Question

A tract of land bordered by a highway along the y-axis, a dirt road along the x-axis, and a river whose path is given by the equation y=4-0.2x^2, where x and y are in hindreds of meters.

The tract is 300m deep along the dirt road The value of the land is constant in any strip parallel to the highway and increases as you move way from the highway, with the value given by v=1000+50x dollars per 10,000 m^2 at the sample point (x,y). Find dW, the worth of a strip. Write an integral that equals the worth of the entire tract.

Answer 1

We can represent the area of the tract with the definite integral:
\[\int\limits_{0}^{3}4-.2x^2dx\]
So we're finding the area from 0 to 3 (0 to 300m), between y and the x-axis. Essentially, we're finding the sum of the areas of an infinite number of squares of diminution y * dx as dx approaches 0. You may already know that but it helps my visual. :) So, to find the value, we need to integrate cubes instead of squares - so we have base (value) * height (y) * width (dx), e.g. area * value. Which makes the integral for the value of the tract:
\[\int\limits\limits_{0}^{3}(4-.2x^2)(1000-50x)dx\]

Answer 2

"diminution" should be "dimension", I think autocorrect got me there...

Answer 3

dont you mean,the area from 0 to 3 between y and the x axis is the sum of areas of an infinite number of rectangles of dimension y dx as dx approaches 0 (for integral y dx) ?
We cant do total area * total value. since area changes with x, and so does the value change with x. so we are doing summing value(x) Area(x) where value(x) means value at x.