Calculate the Riemann sum of the area under the curve of f(x)=9-x^2 between x=-2 and x=3
The answer I come up with is 10/3, but when I solve using integrals, the answer I get is 100/3. Am I doing something wrong?
looks like a decimal place error, check that.
Here is what Worlfram says:
http://www.wolframalpha.com/input/?i=riemann+sum+9-x%5E2+between+x%3D-2+and+x%3D3
To calculate the Riemann sum, we divide the interval between x=-2 and x=3 into smaller subintervals and approximate the area under the curve using rectangles.
Let's start by dividing the interval into subintervals. We can choose any number of subintervals, but for simplicity, let's divide it into 5 equal subintervals. The width of each subinterval is:
Δx = (3 - (-2)) / 5 = 5 / 5 = 1
Now, we need to find the height of each rectangle. We do this by evaluating the function at specific points within each subinterval.
Using the right endpoint of each subinterval, the x-values for each rectangle are:
x1 = -2 + 1 = -1
x2 = -1 + 1 = 0
x3 = 0 + 1 = 1
x4 = 1 + 1 = 2
x5 = 2 + 1 = 3
Now, evaluate the function f(x)=9-x^2 at each x-value:
f(-1) = 9 - (-1)^2 = 9 - 1 = 8
f(0) = 9 - (0)^2 = 9 - 0 = 9
f(1) = 9 - (1)^2 = 9 - 1 = 8
f(2) = 9 - (2)^2 = 9 - 4 = 5
f(3) = 9 - (3)^2 = 9 - 9 = 0
The height of each rectangle is as follows:
h1 = 8
h2 = 9
h3 = 8
h4 = 5
h5 = 0
Now, we can calculate the area of each rectangle as the product of the width and height:
A1 = Δx * h1 = 1 * 8 = 8
A2 = Δx * h2 = 1 * 9 = 9
A3 = Δx * h3 = 1 * 8 = 8
A4 = Δx * h4 = 1 * 5 = 5
A5 = Δx * h5 = 1 * 0 = 0
Finally, sum up the areas of all the rectangles:
Area ≈ A1 + A2 + A3 + A4 + A5 = 8 + 9 + 8 + 5 + 0 = 30/3 = 10
Therefore, the Riemann sum of the area under the curve of f(x)=9-x^2 between x=-2 and x=3, using 5 subintervals, is 10.
The discrepancy between your answer of 10/3 and the answer of 100/3 when solving using integrals could be due to using different methods or potential calculation errors while evaluating the integral. The value 10/3 seems to correspond to a different approximation method, such as the Trapezoidal Rule or Midpoint Rule.