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

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Solution

$\int e^{x^{2}} \cdot e^{x} (2 x^{2} + x + 1) d x = e^{x^{2}} f (x) + c$
Let
$I = \int e^{x^{2}} \cdot e^{x} (2 x^{2} + x + 1) d x \Rightarrow I = \int e^{x^{2} + x} [x (2 x + 1) + 1] d x$
Assuming $x e^{x^{2} + x} = t$
$\Rightarrow [e^{x^{2} + x} + x (2 x + 1) e^{x^{2} + x}] d x = d t$
Therefore,
$I = x e^{x^{2} + x} + c \Rightarrow f (x) = x e^{x} \Rightarrow f^{'} (x) = e^{x} (1 + x) \Rightarrow f^{''} (x) = e^{x} (1 + x) + e^{x} = e^{x} (2 + x)$
For maxima/minima
$f^{'} (x) = 0 \Rightarrow x = - 1 f^{''} (- 1) = e^{- 1} > 0$
Therefore, at $x = - 1$ minima occurs, which is
$f (- 1) = - e^{- 1} = m \Rightarrow - \frac{1}{m} = e ∴ [- \frac{1}{m}] = [e] = 2$

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