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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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:

Similar Questions

The value of the integral $\int \limits_1^{\sqrt{2}+1} \left( \frac{x^2-1}{x^2+1} \right) \frac{1}{\sqrt{1+x^4}} \, dx$ is

The value of $\int_{0}^{\pi} |\sin^3 \theta| \, d\theta$ is

$\int_{0}^{4}|x-2| d x=$

$\int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx = $

The value of $\int_{-2}^{3} |1 - x^2| dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module