How do you use Riemann sums to evaluate the area under the curve of #f(x)= 3 - (1/2)x # on the closed interval [2,14], with n=6 rectangles using left endpoints?

Answer 1

Please see the explanation section below.

will use what I think is the usual notation throughout this solution.

We will approximate. #int_2^14 (3-1/2x)dx#
Note that #f(x) = 3-1/2x# and #a=2# and #b=14#
#n=6# So #Deltax = (b-a)/n = (14-2)/6 =12/6 = 2#
All endpoints: start with #a# and add #Deltax# successively:
#2# #underbrace(color(white)"XX")_(+2)# #4# #underbrace(color(white)"XX")_(+2)# #6# #underbrace(color(white)"XX")_(+2)# #8# #underbrace(color(white)"XX")_(+2)# #10# #underbrace(color(white)"XX")_(+2)# #12# #underbrace(color(white)"XX")_(+2)# #14#
All endpoints: #2#, #4#, #6#, #8#, #10#, #12#, #14#

The subintervals are:

#(2,4)#, #(4, 6)#, #(6,8)#, #(8,10)#, #(10,12)#, #(12,14)#

We have been asked to use the left endpoint of each subinterval.

The left endpoints are:

#2#, #4#, #6#, #8#, #10#, #12#

Now the Riemann sum is the sum of the area of the 6 rectangles. We find the area of each rectangle by

#"height" xx "base" = f("sample point") xx Deltax#
Here we are using left endpoints of subintervals for sample points and #Deltax = 2#. So
#R = f(2)*2+f(4)*2+f(6)*2+ * * * +f(14)*1/2#
#= (f(2)+f(4)+f(6)+ f(8)+f(10)+f(12) +f(14))*2#

Finish the arithmetic to finish.

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Answer 2

To use Riemann sums to evaluate the area under the curve of ( f(x) = 3 - \frac{1}{2}x ) on the closed interval ([2, 14]) with ( n = 6 ) rectangles using left endpoints, we follow these steps:

  1. Calculate the width of each rectangle, ( \Delta x ), which is the total interval width divided by the number of rectangles: ( \Delta x = \frac{14 - 2}{6} = \frac{12}{6} = 2 ).

  2. Determine the left endpoints of each rectangle. Since we're starting at ( x = 2 ) and moving in increments of ( \Delta x = 2 ), the left endpoints are: 2, 4, 6, 8, 10, 12.

  3. Find the height of each rectangle by evaluating the function ( f(x) ) at the left endpoints: ( f(2) = 3 - \frac{1}{2}(2) = 3 - 1 = 2 ), ( f(4) = 3 - \frac{1}{2}(4) = 3 - 2 = 1 ), ( f(6) = 3 - \frac{1}{2}(6) = 3 - 3 = 0 ), ( f(8) = 3 - \frac{1}{2}(8) = 3 - 4 = -1 ), ( f(10) = 3 - \frac{1}{2}(10) = 3 - 5 = -2 ), ( f(12) = 3 - \frac{1}{2}(12) = 3 - 6 = -3 ).

  4. Calculate the area of each rectangle: ( A_i = \text{height}_i \times \Delta x ).

  5. Sum up the areas of all the rectangles to get the approximate area under the curve: ( \sum_{i=1}^{6} A_i ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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