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.
well, the height of the rectangle is given by the equation y = 9 - x^2 and the width is 2x
so area = ?
I know how to get the area, but how do you know the height?
well, think of it this way: on the x-y coordinate plane, the x value tells you how far to the right or left to go, and the y-value tells you how far up or down to go, correct? since we want the height of the rectangle, it naturally follows that we want the y value
so, we're given the equation y = 9 - x^2, so that's our height
but the height of the rectangle is far smaller than the original prabola
nope, not necessarily, remember that x is not a fixed value if we make x something super small, like 0.00001, then the height of the rectangle is almost the height of the parabola the point is that x changes
ok I got this, and about the width, it looks like 4x to me. How is it 2x?
well, think of it this way. let's try splitting the rectangle in half down the y-axis, like this: |dw:1440359170630:dw|
Oh I see :) Yes and as we know negative value does not have any meaning in area.
nope, that's not quite the point I was getting at...
ok go ahead
|dw:1440359429185:dw| would you agree that the distance marked is equal to x?
so, all we have to do now is add the other half of the rectangle back in. |dw:1440359485663:dw|
making our width equal to 2x
oh I see
any other questions?
Everything is good. Thank you for clarifying this to me @Vocaloid