$ \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 ?
On
Let $x$ be a real number, we always have $\lfloor x\rfloor=n$ iff $n\leqslant x<n+1$. Then, over $[1,2[$ we have $\lfloor x\rfloor =1$. Similarly, $\lfloor x^3\rfloor=n$ iff $n\leqslant x^3<n+1$, 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, $$\int_{1}^{2}xe^{\lfloor x\rfloor+\lfloor x^3\rfloor}dx=\sum_{n\in [|1,7|]}\int_{n^{1/3}}^{(n+1)^{1/3}}xe^{1+n}dx=\sum_{n\in [|1,7|]}\left(\frac{(n+1)^{2/3}-n^{2/3}}{2}\right)e^{n+1}\approx 829.0996$$
Edit: corrected.
$1^3=1$ 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:
$$I = \sum_{n=1}^7 \int_\sqrt[3]{n}^\sqrt[3]{n+1}x e^{1+n}\:dx$$