liizzyliizz
  • liizzyliizz
Let f be the function given by f (x) = (x^3)/4 - (x^2)/3 - x/2 + 3cos(x). let R be the shaded region in the second quadrant bounded by the graph of f and let S be the shaded region bounded by the graph of f and line l the line tangent to the graph of f at x=0 . - FIND THE AREA of R ok I just need help, setting up the integral I kind of know what to do (I know the formula for the area) but I know im missing something... can someone help me?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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liizzyliizz
  • liizzyliizz
|dw:1333405813579:dw|
liizzyliizz
  • liizzyliizz
I feel like im supposed to find the intersection, to find the points used for the interval. but im not quite sure what to do after... O.o or if im missing something in between
liizzyliizz
  • liizzyliizz
meep? ):

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More answers

anonymous
  • anonymous
you need to find all x - intercepts
liizzyliizz
  • liizzyliizz
okay is there anything other steps im missing/ forgetting?
anonymous
  • anonymous
um when finding areas below the x axis you have to minus it
anonymous
  • anonymous
\[-\int\limits_{?}^{?}\]
anonymous
  • anonymous
otherwise you'll get a net area
liizzyliizz
  • liizzyliizz
the top would be the intersection right?
anonymous
  • anonymous
what is the line you have above
anonymous
  • anonymous
ah wait at x=0 gotcha
anonymous
  • anonymous
so it's juts asking for region r?
liizzyliizz
  • liizzyliizz
yeah
anonymous
  • anonymous
|dw:1333410353570:dw|
anonymous
  • anonymous
those will be your upper and lower bounds
liizzyliizz
  • liizzyliizz
:) That I understand, it's just that when i try it on the calculator I have trouble finding those points and then I get lost. :x
anonymous
  • anonymous
well one will be 0 as it is asking for only the second quadrant
liizzyliizz
  • liizzyliizz
ok i got the other bound, so then when solving for the integral , the part im a little ehh on is applying the formula, I feel that it would be (something) - f(x) or something like that, and idk how to figure that out. (I don't know if that makes sense)
anonymous
  • anonymous
applying the formula you mean top over bottom
anonymous
  • anonymous
it'd be your function - 0 since the bottom is f(x)=0
liizzyliizz
  • liizzyliizz
that's it? lol i thought I had to do something else O.o maybe im just confusing this with cross sections .-.
anonymous
  • anonymous
|dw:1333411005313:dw|
anonymous
  • anonymous
it's just asking for area correct?
liizzyliizz
  • liizzyliizz
yeah the area of R :) I guess I was just over thinking this o.O
anonymous
  • anonymous
yep.. now idk why they made liks line l or whatever... unless there is like another part where you need ofind area S. in that case you'd have to find the equation of the tangent line at x =0 (f', slope form, yada yada) and then once again top - bottom with the bounds being 0 and whatever the far right point is
liizzyliizz
  • liizzyliizz
yeah that is another part actually :P
anonymous
  • anonymous
yep so in that case you'd just find f', f'(0), which would give you your top and your bottom would be your f(x).
liizzyliizz
  • liizzyliizz
thank you so much :)
anonymous
  • anonymous
np

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