finding area

**alex83** · October 20th 2009, 08:02 AM

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

**calum** · October 20th 2009, 10:21 AM

Originally Posted by alex83

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

Completing the square:

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

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

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

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

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.

**calum** · October 21st 2009, 12:08 AM

Another thought: the area between two functions $f(x)$ and $g(x)$ is the same as the area between the inverse function $f^{-1}(x)$ and the original function $g(x)$ . It will be easier to calculate the area this way, as it requires only one integral.

**HallsofIvy** · October 21st 2009, 03:20 AM

Originally Posted by calum

Another thought: the area between two functions $f(x)$ and $g(x)$ is the same as the area between the inverse function $f^{-1}(x)$ and the original function $g(x)$ . It will be easier to calculate the area this way, as it requires only one integral.

This simply is not true. You may be thinking of the area between [tex]x= f^{-1}(y)[/itex] and $x= g^{-1}(y)$

alex83, to find the graph, I would write $x= y- y^2= y(1- y)$ so the graph crosses the y-axis at (0, 0) and (0, 1). Since the coefficient of $y^2$ 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": $\int_0^2 (2y- y^2)dy$ .

That should be very easy.

**calum** · October 21st 2009, 04:20 AM

Originally Posted by HallsofIvy

This simply is not true. You may be thinking of the area between [tex]x= f^{-1}(y)[/itex] and $x= g^{-1}(y)$

Hmmm...that is what I meant, I got confused because when $g(x) = -x, g^{-1}(x) = -x$ 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 $f^{-1}(x)$ and $g^{-1}(x)$

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