anonymous
  • anonymous
What is the area under the curves of y=3, 2y=5(x)^1/2 and 2y+x=6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
is there a picture attached?
anonymous
  • anonymous
Ah well it's just that graphing this is going to be a pain.. So let's see what we can do
anonymous
  • anonymous
no but I can attach one

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anonymous
  • anonymous
if u can, please do so haha
anonymous
  • anonymous
k here it is
1 Attachment
anonymous
  • anonymous
omg lol more of these webworks... i'm so sick of them haha
anonymous
  • anonymous
sorry
anonymous
  • anonymous
nah it's fine.. actually give the wolframalpha link, they also list the intercepts. it's easier.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=Graph+y%3D3%2C+2y%3D5x^%281%2F2%29%2C+2y%2Bx%3D6
anonymous
  • anonymous
all right great thanks. Here we go..
anonymous
  • anonymous
So here I'm going to split the graph in two at the point where x+2y = 6 and 2y = 5sqrt(x) intersect
anonymous
  • anonymous
okay we did that
anonymous
  • anonymous
That's the upper limit of our first integral
anonymous
  • anonymous
okay
anonymous
  • anonymous
Sorry my browser's having trouble, anyway the upper intersect was 1. Now the lower intersect, which occurs when y =3 and 2y+x=6 intersect.
anonymous
  • anonymous
6+x=6, x = 0. So the first half of that area is: Integral (upper curve - lower curve, 0,1)
anonymous
  • anonymous
how did you get one I got 1.36 doesnt it turn into a quadratic
anonymous
  • anonymous
x+5sqrt(x) = 6? Well I mean take a look, what number plus 5 times it's square root is 6?
anonymous
  • anonymous
okay i get it i never thought of it like that
anonymous
  • anonymous
Keep in mind, we're still calculating the first half. So now, the upper curve minus the lower curve (well they're lines, but whatever) is: 3 - (6-x)/2
anonymous
  • anonymous
So so far, the first half of our area is: Integral (3 - (6-x)/2, 0,1)
anonymous
  • anonymous
Now let's find the 2nd half: This time the lower limit is 1, because we are starting at one, and the upper limit is the intersection of y=3 and y = 5/2sqrt(x)
anonymous
  • anonymous
(5/2)*sqrt(x) = 3 sqrt(x) = 6/5 => x = 36/25
anonymous
  • anonymous
So 2nd integral is again, upper curve minus the lower curve from 1 to 36/25, or Integral (3 - (5/2)*sqrt(x), 1, 36/25)
anonymous
  • anonymous
Calculate each integral and then add them to get 107/300 (lol I cheated, used wolf to compute)
anonymous
  • anonymous
so it is .356666667
anonymous
  • anonymous
Yea that looks right.
anonymous
  • anonymous
k i will try it and tell you if it works
anonymous
  • anonymous
it works thanks sooooo much you have no idea
anonymous
  • anonymous
np.

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