Let x 0 be a fixed real number. Then,the in | vedclass

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(A) Let $I = \int \limits_0^{\infty} e^{-t}|x-t| d t$. Since $x > 0$,we split the integral at $t = x$: $I = \int \limits_0^x e^{-t}(x-t) d t + \int \l

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$x+2 e^{-x}-1$

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(A) Let $I = \int \limits_0^{\infty} e^{-t}|x-t| d t$. Since $x > 0$,we split the integral at $t = x$: $I = \int \limits_0^x e^{-t}(x-t) d t + \int \limits_x^{\infty} e^{-t}(t-x) d t$ For the first integral,using integration by parts: $\int \limits_0^x e^{-t}(x-t) d t = [-(x-t)e^{-t}]_0^x - \int \limits_0^x e^{-t} d t = (0 - (-x)) - [-e^{-t}]_0^x = x - (1 - e^{-x}) = x - 1 + e^{-x}$ For the second integral: $\int \limits_x^{\infty} e^{-t}(t-x) d t = [-(t-x)e^{-t}]_x^{\infty} + \int \limits_x^{\infty} e^{-t} d t = (0 - 0) + [-e^{-t}]_x^{\infty} = 0 - (0 - e^{-x}) = e^{-x}$ Adding both parts: $I = (x - 1 + e^{-x}) + e^{-x} = x + 2e^{-x} - 1$.

Let $x > 0$ be a fixed real number. Then,the integral $\int \limits_0^{\infty} e^{-t}|x-t| d t$ is equal to

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