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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.2012-05-06
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    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?2012-05-06
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    Basic Technical Mathematics with Calculus 9th ed. Allyn J. Washington2012-05-06
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    How did you manage to edit my question to use the symbols?2012-05-06
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    @ChefFlambe: I typed out your image using LaTeX.2012-05-06

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