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

1. anonymous

none of those are curves hehe

2. amistre64

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

3. anonymous

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

4. anonymous

ooops it is sqrt of x and sqrt of -x

5. anonymous

Is y3 still -x + 2?

6. amistre64

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

7. amistre64

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

8. anonymous

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

9. amistre64

we can flip this to its inverse as well; y = x^2; y = -x+2 and integrate from there as well

10. amistre64

x^2 = -x+2 x^2 +x - 2 = 0 x = -2, 1

11. anonymous

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

12. amistre64

the area of -x+2 from -2 to 0 = [S] [-x+2] - [x^2] dx ; [-2,0]

13. amistre64

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

14. amistre64

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

15. anonymous

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

16. anonymous

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

17. 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 :)

18. amistre64

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

19. anonymous

oh ok

20. amistre64

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

21. 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 $\int^2_{-1} x+2-x^2\,dx$

22. anonymous

ok

23. anonymous

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

24. anonymous

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

25. anonymous

but none of the curves touch at -1.

26. amistre64

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

27. amistre64

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

28. amistre64

lovhap; did you get a right answer :)