Decide whether to integrate with respect to x or y?

Question

Sketch the region enclosed by x=2y^2 and x+y=1 . Decide whether to integrate with respect to x or y. Then find the area of the region.

ok, i'm pretty sure that you integrate with respect to x, but i don't know how to find the limits.

Answer 1

To get the limits, just find the intersection points of the two curves (of course the second one is just a line).

You can integrate with respect to either variable, x or y, it is completely up to you. Of course your limits will change since if you are integrating with respect to x you are going from x=lower limit to x=upper limit, whereas if you are integrating with respect to y you are going from y=lower limit to y=upper limit.

Answer 2

Actually, your best bet is to integrate with respect to y. If you want to integrate with respect to x, each equation should written in terms of x. In other words, they should both be solved for y. While you can solve x = 2y^2 for y, there will be two solutions and that makes integration more tedious.

Integrate with respect to y. Start by solving each equation for x.
x = 2y^2 and x = -y+1.
to determine the limits, figure out where 2y^2 = -y + 1
2y^2 + y - 1 = 0
(2y - 1)(y + 1)=0
So, y = 1/2 or y = -1. Find the integral from -1 to 1/2.

The "width" of the shaded area is the difference of the x expressions, greater minus lesser.
For all y between -1 and 1/2, -y + 1 > 2y^2.
So, the bounded area is:
[[integral from -1 to 1/2]](((-y+1)-(2y^2))dy)
=[[integral from -1 to 1/2]]((-2y^2-y+1)dy)
=((-2/3)y^3-(1/2)y^2+y) evaluated from -1 to 1/2
=((-2/3)(1/2)^3 - (1/2)(1/2)^2 + (1/2))
-((-2/3)(-1)^3 - (1/2)(-1)^2 + (-1))
=(1/12)-(1/8)+(1/2)-(2/3)+(1/2)+1
=19/24

Answer 3

Integrating with respect to y is extra obtainable than with respect to x because you'll could do 2 separate integrals with respect to x. hence: the crucial of (y)^2/4 - (7 - 2y)/3 from 2 to 4 = 4

Answer 4

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