Maclaurin Series Term by Term Integration

Question

Approximate the integral from 0 to 1 of (sin(x)/x) dx with an accuracy of 10^-6 using the Maclaurin series of sin, division of that Maclaurin series by x, and term-by-term integration. You will have to figure out how many terms of the series must be used.

Answer 1

∫0¹sinx/xdx = ∫₀¹(x - x³/3! + x⁵/5! - x⁷/7! + x⁹/9! - ...)/x∫dx = ∫₀¹(1 - x²/3! + x⁴/5! - x⁶/7! + x⁸/9! - ...)dx =

(x - x³/(3! ·3) + x⁵/5!·5) - x⁷/7!·7) + x⁹/(9!·9) -....)₀¹ = 1 - 1/(3!·3) + 1/(5!·5) - 1/(7!·7) + 1/(9!·9) - ...

Because we get alternative convergence series: 1) lim n→∞ 1/(n!·n) = 0 and 2) a_{n + 1} = 1/((n+ 1)!·(n+ 1)) <

1/(n!·n) = a_n.

Because error less then first term what we rejecting. !st such a term is 1/(9!·9) = 0.31·10^-6 < 10^-6 but

preciding term 1/(7!·7) = 2.83·10^-5 > 10^-6. Because withn accuracy 10^-6 we have:∫

∫₀¹sinx/xdx = 1 - 1/18 + 1/600 - 1/35280 ≈ 0.946083