Show that \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left[-(x^2+2x+2)\mathrm{e}^{-x}\right]=x^2\mathrm{e

Question

Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\left[-(x^2+2x+2)\mathrm{e}^{-x}\right]=x^2\mathrm{e}^{-x}$.[4] Hence evaluate the exact value of $\displaystyle\int_0^2x^2\mathrm{e}^{-x}\,\mathrm{d}x$.[3]

Accepted Answer

$2-10\mathrm e^{-2}$

Quadratic factor with a negative exponential

Quadratic factor with a negative exponential

Video Solution:

Solution:

Solution locked