Answer
False
Work Step by Step
Consider $ \int^{\infty}_{-\infty}2xdx$
Solve the integral.
$ \int^{\infty}_{-\infty}2xdx=[x^{2}]_{-\infty}^{\infty}$
$=\infty-\infty$
=Indeterminate
Now,
$\lim\limits_{t \to \infty}\int^{t}_{-t}2xdx=\lim\limits_{t \to \infty}[x^{2}]_{-\infty}^{\infty}$
$=\lim\limits_{t \to \infty}0$
$=0$
Therefore, $ \int^{\infty}_{-\infty}2xdx\ne \lim\limits_{t \to \infty}\int^{t}_{-t}2xdx$
Hence, the given statement is false.
