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anonymous
 5 years ago
Find the area bounded by the curves y1 = x , y2 = − x , and y3 = −x + 2
anonymous
 5 years ago
Find the area bounded by the curves y1 = x , y2 = − x , and y3 = −x + 2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0none of those are curves hehe

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1as far as i can tell; those dont bind to each other

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no such area. y2 = x and y3 = x + 2 are parallel

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooops it is sqrt of x and sqrt of x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1y=sqrt(x) y= sqrt(x) y = x+2 will create an area

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1there is no such thing as sqrt(x) in real numbers lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y1=\sqrt{x}, y2=\sqrt{x}, y3=x+2\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we can flip this to its inverse as well; y = x^2; y = x+2 and integrate from there as well

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^2 = x+2 x^2 +x  2 = 0 x = 2, 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0of course there is \[\sqrt{x}\] domain is nonpositive numbers!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the area of x+2 from 2 to 0 = [S] [x+2]  [x^2] dx ; [2,0]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1sqrt(x) is complex imaginary numbers....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^2/2 +2x  x^3/3 ; [2,1]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait... how did you get x^2 if you squared the \[\sqrt{x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how to invert a function?\[f \rightarrow f^{1}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.11/2 + 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
 5 years ago
Best ResponseYou've already chosen the best response.1y = sqrt(x) and y = sqrt(x) inverts to y = x^2 :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1y = x+2; ionverts to y = x+2 lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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 \[\int^2_{1} x+2x^2\,dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got that the 3 bound are 0, 1, 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You want the intersection of \[x^2\] and \[y=x+2\] \[x^2=x+2\] \[x^2x2=0\] \[(x2)(x+1)=0\] \[x=1, x=2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but none of the curves touch at 1.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the intersection of x^2 and x+2 was the original

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its odd that you disagree that sqrt(x) is not a real number.....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1lovhap; did you get a right answer :)
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