= Solution
Integrating first over $z$ multiplies by $e^y$. Split at $y=0$ to handle the <absolute value> correctly:
$$
J=\int_0^1x\left[\int_{-x/2}^0(-y)e^y\,dy+\int_0^xy e^y\,dy\right]dx.
$$
<Integration by parts> gives <antiderivatives> $(1-y)e^y$ and $(y-1)e^y$ for these two inner <integrands>. Their contributions are
$$
1-(1+x/2)e^{-x/2},\qquad (x-1)e^x+1,
$$
respectively. Thus
$$
J=\int_0^1\left[2x+(x^2-x)e^x-(x+x^2/2)e^{-x/2}\right]dx.
$$
For the remaining <integrals>,
$$
\int_0^1(x^2-x)e^x\,dx
=\left[e^x(x^2-3x+3)\right]_0^1=e-3,
$$
and
$$
\int_0^1(x+x^2/2)e^{-x/2}\,dx
=\left[-e^{-x/2}(x^2+6x+12)\right]_0^1
=12-19e^{-1/2}.
$$
Combining them yields
$$
\boxed{J=e-14+19e^{-1/2}}.
$$
The upper $z$ bound is $e^y$, as in the original PDF; an additional truncated version in the OCR must not be used.
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