Practicing Success
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$∫(x-1)e^{-x}dx=$ |
$-xe^x+C$ $xe^x+C$ $xe^{-x}+C$ $-xe^{-x}+C$ |
$-xe^{-x}+C$ |
The correct answer is Option (4) → $-xe^{-x}+C$ $∫e^{-x}(x-1)dx$ $=∫-e^{-x}(x-1)dx$ $=∫\frac{d}{dx}(-xe^{-x})dx=-xe^{-x}+C$ |
