anonymous
  • anonymous
evaluate the following: ( calculus )
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1358278438909:dw|
anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
u can evaluate this easily by using the properties of definite integral \[\int\limits_{3}^{5} \left( 2P(x)+Q(x) \right)dx=\int\limits_{3}^{5} 2P(x) dx+\int\limits_{3}^{5}Q(x)dx\]\[=2\int\limits_{3}^{5}P(x) dx + \int\limits_{3}^{5}Q(x) dx\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
ohh thank you so much! @exraven !
anonymous
  • anonymous
you are welcome :)
anonymous
  • anonymous
|dw:1358278984085:dw|
anonymous
  • anonymous
|dw:1358279110805:dw|
anonymous
  • anonymous
for the last one, you can swap the limits of integration and multiply the whole thing by -1\[\int\limits_{5}^{1} P(x) dx = -\int\limits_{1}^{5} P(x) dx\]
zepdrix
  • zepdrix
|dw:1358279244554:dw|Let's pretend this is what P(x) looks like.
anonymous
  • anonymous
oh, so when you switch the two, you mulply the integration by -1?
anonymous
  • anonymous
ok :) @zepdrix
zepdrix
  • zepdrix
P(-x) will be the same function but reflected over the Y-axis.|dw:1358279336129:dw|
zepdrix
  • zepdrix
So what do you think that area will be? c:
zepdrix
  • zepdrix
Integral represents the area under the curve. This part was 4.|dw:1358279422383:dw|
zepdrix
  • zepdrix
Darn they're suppose to be the same, I didn't draw that left curve very well though lol
anonymous
  • anonymous
@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
  • anonymous
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\[-\int\limits_{3}^{5} Q(u) du\]
zepdrix
  • zepdrix
ya c: good times.
anonymous
  • anonymous
but thank you @exraven . and yup thanks as always @zepdrix
zepdrix
  • zepdrix
Hmm rave came up with -4, maybe I made a silly mistake somewhere.. I better check again.
anonymous
  • anonymous
no the right answer is 4 :) she gave us the answer on the study guide. but just the answers, and no process. :p
zepdrix
  • zepdrix
Oh i see :o
anonymous
  • anonymous
|dw:1358279740263:dw|
zepdrix
  • zepdrix
Oh that's a fun one c:
anonymous
  • anonymous
\[\int\limits_{1}^{5} P(x) dx = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]\[\int\limits_{1}^{5} P(x) dx = -\int\limits_{3}^{1} P(x) dx + \int\limits_{3}^{5} P(x) dx\]solve for \[\int\limits_{3}^{1}P(x) dx\]
anonymous
  • anonymous
could you please explain how you did that..? i am completely lost...
anonymous
  • anonymous
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\[\int\limits_{1}^{5} P(x) dx = A_1 + A_2 = \int\limits_{1}^{3} P(x) dx + \int\limits_{3}^{5} P(x) dx\]
anonymous
  • anonymous
and then, just swap the limit of integration for\[\int\limits_{1}^{3} P(x) dx\]it becomes\[-\int\limits_{3}^{1} P(x) dx\]
zepdrix
  • zepdrix
I'm confused why you're swapping the limits D: hmmm
zepdrix
  • zepdrix
Oh cause that's what we need to solve for :D I read it incorrectly haha
anonymous
  • anonymous
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 \[\int\limits_{3}^{5}p(x)dx\]
anonymous
  • anonymous
are
anonymous
  • anonymous
actually we know\[\int\limits_{3}^{5} P(x) dx\] and \[\int\limits_{1}^{5} P(x) dx\]
anonymous
  • anonymous
oh waiat! could you just subtract \[\int\limits_{3}^{5}p(x)dx\] from \[\int\limits_{1}^{5}p(x)dx\] ?
anonymous
  • anonymous
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?
anonymous
  • anonymous
yes
anonymous
  • anonymous
oh yay thank you guys! :)
anonymous
  • anonymous
you are welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.