Show an exponential function has a valid density. Given:
Let $X$ be exponential with parameter $\lambda$, that is 
$$
f_X(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if }x> 0, \\
0 &\text{for }x\leq 0. \end{cases}
$$
where $\lambda>0$ is called the rate of the distribution.
Question:
Show that this is a valid probability density function (pdf).
Attempt:
Need to show that 
$$
f(x) = \int^{\infty}_{-\infty} \lambda e^{-\lambda x}\,  dx=1\\\\
$$
Thus,
\begin{eqnarray*}
f(x)&=&\int^0_{-\infty} \lambda e^{-\lambda x} \, dx+\int^\infty_0 \lambda e^{-\lambda x} \,  dx\\[8pt]
&=&0+\int^\infty_0 \lambda e^{-\lambda x} \, dx \\[8pt]
&=&\lambda\int^\infty_0 e^{-\lambda x} \, dx \\[8pt]
&=&1
\end{eqnarray*}
The problem is that my professor just told me that I skip a step right before I derive $1$. Any insight?
EDIT:
Given the great feedback, here is the answer:
\begin{eqnarray*}
f(x)&=&\int^{\infty}_{-\infty} \lambda e^{-\lambda x} dx=1\\\\
f(x)&=&\int^{\infty}_{-\infty} f_X(x)dx\\
&=&\int^{0}_{-\infty} 0\,\,\, dx+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\
&=&0+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\
&=&\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\
&=&\int^{\infty}_{0} e^{-\lambda x} \Big(\lambda \,\,\,dx\Big)\\
&=&\int^\infty_0e^{-u} du\\
&=&-e^{-u}\Big|^{\infty}_0\\
&=&1-\lim _{x \to \infty} e^{-u}\\
&=&1\\
\end{eqnarray*}
 A: For more rigor, you could have written
$$\begin{align}
f(x)
&=\int_{-\infty}^{\infty}f_X(x)\ dx\\
&=\int^{0}_{-\infty} 0\ dx+\int^{\infty}_{0} \lambda e^{-\lambda x} dx\\
&=-e^{-\lambda x} \Big|^{\infty}_{0} \\
&=1
\end{align}$$
A: There is an error in this line:
$$
f(x)=\int^0_{-\infty} \lambda e^{-\lambda x} \, dx+\int^\infty_0 \lambda e^{-\lambda x} \, dx. $$
It should say
$$
f(x)=\int^0_{-\infty} 0 \, dx+\int^\infty_0 \lambda e^{-\lambda x} \, dx. $$
Later, where you have
$$
\int_0^\infty \lambda e^{-\lambda x}\,dx,
$$
I would write
$$
\int_0^\infty e^{-\lambda x} \Big(\lambda \,dx\Big) = \int_0^\infty e^{-u}\,du = \cdots[\text{fill in the blanks here}]\cdots =1.
$$
A: Your last equality is correct but to fast:
$$\int_{0 }^\infty \lambda e^{-\lambda x}dx=\lambda\int_0^\infty e^{-\lambda x}=\lambda\left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^\infty =-\left[e^{-\lambda x}\right]_0^\infty =1-\lim_{x\to\infty }e^{-\lambda x}=1.$$
I think that your teacher just wanted you to explain this part. 
