If f(t) = int_{-t}^t frac{e^{-|x|}}{2} dx,the | vedclass

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(A) Given $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$.Since the integrand $g(x) = \frac{e^{-|x|}}{2}$ is an even function (i.e.,$g(-x) = g(x)$),we can

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(A) Given $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$.Since the integrand $g(x) = \frac{e^{-|x|}}{2}$ is an even function (i.e.,$g(-x) = g(x)$),we can write:$f(t) = 2 \int_0^t \frac{e^{-x}}{2} dx = \int_0^t e^{-x} dx$.Evaluating the integral:$f(t) = [-e^{-x}]_0^t = -e^{-t} - (-e^0) = 1 - e^{-t}$.Now,taking the limit as $t \rightarrow \infty$:$\lim_{t \rightarrow \infty} f(t) = \lim_{t \rightarrow \infty} (1 - e^{-t}) = 1 - 0 = 1$.

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$,then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

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