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What is the value of $\int_{0}^{1}(x-1) e^{-x} d x$ ?
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The correct answer is:
$\frac{-1}{\mathrm{e}}$
Given integral is $\mathrm{I}=\int_{0}^{1}(\mathrm{x}-1) \mathrm{e}^{-\mathrm{x}} \mathrm{dx}$
Integrating by parts taking $(\mathrm{x}-1)$ as first function
We get, $\mathrm{I}=\left[(\mathrm{x}-1)\left\{-\mathrm{e}^{-\mathrm{x}}\right\}\right]_{0}^{1}-\int_{0}^{1} 1 .\left(-\mathrm{e}^{-\mathrm{x}}\right) \mathrm{dx}$
$=-(1-1) \frac{1}{\mathrm{e}}+(-1) \mathrm{e}^{0}+\left[-\mathrm{e}^{-\mathrm{x}}\right]_{0}^{1}=-1-\frac{1}{\mathrm{e}}+1=-\frac{1}{\mathrm{e}}$
Integrating by parts taking $(\mathrm{x}-1)$ as first function
We get, $\mathrm{I}=\left[(\mathrm{x}-1)\left\{-\mathrm{e}^{-\mathrm{x}}\right\}\right]_{0}^{1}-\int_{0}^{1} 1 .\left(-\mathrm{e}^{-\mathrm{x}}\right) \mathrm{dx}$
$=-(1-1) \frac{1}{\mathrm{e}}+(-1) \mathrm{e}^{0}+\left[-\mathrm{e}^{-\mathrm{x}}\right]_{0}^{1}=-1-\frac{1}{\mathrm{e}}+1=-\frac{1}{\mathrm{e}}$
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