Well, first graph the graph of f(x)=-1/10x^2 + 3
2. We are going to approximate the area between f and the x-axis from x = 0 to x = 4 using rectangles (the method of Riemann sums). This is not the entire area in the first quadrant, just most of it. Draw four inscribed rectangles of width 1 on the interval [0, 4] on you graph above.
3. What are the heights of each of the four rectangles? What is the total area of the rectangles? This area, although not the same as the area beneath the curve is an approximation for that area called the lower sum
4. The actual area between f and the x-axis on the interval [0, 4] is 28/3 . Why is one area greater?
5. How could you get a better approximation for the area beneath the curve if you still used inscribed rectangles?
Soooo um? I'm not understanding
To answer these questions, we'll follow the steps provided:
2. To approximate the area between the graph of the function and the x-axis, we can use the method of Riemann sums with rectangles. In this case, we'll be using inscribed rectangles with a width of 1 on the interval [0, 4].
So, on the graph of f(x) = -1/10x^2 + 3, draw four rectangles with bases of width 1, starting from x = 0 and ending at x = 4.
3. To find the heights of the four rectangles, we need to evaluate the function at the x-values of each rectangle. Since the rectangles are of width 1, the x-values for the rectangles will be 0, 1, 2, and 3.
Evaluate the function f(x) at these x-values:
f(0) = -1/10(0)^2 + 3 = 3
f(1) = -1/10(1)^2 + 3 = 2.9
f(2) = -1/10(2)^2 + 3 = 2.6
f(3) = -1/10(3)^2 + 3 = 2.1
So, the heights of the four rectangles are 3, 2.9, 2.6, and 2.1, respectively.
To find the total area of the rectangles, we multiply each height by the width (1 in this case) and add up the results:
Area = (3 * 1) + (2.9 * 1) + (2.6 * 1) + (2.1 * 1) = 10.6 square units.
Therefore, the total area of the rectangles is 10.6 square units.
4. The actual area between the graph of the function and the x-axis on the interval [0, 4] is found by calculating the definite integral of the function over that interval. In this case, it is equal to 28/3.
When comparing the total area of the rectangles (10.6 square units) to the actual area (28/3), we find that the actual area is greater. This difference occurs because the rectangles underestimate the total area beneath the curve. The rectangles are inscribed within the curve, which means they do not capture the full extent of the area between the curve and the x-axis.
5. If we want a better approximation for the area beneath the curve using inscribed rectangles, we can decrease the width of the rectangles. By making the width of the rectangles smaller, we can approach a more accurate approximation of the actual area. The more rectangles we use and the smaller their width, the closer our estimate will be to the actual area.