# Integral with Greatest integer function in exponent

$$\int_{1}^{2} xe^{\lfloor x\rfloor +\lfloor x^3\rfloor }\,dx$$

Where $$\lfloor x\rfloor$$ is floor function or greatest integer function

I thought since the limits are from $$1$$ to $$2$$ then I can integrate

$$\int_{1}^{2} xe^{x+x^3}\,dx$$

Then I tried solving via by parts by differentiating $$x$$ and integration the exponential function but that didn't seem to work.

How should i approach this question and other variations involving gif ?

$$1^3=1$$ and $$2^3=8$$, so the integral will need to be split into pieces for each cube root in between. And since $$\lfloor x \rfloor$$ is constant itself on the interval, we only have one sum in the potential double sum:
$$I = \sum_{n=1}^7 \int_\sqrt{n}^\sqrt{n+1}x e^{1+n}\:dx$$
Let $$x$$ be a real number, we always have $$\lfloor x\rfloor=n$$ iff $$n\leqslant x. Then, over $$[1,2[$$ we have $$\lfloor x\rfloor =1$$. Similarly, $$\lfloor x^3\rfloor=n$$ iff $$n\leqslant x^3, then we need break the integral up into pieces of the form $$\displaystyle \int_{n^{1/3}}^{(n+1)^{1/3}}xe^{1+n}dx$$. Then, the sum of these pieces over all $$n$$ in $$[|1,7|]$$. Thus, $$\int_{1}^{2}xe^{\lfloor x\rfloor+\lfloor x^3\rfloor}dx=\sum_{n\in [|1,7|]}\int_{n^{1/3}}^{(n+1)^{1/3}}xe^{1+n}dx=\sum_{n\in [|1,7|]}\left(\frac{(n+1)^{2/3}-n^{2/3}}{2}\right)e^{n+1}\approx 829.0996$$