anonymous
  • anonymous
Consider the area between the graphs x+3y=1 and x+9=y2. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals; Alternatively this area can be computed as a single integral
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
and the question for this statement?
amistre64
  • amistre64
we can integrate them both at the same time, or seperately and add the sums...
anonymous
  • anonymous
I need both please!

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anonymous
  • anonymous
And I need a and b, as well as c
amistre64
  • amistre64
the common line between them is the sum of there parts./2 right? gets an averge?
anonymous
  • anonymous
I also need f(x) and g(x) separately
amistre64
  • amistre64
f(x)=-x/3 +1/3; g(x) = sqrt(x+9)
amistre64
  • amistre64
[S] f(x) dx - [S] g(x) dx [S] f(x) - g(x) dx
anonymous
  • anonymous
The fist thing you have to do is to find the intersection of the two lines. Secondly, you have to find the intersection of the two lines with the x axis. As the third step, you have to integrate the frist integral from the x-coordinate of the first line until the x-cordinate of the intersection point. You have to do the same with the seconde line
amistre64
  • amistre64
it doesnt provide an interval, and so assuming that it just wants the intersections is presumptuous; prolly true, but not accurate
amistre64
  • amistre64
it could want the interval [0,root to the right]
anonymous
  • anonymous
Im not sure how to do that. Sorry
amistre64
  • amistre64
you get something that looks like this
1 Attachment
amistre64
  • amistre64
without a stated interval, you are prolly correct in assuming the left and right intercepts
anonymous
  • anonymous
ahhh ok. give me one second to thing about it
amistre64
  • amistre64
its got issues of being on the other side of the y axis; and youd have to split it into intervals from the common point to the left and right
amistre64
  • amistre64
maybe +9 to get it all on the right side of the y axis?
amistre64
  • amistre64
and my line parts to steep , but its just an artist interpretation lol
amistre64
  • amistre64
kinda like this eh
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anonymous
  • anonymous
Anyway, you have to split the total area in two sides.Carry out integration from intersection of the parable and the x-axis until the intersection of the parable and the line. They you have to integrate from the intersection point to the intersection of the line with the x-axis. That is how I would do that my friend
anonymous
  • anonymous
I got x=1 for c, but it keeps saying that it's wrong.
anonymous
  • anonymous
which c?
anonymous
  • anonymous
the end point on the right
anonymous
  • anonymous
I dont know my friend. I got a little bit confused. Sorry :(

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