How do you use Riemann sums to evaluate the area under the curve of #f(x)= In(x)# on the closed interval [3,18], with n=3 rectangles using right, left, and midpoints?

Answer 1

Please see the explanation section below.

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

Note that #f(x) = lnx# and #a=3# and #b=18#
#n=3# So #Deltax = (b-a)/n = (18-3)/3 =5#
The endpoints: start with #a# and add #Deltax# successively:
#3# #underbrace(color(white)"XX")_(+5)# #8# #underbrace(color(white)"XX")_(+5)# #13# #underbrace(color(white)"XX")_(+5)# #18#

The subintervals then are:

#[3,8]#, #[8,13]#, and #[13,18]#
The left endpoints are: #3#, #8# and #13#.
The right endpoints are #8#, #13#, and #18#
The midpoints may be found by averaging the endpoints. They are: #5.5#, #10.5#, #15.5# (As an alternative method, find the first midpoint and add #Deltax# successively.)
Now the Riemann sum is the sum of the area of 3 rectangles. We find the area of each rectangle by #"height" xx "base" = f("sample point") xx Deltax#

So, using left endpoints, we have

#L = f(3)*5+f(8)*5+f(13)*5#
#= (f(3)+f(8)+f(13))*5#

The arithmetic is left to the student.

Using right endpoints we have

#R = f(8)*5+f(13)*5+f(18)*5#
#= (f(8)+f(13)+f(18))*5#

The arithmetic is left to the student.

Using midpoints we have

#M = f(5.5)*5+f(10.5)*5+f(15.5)*5#
#= (f(5.5)+f(10.5)+f(15.5))*5#

The arithmetic is left to the student.

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Answer 2

To evaluate the area under the curve of ( f(x) = \ln(x) ) on the closed interval ([3,18]) using Riemann sums with ( n = 3 ) rectangles, you can use right, left, and midpoint methods.

  1. Right Riemann Sum: [ \text{Right Riemann Sum} = \sum_{i=1}^{n} f(x_i) \Delta x ] [ = f(x_1) \Delta x + f(x_2) \Delta x + f(x_3) \Delta x ] [ = \ln(x_1) \Delta x + \ln(x_2) \Delta x + \ln(x_3) \Delta x ] [ = \ln(6) \cdot 3 + \ln(9) \cdot 3 + \ln(12) \cdot 3 ]

  2. Left Riemann Sum: [ \text{Left Riemann Sum} = \sum_{i=1}^{n} f(x_{i-1}) \Delta x ] [ = f(x_0) \Delta x + f(x_1) \Delta x + f(x_2) \Delta x ] [ = \ln(x_0) \Delta x + \ln(x_1) \Delta x + \ln(x_2) \Delta x ] [ = \ln(3) \cdot 3 + \ln(6) \cdot 3 + \ln(9) \cdot 3 ]

  3. Midpoint Riemann Sum: [ \text{Midpoint Riemann Sum} = \sum_{i=1}^{n} f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x ] [ = f\left(\frac{x_0 + x_1}{2}\right) \Delta x + f\left(\frac{x_1 + x_2}{2}\right) \Delta x + f\left(\frac{x_2 + x_3}{2}\right) \Delta x ] [ = \ln\left(\frac{3 + 6}{2}\right) \cdot 3 + \ln\left(\frac{6 + 9}{2}\right) \cdot 3 + \ln\left(\frac{9 + 12}{2}\right) \cdot 3 ]

After calculating the values, you can sum them up to find the approximate area under the curve using each method.

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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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