The interval in which the function $g(x) = x^2 e^{-x}$ is increasing is:

Given $g(x) = x^2 e^{-x}$

To find where the function is increasing, find where $g'(x) > 0$.

Using the product rule with $f = x^2$ and $h = e^{-x}$:

$g'(x) = (2x)(e^{-x}) + (x^2)(-e^{-x})$

$g'(x) = 2xe^{-x} - x^2e^{-x}$

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Factor out the common term $e^{-x}$:

$g'(x) = e^{-x}(2x - x^2)$

$g'(x) = e^{-x} \cdot x(2 - x)$

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For the function to be increasing, $g'(x) > 0$:

$e^{-x} \cdot x(2 - x) > 0$

Since $e^{-x} > 0$ for all $x$, the inequality reduces to:

$x(2 - x) > 0$

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The critical points occur when $x(2 - x) = 0$:

$x = 0$ or $x = 2$

Testing the intervals:

For $x < 0$ (test $x = -1$): $(-1)(2 - (-1)) = (-1)(3) = -3 < 0$

For $0 < x < 2$ (test $x = 1$): $(1)(2 - 1) = (1)(1) = 1 > 0$

For $x > 2$ (test $x = 3$): $(3)(2 - 3) = (3)(-1) = -3 < 0$

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The function $g(x) = x^2 e^{-x}$ is increasing on the interval $(0, 2)$.