Given $f(x) = -2xe^{-x}$ and $f'(x) = -2e^{-x} - 2e^{-x}\cdot(-x) = 2e^{-x}(x-1)$. Find the area as 2 sq. units.
Step-by-Step Solution
Key Concept: Integration by parts with exponential functions: use the formula $\int ue^{-x}dx = -ue^{-x} - \int(-e^{-x})du$.
For $f(x) = -2xe^{-x}$, we have $f'(x) = 2e^{-x}(x-1)$. The area is $\int_0^{\infty} 2xe^{-x}dx = 2\int_0^{\infty} xe^{-x}dx = 2[-(e^{-x})x] + 2\int_0^{\infty} e^{-x}dx = 2[0-(−1)] = 2$ sq. units.
Correct Answer: 2