Question

Let $\int e^{x^2}\cdot e^x(2x^2+x+1)dx=e^{x^2}f\left(x\right)+c$, where $c$ is constant of integration. If the minimum value of $f\left(x\right)$ is $m$, then the value of $[-\frac{1}{m}]$ is
(where [.] represents grestest integer function)

Answer 1

$\int e^{x^2}\cdot e^x(2x^2+x+1)dx=e^{x^2}f\left(x\right)+c$
Let
$I=\int e^{x^2}\cdot e^x(2x^2+x+1)dx\Rightarrow I=\int e^{x^2+x}\left[x\right(2x+1)+1]dx$
Assuming $x{e}^{x^2+x}=t$
$\Rightarrow [e^{x^2+x}+x(2x+1\left)e^{x^2+x}\right]dx=dt$
Therefore,
$I=x{e}^{x^2+x}+c\Rightarrow f\left(x\right)=x{e}^x\Rightarrow f^{\prime }\left(x\right)=e^x(1+x)\Rightarrow f^{\prime \prime }\left(x\right)=e^x(1+x)+e^x=e^x(2+x)$
For maxima/minima
$f^{\prime }\left(x\right)=0\Rightarrow x=-1f^{\prime \prime }(-1)=e^{-1}>0$
Therefore, at $x=-1$ minima occurs, which is
$f(-1)=-e^{-1}=m\Rightarrow -\frac{1}{m}=e\therefore [-\frac{1}{m}]=\left[e\right]=2$