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Integrate: $$\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx$$

My attempt: We know that within the interval $[0,1)$, the following is true: $$\{x+1\}=x$$ $$\{x^2+2\}=x^2$$ $$\{x^2+3\}=x^2$$ $$\{x^3+4\}=x^3$$ Therefore $$I_1=\int_{0}^{1}x^3+x^5=5/12.$$

Similarily on the interval $[-1,0)$ we can write: $$\{x+1\}=x+1$$ $$\{x^2+2\}=x^2$$ $$\{x^2+3\}=x^2$$ $$\{x^3+4\}=x^3+1$$ Therefore, $$I_2=\int_{-1}^{0}x^3+2x^2+x^5=2/3-5/12.$$ Hence $$\int_{-1}^{1}(\{x+1\}\{x^2+2\})+(\{x^2+3\}\{x^3+4\})dx=2/3.$$ However the answer is $-2/3.$ I think I've made a mistake in the second transformations. Regardless, please explain where am I going wrong.

Student
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    The answer cannot be negative because the integrand is a sum of products of ${\cdots}$ and ${ y }$ is non-negative for any $y$. The answer is simply wrong. – achille hui Oct 11 '16 at 13:08

2 Answers2

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When you say $\{x\}=x+1$ for $x\in [-1,0) $ you are saying that the fractional part of $-0.3$ is $0.7$. Was that your intended meaning?

Martin Argerami
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If $x\in [-1,1]$, then $x^2,x^3,x^4\in [0,1]$. Now, when we add at least 1 to $y$, where $y\in [-1,1]$ its fractional part will be positive. Thus, you have a positive expression under the integral, so result cannot be negative.

MaliMish
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