intlimits_0^infty [2e^{-x}], dx,where [x] den | vedclass

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(B) We need to evaluate the integral $I = \int\limits_0^\infty [2e^{-x}]\, dx$.Let $f(x) = 2e^{-x}$. As $x$ varies from $0$ to $\infty$,$f(x)$ varies

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$\ln 2$

Vedclass · Answer

(B) We need to evaluate the integral $I = \int\limits_0^\infty [2e^{-x}]\, dx$.Let $f(x) = 2e^{-x}$. As $x$ varies from $0$ to $\infty$,$f(x)$ varies from $2$ to $0$.Specifically,$[2e^{-x}] = 1$ when $1 \le 2e^{-x} For $x > \ln 2$,$0 Thus,$I = \int\limits_0^{\ln 2} 1\, dx + \int\limits_{\ln 2}^\infty 0\, dx = [x]_0^{\ln 2} = \ln 2 - 0 = \ln 2$.

$\int\limits_0^\infty [2e^{-x}]\, dx$,where $[x]$ denotes the greatest integer function,is equal to:

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