How do you approximate the area under #y=10−x^2# on the interval [1, 3] using 4 subintervals and midpoints?

Question

Isabella Ackerman · Accepted Answer

First you divide the interval in four equal parts, and in each part you find the midpoint. These would be:
#[1,1.5]# gives #1.25# as a midpoint
#[1.5,2]# gives #1.75#
#[2,2.5]# gives #2.25#
#[2.5,3]# gives #2.75# Then you plug in the values in the function (I will only do the first)
#y_1=10-1.25^2=8.4375# 
Since each of the subintervals is #0.5# wide, the surface under the first quarter of the interval #[1,3]# can be approximated by:
#A_1=0.5*y_1=0.5*8.4375=4.21875# You do the same for the other three parts and add. (Of course you round the total, as 5 decimals is a bit steep for an approxiation!)  Extra : 
The exact area is #17 2/3# (calculated by using integrals).

Axel Alvarez · Answer

undefined To approximate the area under $ y = 10 - x^2 $ on the interval [1, 3] using 4 subintervals and midpoints, you can follow these steps:

1. Determine the width of each subinterval:
   $$ \Delta x = \frac{b - a}{n} = \frac{3 - 1}{4} = \frac{2}{4} = 0.5 $$

2. Find the midpoints of each subinterval:
   - Subinterval 1: Midpoint = 1 + 0.5/2 = 1.25
   - Subinterval 2: Midpoint = 1.75
   - Subinterval 3: Midpoint = 2.25
   - Subinterval 4: Midpoint = 2.75

3. Evaluate the function $ y = 10 - x^2 $ at each midpoint to get the corresponding y-values:
   - At x = 1.25: $ y = 10 - (1.25)^2 = 8.4375 $
   - At x = 1.75: $ y = 10 - (1.75)^2 = 6.4375 $
   - At x = 2.25: $ y = 10 - (2.25)^2 = 3.4375 $
   - At x = 2.75: $ y = 10 - (2.75)^2 = 0.4375 $

4. Calculate the area of each rectangle formed by the midpoints and the corresponding y-values:
   - Area 1 = $ 0.5 \times 8.4375 = 4.21875 $
   - Area 2 = $ 0.5 \times 6.4375 = 3.21875 $
   - Area 3 = $ 0.5 \times 3.4375 = 1.71875 $
   - Area 4 = $ 0.5 \times 0.4375 = 0.21875 $

5. Add up the areas of all four rectangles to approximate the total area under the curve:
   $$ \text{Total Area} = \text{Area 1} + \text{Area 2} + \text{Area 3} + \text{Area 4} = 4.21875 + 3.21875 + 1.71875 + 0.21875 = 9.375 $$

Therefore, the approximate area under the curve $ y = 10 - x^2 $ on the interval [1, 3] using 4 subintervals and midpoints is 9.375 square units.