For large values of n, the Riemann sum 1/n(sin0 + (pi/2n) + sin(2pi/2n) +sin(3pi/2n)+...sin((n-1)pi/2n))


is an approximation for which of the following integrals?

To determine which integral the given Riemann sum is an approximation for, we can analyze the terms in the sum and look for similarities with the definition of integrals.

The Riemann sum has the form:
1/n (sin(0) + (pi/2n) + sin(2pi/2n) + sin(3pi/2n) + ... + sin((n-1)pi/2n))

Note that the terms within the sum are values of the sine function evaluated at specific angles.

When we take the limit as n approaches infinity, the Riemann sum becomes a Riemann integral. In this case, the integral corresponds to the integral of a function defined by the sine values at equally spaced points within the given range.

To determine the range of integration, we need to find the total interval spanned by the angles in the sine values.

The angles can be expressed as (k * pi) / (2n), where k ranges from 0 to (n-1).

As k varies from 0 to (n-1), the corresponding angles span the interval from 0 to (pi/2).

Hence, the Riemann sum represents an approximation for the integral of the sine function over the interval [0, pi/2]:

∫[0, pi/2] sin(x) dx

Therefore, the given Riemann sum is an approximation for the integral of sin(x) over the interval [0, pi/2].

The Riemann sum you provided is 1/n multiplied by the sum of sin(k(pi/2n)) for k = 0 to n-1. The Riemann sum is an approximation for the integral of a function. To determine which integral this Riemann sum approximates, we can rewrite it in a more general form.

Let's consider the Riemann sum as 1/n multiplied by the sum of sin(k(pi/2n)) for k = 0 to n-1.

We can rewrite the term sin(k(pi/2n)) as sin(k(pi/n) * (1/2)), and recognize that this is equivalent to sin(x) where x = k(pi/n).

So, the Riemann sum can be rewritten as 1/n multiplied by the sum of sin(x) for x = 0 to (n-1)(pi/n), which is equivalent to the sum of sin(x) for x = 0 to pi.

Therefore, the Riemann sum is an approximation for the integral of sin(x) from x = 0 to pi.

So, the Riemann sum you provided is an approximation for the integral of sin(x) from x = 0 to pi.