How do you use Riemann sums to evaluate the area under the curve of #f(x) = 2-x^2# on the closed interval [0,2], with n=4 rectangles using midpoint?

Question

Scarlett Allison · Accepted Answer

The Reimann sum with #n=4# will evaluate to #1.375#.

Since we are using the midpoint method to evaluate the area of 4 rectangles on the interval #[0,2]#, we know that our rectangles will have side lengths #2/4 = 0.5# and #f(x)#. Our rectangles will have midpoints located at; #x = 0.25# #x = 0.75# #x = 1.25# #x = 1.75# Thus our approximation of the area under #f(x) = 2 -x^2# will be: #A = 0.5(f(0.25) + f(0.75) + f(1.25) + f(1.75))# #A = 0.5(1.9375 + 1.4375 + 0.4375 - 1.0625)# #A = 0.5(2.75) = 1.375# We can compare this to the result of an actual integral by evaluating: #int_0^2 2 - x^2 = [2x - 1/3 x^3]_0^2# #= (4 - 8/3) - 0 = 4/3 = 1.33...# So as we can see, the difference is quite small, even for a small number of rectangles like #n =4#.

Isabella Ackerman · Answer

undefined To use Riemann sums to evaluate the area under the curve of $ f(x) = 2 - x^2 $ on the closed interval [0,2] with $ n = 4 $ rectangles using the midpoint method, follow these steps:

1. Determine the width of each rectangle: $ \Delta x = \frac{b - a}{n} $, where $ a = 0 $, $ b = 2 $, and $ n = 4 $. So, $ \Delta x = \frac{2 - 0}{4} = \frac{1}{2} $.

2. Find the midpoint of each subinterval: $ x_i^* = a + \frac{(2i - 1)\Delta x}{2} $, where $ i = 1, 2, \ldots, n $. In this case, $ a = 0 $ and $ \Delta x = \frac{1}{2} $.

$ x_1^* = 0 + \frac{(2 \times 1 - 1) \cdot \frac{1}{2}}{2} = \frac{1}{4} $
   
   $ x_2^* = 0 + \frac{(2 \times 2 - 1) \cdot \frac{1}{2}}{2} = \frac{3}{4} $
   
   $ x_3^* = 0 + \frac{(2 \times 3 - 1) \cdot \frac{1}{2}}{2} = \frac{5}{4} $
   
   $ x_4^* = 0 + \frac{(2 \times 4 - 1) \cdot \frac{1}{2}}{2} = \frac{7}{4} $

3. Evaluate $ f(x) = 2 - x^2 $ at each midpoint to find the heights of the rectangles.

$ f\left(\frac{1}{4}\right) = 2 - \left(\frac{1}{4}\right)^2 = \frac{15}{16} $
   
   $ f\left(\frac{3}{4}\right) = 2 - \left(\frac{3}{4}\right)^2 = \frac{7}{16} $
   
   $ f\left(\frac{5}{4}\right) = 2 - \left(\frac{5}{4}\right)^2 = \frac{1}{16} $
   
   $ f\left(\frac{7}{4}\right) = 2 - \left(\frac{7}{4}\right)^2 = -\frac{9}{16} $

4. Multiply each height by the width of its corresponding rectangle and sum up these products to find the approximate area under the curve:

$ A \approx \Delta x \cdot [f(x_1^*) + f(x_2^*) + f(x_3^*) + f(x_4^*)] $

$ A \approx \frac{1}{2} \cdot \left(\frac{15}{16} + \frac{7}{16} + \frac{1}{16} - \frac{9}{16}\right) $

$ A \approx \frac{1}{2} \cdot \frac{14}{16} $

$ A \approx \frac{7}{8} $

Therefore, the approximate area under the curve of $ f(x) = 2 - x^2 $ on the interval [0,2] with 4 rectangles using the midpoint method is $ \frac{7}{8} $.