How to approximate integral

Question

Pretend S is the integral symbol from 0 to 1. S cosx²dx.

how do I approximate the value w error less than 0.0001?

Answer 1

The Maclaurin series is cos(x) = ∑(-1)ⁿ x²ⁿ/2n!

Then, the Maclaurin series for cos(x²) is ∑(-1)ⁿ x⁴ⁿ/2n!

We can derive this from the cyle cos' = -sin, -sin' = -cos, -cos'= sin, and sin' = cos, and cos(0) = 1, sin(0) = 0

Then, as the integral of xⁿ from 0 to 1 is 1/(n+1), the integral of cos(x²) from 0 to 1 will be

∑(-1)ⁿ/(2n!(4n+1)) Then, we want to approximate this to .0001; as this is an alternating series whose absolute value is monotone decreasing, we just need to find the first term less than .0001

1/(2n!(4n+1)) < .0001

2n!(4n+1) > 10000

When n = 3, 2n!(4n+1) = 9360

When n = 4, 2n!(4n+1) = 685440

Thus, we will sum for n = 0 to n = 3

1 - 1/(2!*5) + 1/(4!*9) - 1/(6!*13) = .9045