Let #f(x) = x^3# and compute the Riemann sum of f over the interval [2, 3], n=2 intervals using midpoints?
The midpoint Riemann sum approximation in this situation is then
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To compute the Riemann sum of ( f(x) = x^3 ) over the interval ([2, 3]) with ( n = 2 ) intervals using midpoints, follow these steps:

Determine the width of each subinterval: [ \Delta x = \frac{b  a}{n} = \frac{3  2}{2} = \frac{1}{2} ]

Calculate the midpoint of each subinterval: [ x_i^* = a + \frac{\Delta x}{2} + i \cdot \Delta x ] [ x_1^* = 2 + \frac{1}{2} \cdot \frac{1}{2} = \frac{5}{4} ] [ x_2^* = 2 + \frac{1}{2} \cdot \frac{3}{2} = \frac{7}{4} ]

Evaluate the function ( f(x) = x^3 ) at each midpoint: [ f(x_1^) = \left(\frac{5}{4}\right)^3 = \frac{125}{64} ] [ f(x_2^) = \left(\frac{7}{4}\right)^3 = \frac{343}{64} ]

Compute the Riemann sum using the formula: [ \text{Riemann Sum} = \Delta x \cdot \left(f(x_1^) + f(x_2^)\right) ]
Substituting the values: [ \text{Riemann Sum} = \frac{1}{2} \cdot \left(\frac{125}{64} + \frac{343}{64}\right) = \frac{468}{128} = \frac{117}{32} ]
So, the Riemann sum of ( f(x) = x^3 ) over the interval ([2, 3]) with ( n = 2 ) intervals using midpoints is ( \frac{117}{32} ).
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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.
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