Answer
The coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {1,\frac{1}{4}} \right)$.
Work Step by Step
The infinite lamina bounded by the $x$- and $y$-axes and the graph of $y = {{\rm{e}}^{ - x}}$ is located in the first quadrant. We define ${\cal D}$ as a vertically simple region:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le x < p,0 \le y \le {{\rm{e}}^{ - x}}} \right\}$
and let $p \to \infty $.
Evaluate the area:
$A = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p \mathop \smallint \limits_{y = 0}^{{{\rm{e}}^{ - x}}} {\rm{d}}y{\rm{d}}x$
$ = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p \left( {y|_0^{{{\rm{e}}^{ - x}}}} \right){\rm{d}}x$
$ = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p {{\rm{e}}^{ - x}}{\rm{d}}x$
$ = - \mathop {\lim }\limits_{p \to \infty } \left( {{{\rm{e}}^{ - x}}|_0^p} \right)$
$ = \mathop {\lim }\limits_{p \to \infty } \left( {1 - {{\rm{e}}^{ - p}}} \right) = 1$
Evaluate the average of $x$-coordinate:
$\bar x = \frac{1}{A}\mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p \mathop \smallint \limits_{y = 0}^{{{\rm{e}}^{ - x}}} x{\rm{d}}y{\rm{d}}x$
$ = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p x\left( {y|_0^{{{\rm{e}}^{ - x}}}} \right){\rm{d}}x$
$ = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p x{{\rm{e}}^{ - x}}{\rm{d}}x$
Write $u=x$ and $dv = {{\rm{e}}^{ - x}}dx$. So, $du = dx$ and $v = - {{\rm{e}}^{ - x}}$. Using Integration by Parts Formula (Section 8.1),
$\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$
we get
$\bar x = \mathop {\lim }\limits_{p \to \infty } \left( { - x{{\rm{e}}^{ - x}}|_0^p + \mathop \smallint \limits_{x = 0}^p {{\rm{e}}^{ - x}}{\rm{d}}x} \right)$
$ = \mathop {\lim }\limits_{p \to \infty } \left( { - p{{\rm{e}}^{ - p}} - \left( {{{\rm{e}}^{ - x}}|_0^p} \right)} \right)$
$ = \mathop {\lim }\limits_{p \to \infty } \left( { - p{{\rm{e}}^{ - p}} - {{\rm{e}}^{ - p}} + 1} \right)$
$ = - \mathop {\lim }\limits_{p \to \infty } \frac{p}{{{{\rm{e}}^p}}} - \mathop {\lim }\limits_{p \to \infty } {{\rm{e}}^{ - p}} + 1$
Using L'Hôpital's Rule on the first limit at the right-hand side gives
$\bar x = - \mathop {\lim }\limits_{p \to \infty } \frac{1}{{{{\rm{e}}^p}}} - \mathop {\lim }\limits_{p \to \infty } {{\rm{e}}^{ - p}} + 1 = 1$
Evaluate the average of $y$-coordinate:
$\bar y = \frac{1}{A}\mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p \mathop \smallint \limits_{y = 0}^{{{\rm{e}}^{ - x}}} y{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p \left( {{y^2}|_0^{{{\rm{e}}^{ - x}}}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{x = 0}^p {{\rm{e}}^{ - 2x}}{\rm{d}}x$
$ = - \frac{1}{4}\mathop {\lim }\limits_{p \to \infty } \left( {{{\rm{e}}^{ - 2x}}|_0^p} \right)$
$ = - \frac{1}{4}\mathop {\lim }\limits_{p \to \infty } \left( {{{\rm{e}}^{ - 2p}} - 1} \right) = \frac{1}{4}$
So, the coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {1,\frac{1}{4}} \right)$.
