Question

Determine which of the following regions has an area equal to the given limit without evaluating the limit

n
lim Σ π/4n tan (iπ/4n)
n -> [infinity] i = 1

A. The area of the region under the graph of y =tan(z) on the interval [0.7/4).
B. The area of the region under the graph of y = sec(z) on the interval [0.7/4].
C. The area of the region under the graph of y =tan(x) on the interval (0,7/8].
D. The area of the region under the graph of y = sec(x) on the interval [0, 7/8].

Answer 1

The given limit describes a Riemann sum of the function y = tan(x) as n approaches infinity, which corresponds to the integral of tan(x) from 0 to π/4. Therefore, the region with an area equal to the given limit is the region under the graph of y = tan(x) on the interval [0, π/4). The correct option is A.

Step-by-step explanation:

We are given the limit nlim Σ (π/4n) tan (iπ/4n) as n approaches infinity and i runs from 1 to n. This is a limit of a Riemann sum for the integral of the function y = tan(x) over a certain interval.

The Riemann sum can be expressed as: nlim Σ (Δx) f(xi), where Δx is the width of each subinterval and f(xi) is the function value at a point in the i-th subinterval. As n approaches infinity, Δx approaches 0, and the Riemann sum approaches the definite integral of y = f(x) from the lower to the upper limit of the interval.

In the given limit, Δx = (π/4n) and f(x) = tan(x), with the points xi taken at i(π/4n) for i = 1 to n. This corresponds to the interval from 0 to π/4. Therefore, the area under the curve described by the given limit is the same as the area under the graph of the function y = tan(x) from 0 to π/4.

Hence, the correct option is A. The area of the region under the graph of y = tan(x) on the interval [0, π/4).