Integrate using first three terms of appropriate series... $\int_0^1 \sin x ~dx.$
So I use $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}$ for my three terms and if I integrate just that I get the answer which is $.3103$.
However the solution book is showing a negative is taken outside the integral then just sorta disappears.
Am I missing something here or is this a typo? (What follows is what is in the book):
$\begin{align*} \int_0^1\sin x^2\,dx &= \int_0^1\left(x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!}\right)\,dx\\ &= -\int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx\\ &= \left.\left(\frac{1}{3}x^3 - \frac{x^7}{42} + \frac{x^{11}}{1320}\right)\right|_0^1\\ &= \frac{1}{3} - \frac{1}{42} + \frac{1}{1320} = 0.3103. \end{align*}$