How do you Find the Riemann sum for #f(x)=x^3# on the interval #[0,5]# using right endpoints with #n=8#?

Question

Charles Anctil · Accepted Answer

Let us find #Delta x#. #Delta x={b-a}/n={5-0}/8=5/8# Now, we can find the right sum #R_8#. #R_8=[f(5/8)+f(10/8)+f(15/8)+f(20/8)+f(25/8)+f(30/8)+f(35/8)+f(40/8)]cdot Delta x# #=[(5/8)^3+(10/8)^3+(15/8)^3+(20/8)^3+(25/8)^3+(30/8)^3+(35/8)^3+(40/8)^3]cdot 5/8# #approx 197.75#

Charles Arocha · Answer

undefined To find the Riemann sum for $ f(x) = x^3 $ on the interval $[0,5]$ using right endpoints with $ n = 8 $, follow these steps:

1. Determine the width of each subinterval:
   $ \Delta x = \frac{b - a}{n} $, where $ a = 0 $ (the lower limit of the interval), $ b = 5 $ (the upper limit of the interval), and $ n = 8 $.

$ \Delta x = \frac{5 - 0}{8} = \frac{5}{8} $

2. Determine the right endpoints of each subinterval:
   $ x_i = a + i \cdot \Delta x $, where $ i = 1, 2, \ldots, n $.

$ x_i = 0 + i \cdot \frac{5}{8} $ for $ i = 1, 2, \ldots, 8 $.

3. Evaluate the function $ f(x) = x^3 $ at each right endpoint.

4. Multiply each function value by the width of the corresponding subinterval.

5. Sum up all the products obtained in step 4.

Here's the calculation:

- For $ i = 1 $:
  $ x_1 = \frac{5}{8} $
  $ f(x_1) = \left(\frac{5}{8}\right)^3 = \frac{125}{512} $

- For $ i = 2 $:
  $ x_2 = \frac{5}{8} \times 2 = \frac{5}{4} $
  $ f(x_2) = \left(\frac{5}{4}\right)^3 = \frac{125}{64} $

Continue this process until $ i = 8 $.

Then, calculate the Riemann sum:

$$ R = \frac{5}{8} \left( \frac{125}{512} + \frac{125}{64} + \ldots + \text{all other terms} \right) $$