# How do you find the Riemann sum for #f(x) = 4 sin x#, #0 ≤ x ≤ 3pi/2#, with six terms, taking the sample points to be right endpoints?

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To find the Riemann sum for ( f(x) = 4 \sin(x) ) on the interval ( 0 \leq x \leq \frac{3\pi}{2} ) with six terms, taking the sample points to be right endpoints, follow these steps:

- Divide the interval ( [0, \frac{3\pi}{2}] ) into six subintervals of equal width.
- Determine the width of each subinterval: ( \Delta x = \frac{3\pi/2 - 0}{6} = \frac{\pi}{4} ).
- Identify the right endpoints of the subintervals.
- Evaluate ( f(x) = 4 \sin(x) ) at each right endpoint.
- Multiply the function values by the width of the subintervals and sum them up.

The Riemann sum can be expressed as:

[ \text{Riemann sum} = \Delta x \sum_{i=1}^{n} f(x_i) ]

Where ( \Delta x ) is the width of each subinterval, and ( f(x_i) ) is the value of the function ( f(x) ) at the right endpoint ( x_i ) of the ( i )th subinterval.

Given ( f(x) = 4 \sin(x) ), you evaluate it at the right endpoints of each subinterval:

[ x_1 = \frac{\pi}{4}, \quad x_2 = \frac{\pi}{2}, \quad x_3 = \frac{3\pi}{4}, \quad x_4 = \pi, \quad x_5 = \frac{5\pi}{4}, \quad x_6 = \frac{3\pi}{2} ]

Then, calculate ( f(x_i) ) for each ( x_i ):

[ f\left(\frac{\pi}{4}\right), \quad f\left(\frac{\pi}{2}\right), \quad f\left(\frac{3\pi}{4}\right), \quad f(\pi), \quad f\left(\frac{5\pi}{4}\right), \quad f\left(\frac{3\pi}{2}\right) ]

After evaluating ( f(x_i) ) for each ( x_i ), multiply each value by ( \Delta x = \frac{\pi}{4} ), and sum up the results to find the Riemann sum.

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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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