Answer
See explanation
Work Step by Step
$\begin{align*}
M_Y(t) &= E(e^{Yt}) \\
&= \int_{-\infty}^\infty e^{yt}f_Y(y)\ dy \\
&= \int_0^\infty e^{yt}f_Y(y)\ dy & \text{[ since }\ f_Y(y) = 0,\ y\lt 0\ ] \\
&= \int_0^\infty e^{yt}\cdot\left(ye^{-y}\right)\ dy & \text{[ since }\ f_Y(y) = ye^{-y},\ y\ge 0\ ] \\
&= \int_0^\infty ye^{-y+yt}\ dy \\
&= \lim_{L\ \to\ \infty} \int_0^L ye^{-y(1-t)}\ dy \\
& \\
& \qquad\text{Use integration by parts:} \\
& \qquad \begin{array}{l|l} u = y & dv = e^{-y(1-t)} \\ \hline du = dy & \displaystyle v = \frac{e^{-y(1-t)}}{-(1-t)} \end{array} \\
& \\
&= \lim_{L\ \to\ \infty} \left( ye^{-y(1-t)}\ \biggr\vert_0^L - \int_0^L \frac{e^{-y(1-t)}}{-(1-t)}\ dy \right)\\
&= \lim_{L\ \to\ \infty} \left( ye^{-y(1-t)}\ \biggr\vert_0^L - \frac{e^{-y(1-t)}}{(1-t)^2}\ \biggr\vert_0^L \right)\\
&= \lim_{L\ \to\ \infty} \left[\left( \frac{L}{e^{L(1-t)}} - 0 \right) - \left(\frac{e^{-L(1-t)}}{(1-t)^2} - \frac{1}{(1-t)^2} \right)\right] \\
& \\
& \quad\quad {\small t<1,} \\
& \quad\quad \small
\begin{array}{c|c}
\lim\limits_{L\ \to\ \infty} \frac{L}{e^{L(1-t)}} = \frac{\infty}{\infty} & \lim\limits_{L\ \to\ \infty} \frac{e^{-L(1-t)}}{(1-t)^2} \\
\stackrel{\text{L.H.}}{=} \lim\limits_{L\ \to\ \infty} \frac{1}{(1-t)e^{L(1-t)}} = \frac{1}{e^\infty} & = \frac{e^{-\infty}}{(1-t)^2} \\
=\frac{1}{\infty}= 0 & =0
\end{array} \\
& \\
&= \left( 0 -0\right) - \left( 0 - \frac{1}{(1-t)^2}\right) \\
\color{blue}{M_X(t)}\ &\color{blue}{= \frac{1}{(1-t)^2},\ t <1}
\end{align*}$
