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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

none of those are curves hehe
as far as i can tell; those dont bind to each other
no such area. y2 = -x and y3 = -x + 2 are parallel

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question