How do you use a Riemann Sum with n = 4 to estimate #ln3 = int (1/x)# from 1 to 3 using the right endpoints and then the midpoints?

Answer 1
Riemann Sum with n = 4 to estimate #int (1/x)# from 1 to 3 using the right endpoints and then the midpoints
#f(x) = 1/x#
#[a,b] = [1,3]#
#n=4#
#Delta x = (b-a)/n = (3-1)/4 = 1/2#
The endpoints of the subintervals are found by beginning at #a# and successively adding #Deltax# until we get to #b#.

The subintervals are:

#[1,3/2] [3/2,2] [2,5/2] [5/2,3]#
The right endpoints are #x_1=3/2, x_2-2, x_3=5/2, x_4=3#
#R_4 = f(x_1)Deltax + f(x_2)Deltax+ f(x_3)Deltax+ f(x_4)Deltax#

Plug in the numbers and do the arithmetic.

The midpoints of the intervals can be found by averaging the endpoints or by finding the first midpoint and successively adding #Deltax#
#m_1 = 5/4, m_2 = 7/4, m_3 = 9/4, m_4=11/4#
#M_4 = f(m_1)Deltax + f(m_2)Deltax+ f(m_3)Deltax+ f(m_4)Deltax#

Plug in the numbers and do the arithmetic

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Answer 2

To estimate (\ln(3) = \int_{1}^{3} \frac{1}{x} , dx) using a Riemann sum with (n = 4) and right endpoints, you partition the interval ([1, 3]) into (4) subintervals of equal width. Then, you evaluate the function (\frac{1}{x}) at the right endpoint of each subinterval and multiply the function value by the width of the subinterval. Finally, you sum up all these products to get the estimate.

For the midpoints, you follow a similar process but instead evaluate the function at the midpoint of each subinterval.

Here are the steps:

  1. Partition the interval ([1, 3]) into (4) subintervals of equal width: ([1, 1.5]), ([1.5, 2]), ([2, 2.5]), ([2.5, 3]).

  2. For right endpoints: Evaluate (\frac{1}{x}) at the right endpoint of each subinterval: (f(1.5)), (f(2)), (f(2.5)), (f(3)). Then multiply each function value by the width of the subinterval: (f(1.5) \cdot 0.5), (f(2) \cdot 0.5), (f(2.5) \cdot 0.5), (f(3) \cdot 0.5). Sum up these products.

  3. For midpoints: Evaluate (\frac{1}{x}) at the midpoint of each subinterval: (f(1.25)), (f(1.75)), (f(2.25)), (f(2.75)). Then multiply each function value by the width of the subinterval: (f(1.25) \cdot 0.5), (f(1.75) \cdot 0.5), (f(2.25) \cdot 0.5), (f(2.75) \cdot 0.5). Sum up these products.

Performing these calculations will give you the estimates for (\ln(3)) using right endpoints and midpoints with (n = 4) subintervals.

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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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