oh i see :o
Oh that's a fun one c:
\[\int\limits_{1}^{5} P(x) dx = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]\[\int\limits_{1}^{5} P(x) dx = -\int\limits_{3}^{1} P(x) dx + \int\limits_{3}^{5} P(x) dx\]solve for \[\int\limits_{3}^{1}P(x) dx\]
could you please explain how you did that..? i am completely lost...
okay, first of all suppose P(x) is like this|dw:1358280041569:dw|notice that I divide the area from x = 1 to x = 5 into two parts, and that's why I can write it like this\[\int\limits_{1}^{5} P(x) dx = A_1 + A_2 = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]
and then, just swap the limit of integration for\[\int\limits_{1}^{3} P(x) dx\]it becomes\[-\int\limits_{3}^{1} P(x) dx\]
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
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 \[\int\limits_{3}^{5}p(x)dx\]
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 \[\int\limits_{1}^{5}p(x)dx\] ?
so like.. \[\int\limits_{1}^{5}p(x)dx - \int\limits_{3}^{5}p(x)dx = -1-3 = -4\] 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?
oh yay thank you guys! :)
you are welcome :)

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