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Let $\mathrm{x}>0$ be a fixed real number. Then the integral $\int_{0}^{\infty} e^{-t}|x-t| d t$ is equal to.
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Verified Answer
The correct answer is:
$x+2 e^{-x}-1$
$\int_{0}^{x} \mathrm{e}^{-t}|(\mathrm{x}-\mathrm{t})| \mathrm{dt}$
$=\int_{0}^{\mathrm{x}} \mathrm{e}_{11}^{-1}(\mathrm{x}-\mathrm{t}) \mathrm{dt}+\int_{\mathrm{x}}^{x} \mathrm{e}_{11}^{-1}(\mathrm{t}-\mathrm{x}) \mathrm{dt}$
$\left\{-(\mathrm{x}-\mathrm{t}) \mathrm{e}^{-t}-(-1) \mathrm{e}^{-\mathrm{t}}\right\}_{0}^{\mathrm{x}}-\left\{-(\mathrm{x}-\mathrm{t}) \mathrm{e}^{-\mathrm{t}}+\mathrm{e}^{-t}\right)_{\mathrm{x}}$
$=\mathrm{x}+2 \mathrm{e}^{-x}-1$
$=\int_{0}^{\mathrm{x}} \mathrm{e}_{11}^{-1}(\mathrm{x}-\mathrm{t}) \mathrm{dt}+\int_{\mathrm{x}}^{x} \mathrm{e}_{11}^{-1}(\mathrm{t}-\mathrm{x}) \mathrm{dt}$
$\left\{-(\mathrm{x}-\mathrm{t}) \mathrm{e}^{-t}-(-1) \mathrm{e}^{-\mathrm{t}}\right\}_{0}^{\mathrm{x}}-\left\{-(\mathrm{x}-\mathrm{t}) \mathrm{e}^{-\mathrm{t}}+\mathrm{e}^{-t}\right)_{\mathrm{x}}$
$=\mathrm{x}+2 \mathrm{e}^{-x}-1$
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