Exercise 31, Section 8.8 To Evaluate \(\int_{0}^{\sqrt\pi} sin(x^2) dx\), we need to find the Taylor series for \(sin(x)\) first.
\[ sin(x)= \sum_{n=0}^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\] \[ sin(x^2)=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+...\] \[\int_{0}^{\sqrt\pi} sin(x^2) dx=\int_{0}^{\sqrt\pi} x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+...dx\]
f <-function(x) {sin(x^2)}
integrate(f, lower=0, upper=sqrt(pi))## 0.8948315 with absolute error < 9.9e-15g <-function(x) {x^2-x^6/factorial(3)+x^10/factorial(5)-x^14/factorial(7)}
integrate(g, lower=0, upper=sqrt(pi))## 0.8877069 with absolute error < 9.9e-15