##8.8. Taylor Series ### Exercise 31 P.496
In Exercises 31 – 32, approximate the value of the given definite integral by using the first 4 nonzero terms of the integrand’s Taylor series. \(\int_{0}^{\sqrt{\pi}}{sin(x^2)dx}=\int_{0}^{\sqrt{\pi}}{\sum_{n=0}^{\infty}{\frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}}dx}\\ \\{\quad\quad\quad\quad\quad\quad}=\sum_{n=0}^{\infty}{{\int_{0}^{\sqrt{\pi}}}\frac{(-1)^n(x)^{4n+2}}{(2n+1)!}}dx\\ \\{\quad\quad\quad\quad\quad\quad}=\sum_{n=0}^{\infty}{\left[\frac{(-1)^n(x)^{4n+3}}{(4n+3)(2n+1)!}\right]_{0}^{\sqrt{\pi}}}\\ \\{\quad\quad\quad\quad\quad\quad}=\sum_{n=0}^{\infty}{\left[\frac{(-1)^n(\sqrt{\pi})^{4n+3}}{(4n+3)(2n+1)!}\right]}\)
taln <- 0
n <- 100
for (i in 1:n) {
taln <- taln + (((-1)^i)*(sqrt(pi))^(4*i+3))/((4*i+3)*factorial(2*i+1))
}value = -0.531464