Question

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

Answer 1

none of those are curves hehe

Answer 2

as far as i can tell; those dont bind to each other

Answer 3

no such area. y2 = -x and y3 = -x + 2 are parallel

Answer 4

ooops it is sqrt of x and sqrt of -x

Answer 5

Is y3 still -x + 2?

Answer 6

y=sqrt(x)
y=- sqrt(x)
y = -x+2 will create an area

Answer 7

there is no such thing as sqrt(-x) in real numbers lol

Answer 8

\[y1=\sqrt{x}, y2=-\sqrt{x}, y3=-x+2\]

Answer 9

we can flip this to its inverse as well;

y = x^2; y = -x+2 and integrate from there as well

Answer 10

x^2 = -x+2

x^2 +x - 2 = 0

x = -2, 1

Answer 11

of course there is \[\sqrt{-x}\] domain is non-positive numbers!

Answer 12

the area of -x+2 from -2 to 0 =

[S] [-x+2] - [x^2] dx ; [-2,0]

Answer 13

sqrt(-x) is complex imaginary numbers....

Answer 14

-x^2/2 +2x - x^3/3 ; [-2,1]

Answer 15

wait... how did you get x^2 if you squared the \[\sqrt{x}\]

Answer 16

do you know how to invert a function?\[f \rightarrow f^{-1}\]

Answer 17

-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 :)

Answer 18

y = sqrt(x) and y = -sqrt(x) inverts to y = x^2 :)

Answer 19

oh ok

Answer 20

y = -x+2; ionverts to y = -x+2 lol

Answer 21

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
\[\int^2_{-1} x+2-x^2\,dx\]

Answer 22

ok

Answer 23

i got that the 3 bound are 0, 1, 4

Answer 24

You want the intersection of \[x^2\] and \[y=x+2\]
\[x^2=x+2\]
\[x^2-x-2=0\]
\[(x-2)(x+1)=0\]
\[x=-1, x=2\]

Answer 25

but none of the curves touch at -1.

Answer 26

the intersection of x^2 and -x+2 was the original

Answer 27

its odd that you disagree that sqrt(-x) is not a real number.....

Answer 28

lovhap; did you get a right answer :)