Answer
The statement is False, so
$$
\int_{-\infty}^{\infty} f(x) d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t}f( x) d x
$$
Work Step by Step
The statement is false, so
$$
\int_{-\infty}^{\infty} f(x) d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t}f( x) d x
$$
Now, we give an example that disproves the statement.
Let $f(x)=x$
$$
\int_{-\infty}^{\infty} x d x \ne \lim _{t \rightarrow \infty} \int_{-t}^{t} xd x
$$
(a) The given integral
$$
\int_{-\infty}^{\infty} x d x
$$
is an improper integral, hence
$$
\int_{-\infty}^{\infty} x d x =\int_{-\infty}^{0} x d x+\int_{0}^{\infty} x d x
$$
and
$$
\begin{split}
\int_{0}^{\infty} x d x & = \lim _{t \rightarrow \infty} \int_{0}^{t} x d x\\
& =\lim _{t \rightarrow \infty}\left[\frac{1}{2} x^{2}\right]_{0}^{t}
\\
& =\lim _{t \rightarrow \infty}\left[\frac{1}{2} t^{2}-0\right]
\\
& = \infty,
\end{split}
$$
so the integral
$$
\int_{0}^{\infty} x d x
$$
is divergent, and hence,
$$
\int_{-\infty}^{\infty} x d x
$$
is divergent.
(b)
$$
\begin{split}
\int_{-t}^{t} x d x & = \left[\frac{1}{2} x^{2}\right]_{-t}^{t} \\
& =\left[\frac{1}{2} t^{2}-\frac{1}{2} t^{2}\right]
\\
& = 0,
\end{split}
$$
so
$$
\lim _{t \rightarrow \infty} \int_{-t}^{t} x d x=0
$$
Therefore, from (a) and (b) we get
$$
\lim _{t \rightarrow \infty} \int_{-t}^{t} x d x \ne \int_{-\infty}^{\infty} x d x
$$