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 lefthand endpoints as sample points?
See below.
The sum we seek is
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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 lefthand endpoints as sample points:

Calculate the width of each subinterval: [ \Delta x_i = x_{i}  x_{i1} ]

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

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

Compute the Riemann sum using the lefthand 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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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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