What is the smallest value of n for which the Trapezoidal Rule approximation of ∫π0cos2x dx has an error less than 0.1?

Answer 1

To find the smallest value of n for which the Trapezoidal Rule approximation of ∫π0cos^2x dx has an error less than 0.1, we can use the error formula for the Trapezoidal Rule:

Error = -(b-a)^3/12n^2 * f''(c)

In this case, a = 0, b = π, and f(x) = cos^2(x). The second derivative of cos^2(x) is -2cos^2(x) - 2sin^2(x).

We want the error to be less than 0.1, so:

0.1 > π^3 / 12n^2 * max(|-2cos^2(c) - 2sin^2(c)|)

Since the maximum value of |-2cos^2(c) - 2sin^2(c)| is 2 for all c, we have:

0.1 > π^3 / 6n^2

n^2 > π^3 / 0.6

n > √(π^3 / 0.6)

Calculating this value, we get n > 4.974.

Therefore, the smallest value of n for which the Trapezoidal Rule approximation of ∫π0cos^2x dx has an error less than 0.1 is n = 5.