$ \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 ?
Best Answer
$1^3=1$ 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:
$$I = \sum_{n=1}^7 \int_\sqrt[3]{n}^\sqrt[3]{n+1}x e^{1+n}\:dx$$