3
$\begingroup$

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):

$\begin{align*} \int_0^1\sin x^2\,dx &= \int_0^1\left(x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!}\right)\,dx\\ &= -\int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx\\ &= \left.\left(\frac{1}{3}x^3 - \frac{x^7}{42} + \frac{x^{11}}{1320}\right)\right|_0^1\\ &= \frac{1}{3} - \frac{1}{42} + \frac{1}{1320} = 0.3103. \end{align*}$

  • 0
    @ChefFlambe: I typed out your image using LaTeX.2012-05-06

1 Answers 1

3

Yes, the minus sign is a typo and should not be there.

Note as well that you don't actually have equalities. Since $\sin x\approx x - \frac{x^3}{3!} + \frac{x^5}{5!},$ and we don't have actual equality, we have that $\int_0^1\sin(x^2)\,dx \approx \int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx.$ Likewise, in the final line we actually have $\frac{1}{3} - \frac{1}{42} + \frac{1}{1320}\approx 0.3103,$ not equality.