Question

Calculus question: Use the Midpoint Rule with n = 4 to approximate the area of the region bounded by the curves y=(16-x^3)^(1/3), y=x and x=0

Answer 1

We would start with four subintervals of equal length from x=0 to x=2, where y=(16-x^3)^(1/3) intersects with y=x.

Answer 2

This problem is basically asking you to sum up the total area of 4 rectangles in between the two functions, where their width is 0.5 because (2-0)/4=0.5.

Answer 3

That just leaves you with the height of each one, which is determined by the difference between the two functions ((16-x^3)^(1/3)-x) at their respective x values, the midpoints.

Answer 4

Let f(x) be y=(16-x^3)^(1/3), and your answer is 0.5*[(f(0.25)-0.25)+(f(0.75)-0.75)+(f(1.25)-1.25)+(f(1.75)-1.75)]

Answer 5

okay, I forgot to use the midpoint rule. Got it now.

Answer 6

Great!