It's about to hit the floor

I would first like to thank Pi Han Goh for telling me that fractional function is different for positive and negative x! Without Pi Han i could never have posted this problem!

First we divide the integral into two parts, $\int_{-1}^{0}\{x^2+x-5\}dx+\int_{0}^{1}\{x^2+x-5\}dx$ .This is done as the definition of fractional function is different when $x$ is negative and positive.Now,we rewrite the integrals as, $\int_{-1}^{0}\{x^2+x-5\}dx=\int_{-1}^{0}((x^2+x-5)-\lceil{x^2+x-5}\rceil)dx$ .And, $\int_{0}^1\{x^2+x-5\}dx=\int_{0}^{1}((x^2+x-5)-\lfloor{x^2+x-5}\rfloor)dx$ .The first part of both the integrals can be calculated quite easily,hence we leave that and move on to the second part of the first integral,since $-1<x<0,0<(x+1)<1\Longrightarrow \lceil{x(x+1)-5}\rceil=-5$ .Now,we can easily calculate the second part of the first integral.Now,we move on to the second part of the second integral, $\int_{0}^{1}\lfloor{x^2+x-5}\rfloor dx=\int_{0}^{1}(\lfloor{x^2+x}\rfloor-5)dx$ .Now,we just have to calculate the first part as the second part is quite easy,so $x^2+x=1\ or\ x^2+x-1=0$ has only one solution in our interval and that is $\dfrac{\sqrt{5}-1}{2}$ which implies that when $x$ ranges from $0$ to this value the integral part would be $0$ hence we just calculate it from $\dfrac{\sqrt{5}-1}{2}$ to $1$ . $\int_{\frac{\sqrt{5}-1}{2}}^{1}\lfloor{x^2+x-1}\rfloor=\int_{\frac{\sqrt{5}-1}{2}}^{1}{1dx}$ .Now it is just a matter of calculations before we reach our answer.And done!