anonymous
  • anonymous
Find the area bounded by the curves y1 = x , y2 = − x , and y3 = −x + 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
none of those are curves hehe
amistre64
  • amistre64
as far as i can tell; those dont bind to each other
anonymous
  • anonymous
no such area. y2 = -x and y3 = -x + 2 are parallel

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anonymous
  • anonymous
ooops it is sqrt of x and sqrt of -x
anonymous
  • anonymous
Is y3 still -x + 2?
amistre64
  • amistre64
y=sqrt(x) y=- sqrt(x) y = -x+2 will create an area
amistre64
  • amistre64
there is no such thing as sqrt(-x) in real numbers lol
anonymous
  • anonymous
\[y1=\sqrt{x}, y2=-\sqrt{x}, y3=-x+2\]
amistre64
  • amistre64
we can flip this to its inverse as well; y = x^2; y = -x+2 and integrate from there as well
amistre64
  • amistre64
x^2 = -x+2 x^2 +x - 2 = 0 x = -2, 1
anonymous
  • anonymous
of course there is \[\sqrt{-x}\] domain is non-positive numbers!
amistre64
  • amistre64
the area of -x+2 from -2 to 0 = [S] [-x+2] - [x^2] dx ; [-2,0]
amistre64
  • amistre64
sqrt(-x) is complex imaginary numbers....
amistre64
  • amistre64
-x^2/2 +2x - x^3/3 ; [-2,1]
anonymous
  • anonymous
wait... how did you get x^2 if you squared the \[\sqrt{x}\]
anonymous
  • anonymous
do you know how to invert a function?\[f \rightarrow f^{-1}\]
amistre64
  • amistre64
-1/2 + 2 - 1/3 is one part of it: 7/6 -4/2 - 4 + 8/3 = -10/3 -10/3 - 7/6 = -27/6 area = 27/6....if i did it right :)
amistre64
  • amistre64
y = sqrt(x) and y = -sqrt(x) inverts to y = x^2 :)
anonymous
  • anonymous
oh ok
amistre64
  • amistre64
y = -x+2; ionverts to y = -x+2 lol
anonymous
  • anonymous
i disagree. no one says -x is negative. In any cases if you draw the picture and then rotate 90 degrees clockwise you see \[y=x^2\] and \[y=x+2\] they intersect at -1 and 2 so integrate \[\int^2_{-1} x+2-x^2\,dx\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
i got that the 3 bound are 0, 1, 4
anonymous
  • anonymous
You want the intersection of \[x^2\] and \[y=x+2\] \[x^2=x+2\] \[x^2-x-2=0\] \[(x-2)(x+1)=0\] \[x=-1, x=2\]
anonymous
  • anonymous
but none of the curves touch at -1.
amistre64
  • amistre64
the intersection of x^2 and -x+2 was the original
amistre64
  • amistre64
its odd that you disagree that sqrt(-x) is not a real number.....
amistre64
  • amistre64
lovhap; did you get a right answer :)

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