# Maclaurin Series

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*}

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Definitely a typo. –  Antonio Vargas May 6 '12 at 20:45
Yes, it's a typo. And I'm not wild about them claiming the first integral is equal to the second (it's not, it's just an approximation) or that $\frac{1}{3}-\frac{1}{42}+\frac{1}{1320}$ is exactly equal to $0.3103$ (again, that's just an approximation). What book is this? –  Arturo Magidin May 6 '12 at 20:55
Basic Technical Mathematics with Calculus 9th ed. Allyn J. Washington –  Chef Flambe May 6 '12 at 21:02
How did you manage to edit my question to use the symbols? –  Chef Flambe May 6 '12 at 21:07
@ChefFlambe: I typed out your image using LaTeX. –  Arturo Magidin May 6 '12 at 21:16

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.