Suppose #f(x)= 2x^(2)-1#, how do you compute the Riemann sum for f(x) on the interval [-1,5] with partition {-1,2,4,5} using the left-hand endpoints as sample points?

Answer 1

See below.

The left endpoints are #-1#, #2#, and #4#.
The various #Deltax#'s are
#Deltax_1 = 2-(-1) = 3#, #Deltax_2 = 4-(2) = 2#, #Deltax_3 = 5-(4) = 1#

The sum we seek is

#f(-1)Deltax_1 + f(2)Deltax_2+f(4)Deltax_3 = (1)(3)+(7)(2)+(31)(1) = 48#
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Answer 2

To compute the Riemann sum for ( f(x) = 2x^2 - 1 ) on the interval ([-1, 5]) with the partition ({-1, 2, 4, 5}) using the left-hand endpoints as sample points:

  1. Calculate the width of each subinterval: [ \Delta x_i = x_{i} - x_{i-1} ]

  2. Determine the left-hand endpoints of each subinterval: [ x_i = -1, \ x_{i+1} = 2, \ x_{i+2} = 4, \ x_{i+3} = 5 ]

  3. Evaluate the function at each left-hand endpoint: [ f(x_i) = f(-1), \ f(x_{i+1}) = f(2), \ f(x_{i+2}) = f(4), \ f(x_{i+3}) = f(5) ]

  4. Compute the Riemann sum using the left-hand endpoints: [ \text{Riemann sum} = \sum_{i=1}^{n} f(x_i) \Delta x_i = f(-1) \Delta x_1 + f(2) \Delta x_2 + f(4) \Delta x_3 + f(5) \Delta x_4 ]

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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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