Answer
Apply the definition of the expectation of a random variable then evaluate the improper integral obtained. See details below.
Work Step by Step
$\begin{align*}
E(Y) &= \int_\mathbb{R} y\cdot f_Y(y)\ dy \\
&= \int_{-\infty}^\infty y\cdot f_Y(y)\ dy \\
&= \int_0^\infty y\cdot\left(\lambda e^{-\lambda y}\right)\ dy \qquad \text{since}\ f_Y(y)=\lambda e^{-\lambda y},\ y\gt 0 \\
& \qquad\qquad \\
& \qquad\qquad \text{Integration by parts:} \\
& \qquad\qquad \begin{array}{c|c} u = y & dv = \lambda e^{-\lambda y}\ dy \\ \hline du = dy & v = -e^{-\lambda y} \end{array} \\
& \qquad\qquad \begin{array}{rcl} \displaystyle \int y\cdot\left(\lambda e^{-\lambda y}\right)\ dy &=& \displaystyle \int u\ dv \\ &=& \displaystyle uv - \int v\ du \\ &=& \displaystyle -ye^{-\lambda y} - \int -e^{-\lambda y}\ dy \\ \displaystyle\int y\cdot\left(\lambda e^{-\lambda y}\right)\ dy&=& \displaystyle -\frac{y}{e^{\lambda y}} + \frac{1}{\lambda} \int \lambda e^{-\lambda y} \end{array} \\
& \qquad\qquad \\
&= -\frac{y}{e^{\lambda y}}\biggr\vert_0^\infty + \frac{1}{\lambda} \underbrace{\int_0^\infty \lambda e^{-\lambda y}}_{\int_\mathbb{R}({\rm pdf\ of}\ Y)\ dy \;=\; 1} \\
&= \biggl( -\frac{\infty}{e^{\lambda \cdot \infty}} + \frac{0}{e^{\lambda\cdot 0}}\biggr) + \frac{1}{\lambda}(1) \\
&= \biggl(\underbrace{-\frac{\infty}{\infty}}_{\scriptsize\rm L'Hospital's\ case} + \frac{0}{1} \biggr) + \frac{1}{\lambda} \\
&= \lim_{t\ \to\ \infty} -\frac{1}{\lambda e^{\lambda t}} + 0 + \frac{1}{\lambda} \\
&= -\frac{1}{\lambda e^{\infty}} +0 + \frac{1}{\lambda} \\
&= -\frac{1}{\infty} -0 + \frac{1}{\lambda} \\
&= -0 + 0 + \frac{1}{\lambda} \\
\color{blue}{E(Y)}\ &\color{blue}{= \frac{1}{\lambda}}
\end{align*}$
