sketch and find the area between y=-x,& x=y-y^2

im having trouble drawing the (x=y-y^2) any help pls

thnx

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- Oct 20th 2009, 08:02 AMalex83finding area
sketch and find the area between y=-x,& x=y-y^2

im having trouble drawing the (x=y-y^2) any help pls

thnx - Oct 20th 2009, 10:21 AMcalum
Completing the square:

$\displaystyle -x = (y^2-y+\frac{1}{4}) - \frac{1}{4}$

$\displaystyle -x + \frac{1}{4} = (y-\frac{1}{2})^2$

$\displaystyle y-\frac{1}{2} = \pm \sqrt{-x + \frac{1}{4}}$

$\displaystyle y = \frac{1}{2} \pm \sqrt{-x + \frac{1}{4}}$

You should be able to sketch that and find the area.

Remember that the positive and negative square roots are different functions: you may need to divide the area up into several integrals. - Oct 21st 2009, 12:08 AMcalum
Another thought: the area between two functions $\displaystyle f(x)$ and $\displaystyle g(x)$ is the same as the area between the inverse function $\displaystyle f^{-1}(x)$ and the original function $\displaystyle g(x)$. It will be easier to calculate the area this way, as it requires only one integral.

- Oct 21st 2009, 03:20 AMHallsofIvy
This simply is not true. You may be thinking of the area between [tex]x= f^{-1}(y)[/itex] and $\displaystyle x= g^{-1}(y)$

alex83, to find the graph, I would write $\displaystyle x= y- y^2= y(1- y)$ so the graph crosses the y-axis at (0, 0) and (0, 1). Since the coefficient of $\displaystyle y^2$ is negative, it is a parabola opening to the left going through those points.

More important is the fact that y= -x is the same as x= -y so the two graphs intersect when x= -y= y- y^2 or, adding y to both sides, 2y- y^2= y(2-y)= 0 so y= 0 or y= 2. The two graph intersect at (0,0) and (-2, 2).

Yes, you could do this integrating in terms of x, one integral from -2 to 0, the other from 0 to 1/4.

But it might be simpler to do the integral in terms of y (perhaps that was what calum was thinking of in his second post). Drawing a**horizontal**line from from x= -y to x= y- y^2, we see that the height of a rectangle (in the Riemann sum) is (y- y^2)- (-y)= 2y- y^2 and that the "width" is dy. The area of any such rectangle is (2y- y^2)dy and so the area is their "sum": $\displaystyle \int_0^2 (2y- y^2)dy$.

That should be very easy. - Oct 21st 2009, 04:20 AMcalum
Hmmm...that is what I meant, I got confused because when $\displaystyle g(x) = -x, g^{-1}(x) = -x$ ie the same, so for this particular problem this statement works, but not in general.

What I mean to say is if you have two functions f(x) and g(x), the area between the functions is equal to the area between the two inverse functions $\displaystyle f^{-1}(x)$ and $\displaystyle g^{-1}(x)$