y is positive from x = -sqrt8 to +sqrt8
The area is Integral (8-x^2) dx
from x = -sqrt8 to +sqrt8
A rectangle with its base on the x axis is to be inscribed under the positive portions of the graph of y=8-x^2 . Write an equation for the area of the rectangle which uses only the variable x.
2 answers
eh? We're not looking for the area under the parabola. We want the area of a rectangle inscribed under it.
Let the width of the rectangle be 2x. That is, x is the distance from (0,0) to a bottom corner of the rectangle.
the area is thus
f(x) = 2xy = 2x(8-x^2)
The domain of f is obviously [-√8,√8]
Let the width of the rectangle be 2x. That is, x is the distance from (0,0) to a bottom corner of the rectangle.
the area is thus
f(x) = 2xy = 2x(8-x^2)
The domain of f is obviously [-√8,√8]
1 answer